Kinetics. Chemical kinetics contends with a number of issues. Specfically, it seeks to do the following:

Transcription

1 Kinetics Chemical kinetics contens with a number of issues. pecfically, it seeks to o the following: Detere an preict the rate of chemical reactions Detere the reaction mechanism. With the previous goal, to etere specific atomic an molecular properties of unstable species Develop theoretical moels to ai in our unerstaning of the reaction process an to be able to preict an optimize reaction mechanisms an rates. First, let us efine some basic terms an concepts... For the general reaction: aa + bb -> cc + D the rate of the reaction can be efine by the isappearance or appearance of any of the reactants or proucts by RATE a t A b t B c t C t D Note that the ifferentials are ivie by the stochiometric coefficients to insure uniform rate values. For example, in the reaction: H + O -> H O oxygen isappears at half the rate of the isappearance of hyrogen or appearance of water. A rate law is a statement of the relationship between the reaction rate an the material concentration (or pressure). We fin, in general.. RATE ka m B n C o D p where k is efine as the specific rate constant an the power coefficients m,n,o an p are the rate orers w.r.t A,B,C an D, respectively. The overall orer of the reaction is a sum of the iniviual orers of the reaction, that is, overall orer m + n + o + p. Finally, we must efine the concept of an "elementary step" or "elementary reaction" In most cases, an observe reaction is a result of a variety of iniviual reactions that often involve multiple an short live intermeiates. For elementary reactions, the rate law may be written irectly as efine above. In all other cases, the experimental information must be known or the mechnism must be known in orer to prouce the rate law. In aition, just because we know the rate law, this oes not imply that we know the mechanism.

2 Example: A propose mechanism for the formation of HBr from elemental broe an hyrogen is: Overall Observe Reaction: H + Br -> HBr Propose Mechanism: Rates for Each tep Br -> Br Rate k {Br ) Br + H -> HBr + H Rate k (Br)(H ) H + Br -> HBr + Br Rate k 3 (H)(Br ) H + HBr -> H + Br Rate k 4 (H)(HBr) Br + Br -> Br Rate k 5 (Br) Each of these steps are elementary an the rate laws for each must be known in orer to be able to combine an prouce the experimentally etere rate law given as: Rate k k k 3 H Br k 5 k 4 ( HBr) k 3 Br 3 It is thus necessary to investigate the kinetics of a series of elementary reactions. Knowlege of the simple reactions will assist us later in etering the overall mechansims. Our approach will be as follows: Exae how elementary step reactions occur. Exae experimental methos to etere what the overall rate law is observe to be. Propose a mechanism an use ifferential methos to attempt to reprouce the observe rate law.

3 3 These reactions follow the general rate law... A R ka n t Concerte Reactions This rate efinition gives us a rate when we know A an n. Better woul be a rate law that gives us concentration an time. To get this uses the methos of CALCULU which is not part of this course iscussion. However, the result is known as an integrate rate law. For reaction systems that have a single component, A, the results are. A A o t A k t A n or 0 A n n A o ( n ) kt for n For n, we have A A o t A k t A or ln ( A ) ln A o kt for n 0 Note that the resulting equations have time explicitly This implies that we can look at the time that has elapse an be able to etere the concentration of species A. For example, if n, or a secon orer reaction, then the integrate rate law becomes. ( ) kt or A A o A A o kt which is commonly presente in general chemistry courses. Note that each of these expressions are presente in linear form: that is ion this case, the form a a straight line. For the example, it is implie that a plot of /A vs t will present a straight line that has a postive k as the slope. For the first orer example, a plot of ln(a) vs t will give a straight line having a slope of -k. This forms the basis for a metho of etere the reaction orer calle the trial plotting metho. However, this metho is ifficult to to get goo results unless oe has very goo ata over a long time. A homework problem illustrates this.

