Chapter 12 Environmental Chemical Reactions and Transformations

Environmental Chemical Reactions and Kinetics1.kinetics  order of a reaction  Arrhnenius temperature -rate const  rate limiting steps  steady-state approximation  Hammett relationships and rate constants  kinetic simulations1 Kinetics: the rate law of a reaction must be verified by experimentation1st order reactions A ---> B-d [A] /dt = krate [A]- d [A]/[A] = kratedtA,t  t ln[A] A, t 0   k t-kt [A]t= [A]0 e typically to get a 1st order fit, 70% of A needs to react if the rate constant is independent of concentration2 Some time vs conc. dataHr Conc [A] Ln[A] 0 2.718 1 0.3 2.117 0.75 0.6 1.649 0.50 0.9 1.284 0.25 1.2 1.000 0.00 1.5 0.779 -0.25Time vs. concentration32.52 m p p n i 1.5. c n o c 10.50 0 0.5 1 1.5 2 tim e in hours3 1st order plot 1.210.80.6 ] A [ 0.4 n l 0.20 0 0.5 1 1.5 2 -0.2-0.4 time in hours t1/2 = time it takes for [A] to decrease by a factor of 2 ln [A]/[A]o=-k t1/2 ; ln2 /k =t1/2 life times are define slightly differently ln e = k 1/k = 4 1/k = time scale for the reactionPseudo first order rate constantsA. The reaction of benzyl chloride to produce benzyl alcohol water reacts with benzyl chloride+ - CH2 -Cl + H2O CH2-OH + H + Cl d[benz-Cl]/dt= -k [benz-Cl] [H2O] k here is a second order rate const. in L mol-1sec-1If we assume that the reaction is run in dilute solutions as we could say that the [H2O] is constant kpseudo = k[H2O] d[benz-Cl]/dt= -kpseudo [benz-Cl]B. Consider the reaction of methyl mercaptan in water to produce dimethyl disulfide5 2CH3 SH + ½ O2 ----> H3 C-S-S-CH32 1/2 d[CH3 SH]/dt = k [CH3 SH] [O2] if we assume that O2 is constant2 d[CH3 SH]/dt = kpseudo [CH3 SH]For the special case of [A] reacting with [A]A + A  productsSince A is reacting with A d[A] d[A]   k[A]2 and kdt dt [A]21  At  kt [A] Ao1 1  kt [A]t [A]o6 for this type of simple second order reaction, plot of 1/[A]t vs. t gives a straight line with a slope of...??? and an intercept of....? the half-life is when [A]t = ½ [A]o 1  kt1/ 2 [A]o7 More on second order reactions if A + B----> products and x is the amount of A and B reacted the differential equation that describes a second order rate law for the change in x with respect to time dx/dt = - k[A] [B] or dx/dt = - k [Ao-x] [Bo-x]1 x dx dx t (  )k dt 0 [Ao ]  [Bo ] 0 [Ao ]  x [Bo ]  x this has an exact solution1 [B ]([A ]x) ln o o   k t [Ao][Bo] [Ao]([Bo]x) so if we measure the amount reacted, x, over time and plot the left side of the equation vs time, the rate constant, k can be measured8 Time -pinene O3 (hh.hh) (ppm) (ppm) 9.15 0.820 0.600 9.23 0.406 0.435 9.50 0.158 0.223 9.83 0.097 0.149 1 [Bo]([Ao]x) ln   k t ???? [Ao][Bo] [Ao]([Bo]x)9 Rate Constants and Temperature1850 - Wilhelmy related rate constant to temperatureDT 1862 - Bertholet proposed k = A e1889 - Arrhenius showed that the rate constant increases exponentially with 1/temp. lnK  U 1884 - van’t Hoff Equation c    2  T  p RT where Kc= “concentration equilibrium constant” U = ‘standard internal energy change andKc = k1/k-1 so 10 dlnk dlnk U 1  1  dT dT RT2 dlnk dlnk U 1  1  dT dT RT2 van’t Hoff proposed two energy factors, so that U = E1 - E-1It follows that: ln k1 = - E1/RT +const-E /RT or k1 = A e 1 (Arrhenius Equation)There are other equations relating temp and rate constants1898 van’t Hoff rate equation usually give very good empirical fits 2 k  ATme(BDT )/T11 The Arrhenius equation is most widely used because it provides insight into how reactions proceed-Ea/RT k1 = A e ln k1 = ln A - Ea/RTEa (1/T2-1/T1)/R krate (Ti/T2) = e o krate relative to 25 C x 100 (%) Ea 10oC 20oC 30oC Avg KJ/mol increase in krate 40 42 76 130 1.8 50 34 71 139