4 4 Example: The imerization of C 4 H 6 -> C 8 H is known to be secon orer. Create a secon orer rate plot using the ata provie. t i A i M 0.M 0.04M 0.09M Rate plot of /A vs. time /A (/M) time () When we begin a stuy of the mechanisms, we will be intereste in the most common type of reaction, that being a bimolecular collision. Therefore, let us take a moment to exae this system from a phenomenological kinetic view. econ Orer: A + B -> P In contrast to the secon orer A + A -> P, there is a secon type where the two reactants are ifferent. IN this instance, the rate i efione as: R kab In this instance the reactants are first orer each but overall the reaction is secon orer. Treatment of this system involves more methos using CALCULU an the etails won't be presente here. I only present the result that the integrate rate law becomes: ln B A ln B o A o B o A o kt

5 5 Example: Ethyl Ethanoate + NaOH -> oium Acetate + Ethanol M mol L Ester o.5m NaOH o.055m i 0 7 t 0 t i 0sec 0sec 40sec 60sec 80sec 00sec 0sec 40sec Ester i 0.5M 0.353M 0.53M 0.84M 0.95M M 0.060M 0.038M NaOH i 0.055M M M 0.034M M M 0.0M M Ester i NaOH i ln Ester i NaOH i ln Ester i NaOH i y ln i Ester i NaOH i mm slope ty t i ( ) s But since the slope kester o NaOH o mm k k 0.00 Ester NaOH 0 0 Ms It's important to note that if the concentrations of the reactants are equivalent, then the reaction reuces to the concerte reaction A ---> Proucts an the associate rate law given above.

6 6 Half-life an Relaxation Time of a Reaction Because reactions typically slow as the reactants ecrease, it is convenient to efine a half life, as the time it takes for a reactant to rop to / of it's original value. This is particularly appropriate for first orer reactions as we will see. For first orer reactions... At t /, A A o / therefore.. kt halflife ln( A) ln A o ln A o so ln therefore t halflife k A o A o k t halflife Note that for first orer kinetics, the half life is inepenent of initial concentration. For any other concerte reaction of orer, n, we have.. n t halflife n ( n ) A o k Relaxation time... This efinition is similar to the half-life except that the time efine as the relaxation time is that time for which the reactant concentration rops to /e of the initial value. It's usefulness in first orer reactions is evient whe we see that. k τ ln( A) ln A o ln A o so ln e A o e A o kτ.0 therefore τ k Thus, we measure the relaxation time an it's inverse is the rate constant irectly for a first orer reaction.

7 7 Experimental Deteration of a Rate Law Ultimately, for a reaction of interest, it will be necessary to etere the reaction rate law. That in concert with theoretical eterations will lea to the evelopment of a reaction mechanism. Trial plotting methos: These methos are best illustrate by the previous examples. Data is taken in the form of concentrations or pressures an time. Various orer plots are mae an linearity is etere. Caution: This works if ata over a significant perio of time is taken. Metho of initial velocities: For any reaction, the only time we know the concentrations of all species unequivocally is at t0. Thus, the reaction rate is taken at the very beginning of a reaction. Reactant concentrations are then altere an the rate teken again. Orers are then etere along with the specific rate constant. Example: NO + H -> N + H O Exp. P(H )/kpa P(NO)/kPa Initial Rate (kpa/s) Result: R k(h )(NO) k.58 x 0-6 kpa - s - Pseuo-orer methos - (flooing) In this approach, reactants other than one of interest are increase in concentration until they effectively remain constant over the course of the reaction. Then the orer of the remaining reactant may be exae inepenently. As an example, consier the reaction A + B -> Proucts. The preliary rate law may be written as: R ka m B n Now the reaction is carrie out uner the conitions that the concentration of A -> large. In this instance, A becomes a constant in this reaction an the rate law becomes: R k B n where k ka m Now the orer of the reaction w.r.t. B can be etere conveniently.

8 8 Example: Using the value for the rate constant etere in the reaction of ethyl acetate an NaOH, exae the concentrations of the species if the initial concentration of NaOH is.0 M. From above... Ester o.055m NaOH o.0m k 0. liter molesec t 0s s 0s NaOH() t M Ester() t M Notice how the Ester concentration rops significantly while the NaOH concentration stays relatively constant. Plotting ln(ester) vs t yiels... ln Ester() t M t me slope( ty ) me 0.6 s Now, recall that this slope is multiplie by [NaOH]. Diviing... me k k 0. NaOH o Ms as before

9 9 Pseuo-orer Methos with Initial Rate In the previous example, a sample was flooe with a reactant an subsequently, trial plotting was use in orer to exae the rate law. If it is possible to isolate a single reactant, then, we may use an initial rate metho to fin the orer an rate constant. To wit: For a pseuo-orer conition, we may write that R k ps C n where k ps is the pseuo orer rate coefficient as escribe above an C is the concentration. Taking the log of both sies an re-arranging gives.. ln( R) nln( C) lnk ps In this instance, plotting the log of the rate vs. the log of the concentration yiels both the orer an the pseuo-orer rate coefficient from evaluation of the slope an the intercept. Example: For the reaction between gas phase Ethylene an Broe, the following pressures were use for Ethylene at an initial broe pressure of.00 bar. The rate of the prouction of ethylene bromie was followe at the beginning of each reaction concentration. The initial rates are plotte below. m n For this reaction, we can write the rate law as: R kp Br P Eth ince we are initially flooing the system with broe, the reaction becomes pseuo orer an we write... n R k ps P Eth where k ps k true Br m Applying out expression above, we plot the ln(rate) vs ln(ethylene)

10 0 Ethylene initi 0.00bar 0.035bar 0.050bar 0.075bar 0.00bar R initi 0. bar 0.65 bar.3 bar.97 bar 5.8 bar Br.init.00bar slope b n n.00 intercept b 6.75 k ps_ethyl e b k ps_ethyl bar bars Natural log of Initial Rate Initial Rates Plot Natural log of Ethylene Initial concentration In a similar analysis, the system was flooe with Ethylene an graphical analysis reveale. m an k ps_br 5.8 bars we can now use this information to calculate k true since.. k ps_ethyl k true Br m k true Br _init k ps_ethyl thus k true k true Br.init 3 s( bar)

11 Half-life measurements: Now that we have a general equation for the half-life of a reaction as presente above, we can use it to etee a reaction orer. Recalling n n n t halflife A o n ( n ) k ( n ) A o k Taking the logarithm of both sies an rearranging prouces: lna o ln t halflife n ln n ( n) lna o ln n ( n ) k ( n ) k Thus, a plot of ln(t halflife ) vs ln(a o ) shoul give a straight line with a slope of (-n).

12 i 0 0 Example: Consier the reaction: N O - > N + / O. NO i atm.5atm.0atm 4.0atm 5.0atm 0.0atm.0atm 5.0atm 0.0atm 30.0atm 50.0atm t half i ms Natural log of half life Natural log of Initial NO Pressure 3.97 ince the slope - n, the orer, n -slope slp n slp n Now, the rate constant can be calculate from the intercept as: Int ln n ln n ( n ) k n ln( k) or k exp ln n n Int n n exp( Int) From plot b n k n exp( b) ms k ms atm atm

13 3 Temperature-Depenence of the Rate Constant. It was observe very early that the rate of a reaction is epenent on the temperature. Bycareful empirical evaluation, vante Arrhenius iscovere that the rate constant is relate to the temperature via... E a RT k Ae or ln( k) E a RT Constant This is illustrate as follows: The reaction.. HI -> H + I. was stuie over several temperatures. Rate constants were calculate an are presente as follows: J i 0 8 R x 0 y 0 kj 000J molk T i k i 556K 575K 69K 647K 666K 683K 700K 76K 78K Lmol s.0 6 Lmol s Lmol s Lmol s.90 4 Lmol s Lmol s.60 3 Lmol s Lmol s Lmol s Natural log k Arrhenius Plot /T slp K E a ( slpr) E a kj mol Activation Energy Note: This is for the ecomposition of moles of HI! To get the ecomposition energy per mole, ivie by

14 4 Approximate Methos - Mechanisms With some juicious exaation of reaction concepts, some approximations can be mae that simplify the solutions of the rate expressions for complex reactions teay-tate Approximation: In this approach, an intermeiate is thought to rise to a concentration initially. As the concentration buils, reaction channels leaing to proucts increase in rate. (ince their rate epens on the intermeiate concentration) Thus, the intermeiate reaches a steay-state concentration. Consier, for example, the reaction A ->C via the following mechanism: A <> B B -> C If B is assume to reach a steay state, then t B 0, an the reaction is evaluate as follows: The overall rate of the reaction can be expresse as the rate of prouction of proucts, that is Rate T P k B assug all steps are elementary Accoring to the approximation t B 0 k A k r B k B thus B k A k r k substituting into the original rate expression k A Rate k k r k k k A k r k Intermeiate Fast-Equilibrium Approximation: This approach is useful when the intermeiate buils up to an initial concentration an waits for the slow following reaction to occur. In such cases, the reaction sets up an equilibrium with intermeiate. Consier the following: A + B <> C C -> P If the prouct forg reaction C -> P is the slow step, then A, B an C are in equilibrium, thus

15 5 k AB k r C thus C k AB k r The overall rate of prouction of P eteres the overall reaction rate, thus Rate k C substituting from above Rate k k AB k r ome examples: NO O NO The observe rate law for this reaction is: Rate kno O The propose mechanism is as follows: NO + O <> NO 3 NO 3 + NO -> NO (fast-equilibrium) (slow) The secon step is propose to be rate etering, thus k k NOO k NO 3 or NO 3 NOO k now Rate k 3 NO 3 NO k k 3 NO O k (Note: Approx with k 3 NO < k gives the same result.)

16 6 Combination Reaction Example: Consier: I -> I If this were an elementary step, the rate law coul be written as Rate ki Experimentally, it is foun Rate exp ki M where M is some NONreactive species! Why?! The explanation has to o with energy issipation requirement for combination to occur. We must have I + I + M -> I + M* where M* is energetically excite. The problem is that 3 boy collisions are efinitively rare. Consier, then, two propose mechanisms Mechanism A I + I <> I * I * + M -> I + M* Mechanism B I + M <> I---M where I---M is a "loose" complex I---M + I -> I + M* Let us test the two mechanisms beginning with B Rate t I k 3 I MI but, if I---M is assume to be in teay tate, then t I M 0 k IM k I M k 3 I MI thus I M k IM k 3 I k substituting into the rate expression... t I k k 3 I M k 3 I k

17 7 The observe rate law is epenent on relative magnitues of k's an experimental conitions. Thus, for example, at low I concentration k k 3 I an t I k k 3 I M ki M which fits experimentally observe rate law. k Now for mechanism A t I % k 3 I M applying approx t I % 0 k I % % k I k 3 I M thus % k I I substituting... k 3 M k t I k k 3 I M k 3 M k if M concentration is low, then t I k k 3 I M as before k It appears that both mechanisms support the experimentally observe mechanism, so which is most likely? Often, experimental etails are neee to eluciate a particular mechanism. Note the effect of the ientity of M M k polarizability (x0-30 m 3 ) Helium 0.3 x Argon Benzene Toluene Clearly, epenence of k on polarizability favors mechanism B

18 8 Enzymes Many biological reactions involve enzymatic action. The prupose of this section is to exae the nature of the interactions of enzymes an substrates an eluciate a mechanism in normal an abnormal function. ince enzymes are not use up in a reaction, it was realize that they act as catalysts. It was further propose that enzymes couple with substrates to form a complex. The complex is then issociate as the enzyme an prouct. This propose mechnism is illustrate by the following mechanism... E + E E -> E + P A typical approach to this system is to treat the complex as being in a "steay-state". Thus.. v k E which is the velocity of the reaction. The complex is at a steay concentration, thus... 0 k E k r E k E solving for E E k r k k E The expression, k r k, is a constant compose of ifferent iniviual rate constants an is k commonly calle the "Michaelis constant" an symbolize,. Thus... E E At this point, it is usual practice to solve for the complex concentration an resubstitute back into the original velocity expression. However, in this instance the actual free enzyme concentration, E, is ifficult to follow in that it usually occurs in small amounts an much of it is complexe with the substrate. Therefore, in this instance we will first recognize that the total free enzyme concentration is the original concentration us that which is complexe, or... E o E E or substituiting the previous expression... E o E E E E E o

19 9 Finally, substituting back into the velocity expression yiels... v k E o Reaction Rate vs. ubstrate Concentration 8 v max Reaction Rate ubstrate Concentration Note: Initially, the velocty is epenent on substrate concentration. At high concentrations, the velocity becomes constant. The limiting rate at high is k E o. The rate of increase initially is inversely proportional to. With regar to limiting velocity...if -> high... v k E o k E o _large k E o k E o v max k is calle the "turnover rate" for this reaction an is the maximum amount of substrate that can be converte per unit time,

20 0 These observations lea us to take the inverse of both sies of the rate expression. Doing so yiels... v k E o v max v max v max Thus a linear expression results!. Thus a plot of /v vs / will yiel a straight line. This is known as a Lineweaver-Burk plot Example: The ecomposition of hyrogen peroxie, H O is catalyze by the enzyme catalase. Initial rates for varying concentrations of peroxie substrate are given for an initial catalase concentration of 4.0 x 0-9 mol/l. H O (mol/l) Initial Rate (mol/l-s) Calculate the maximum reaction rate an the Michaelis constant. perox i rate i.00m.00m.005m.0038 M s.0067 M s M s inverse reaction rate Linweaver-Burk Plot for Peroxie Decomposition inverse peroxie concentration b 0.07 s m 3 mol v max v max M b s slp s ( slp) v max 0.06M

21 Inhibitory mechanisms an others may be expresse in terms of the non-inhibite rate law. Consier, for example, the following case: In this mechanism, the inhibitor present competes for the enzyme. Following the kinetics erivation as before... E + E E -> E + P an now E + I EI As before, we initially approach this system treating the complex as being in a "steay-state". v k E The complex is at a steay concentration, thus... k r k E E 0 k E k r E k E or E k where the expression as efine previously has been use. The RH expression will be use below. The last reaction reuces the free enzyme by complexation. We will assume that an equilibrium state is present with the inhibitor, thus K I EI or E EI EI K I I where K I is the inhibitor equilibrium constant. ave this expression, we will nee it in a moment. At this point, we have solve for the free enzyme concentration, E in terms of the normal enzyme complexation constant, an the inhibitor complexation constant, K I.Both are equilibrium conitions an both are vali, thus EI E E K I an, thus EI I K I I E Now, substituting... E I E o E E EI E E an solving for E K I

22 E E o K I I E o I K I E o "" where "" I K I Finally, the rate may be written as: v k E o " " v max " " A Linweaver-burk rearrangement yiels v v max "" v max Note that comparing to the uninhibite LB rearrangement shows that the intercept will remain unchange, however, the slope will be altere. The new slope allows the calculation of K I is is known from an analysis of an uninhibite reaction. This particular inhibition is known as a competitive inhibition. Other mechanisms exist, some of which are elineate below.

23 3 Inhibitory Mechanisms: It is common in enzyme reactions to have some type of interference. There are several types an they each alter the mechanism to some egree. ome examples are given along with the rate laws. Competitive Inhibition: In this mechanism, the inhibitor competes for the substrate for the same enzymatic complex site preventing the substrate from bining. v max v Result: L-B plot has same y-intercept, but ifferent slope I K I I New slope > slope v max Uncompetitive Inhibition: In this instance, the inhibitor bins to the enzyme-substrate complex preventing prouct formation. K I v max v Result: L-B plot has same slope but ifferent intercept I K I I New intercept > intercept K I v max Noncompetitive Inhibition: Here, the inhibitor presumably bins at a ifferent site than the substrate. This alters the shape an function of the enzyme renering inactive. v max v Result: L-B plot has ifferent slope an intercept I K I I K I I New slope > slope v max K I New intercept > intercept I K I v max

24 4 Example: The following ata is for the enzymatic estruction of thiae with an without the inhibitor, o-ao-4-methylthiazolium chlorie. Detere the type of inhibition involve. Th i 0.0mM 0.5mM 0.50mM.00mM.00mM v uninhibitei M M.0 6 M.60 6 M M v inhibitei M M M M.90 6 M Linweaver-Burk Plot for Inbite an Uninhibite Reactions uninhibite ata inhibite ata inhibite fit uninhibite fit /Rate /ubstrate This plot has the same intercept but ifferent slope, thus o-ao-4-methylthiazolium chlorie inhibits competitively.

25 5 Example: The hyrolysis of sucrose by the enzyme invertase was followe by measuring the initial rate of change in polarimeter reaings at various initial concentrations of sucrose. The inhibitor, urea was then ae are rates measure. Detere the type of inhibition that is taking place. i 0 5 ucrose i rate i rate inhibitei 0.09M M M 0.7M 0.75M 0.34M 0.8 M 0.65 M 0.3 M M 0.9 M 0.54 M M 0.67 M 0.37 M 0.9 M 0.37 M 0.88 M 0.8 Linweaver-Burk Plot for Inbite an Uninhibite Reactions uninhib ata inhibite ata uninhibite fit inhibite fit /Rate /ubstrate The inhibite plot shows a ifferent slope an intercept, thus noncompetitive inhibition is taking place.

26 6 Alternative Plotting Methos: The Linweaver-Burk metho is somewhat problematic in that it is a ouble-inverse plot. Inverse vales emphasize errors at small values ramatically. There are several plotting methos esigne to circumvent this problem, however, I'll focus on one, the Eaie-Hofstee plotting metho. Beginning with the inverse L-B expression.. v multiplying through by v an v max an rearranging yiels v max v max v v v max Thus, a plot of v vs. v/ yiels a straight line as well. using the ata from above... Example: The ecomposition of hyrogen peroxie, H O catalyze by the enzyme catalase. Initial catalase concentration 4.0 x 0-9 mol/l. H O (mol/l) Initial Rate (mol/l-s) Calculate the maximum reaction rate an the Michaelis constant. i 0 perox i rate i x 0 y 0.00M.00M.005M.0038 M s.0067 M s M s rate i 3 rate i x y rate i perox i i i v max intercept( xy ) v max M s rate i perox i corr( xy ) slope( xy ) 0.054M

27 7 For the inhibitor plots... Example: The hyrolysis of sucrose by the enzyme invertase inhibite by urea. Detere the type of inhibition that is taking place. i 0 5 ucrose i rate i rate inhibitei 0.09M M M 0.7M 0.75M 0.34M 0.8 M 0.65 M 0.3 M M M 0.9 M 0.54 M 0.67 M x uni x ini rate i ucrose i rate inhibitei ucrose i 0.37 M 0.9 M 0.37 M 0.88 M 0.04 slope x un rate slope x in rate inhibite M M 0.45 M intercept x un rate 0.45 M intercept x in rate inhibite 0 rate i rate inhibitei 5 un( xx) in( xx) x uni x ini xxxx The inhibite plot shows a ifferent slope an intercept, thus noncompetitive inhibition is taking place.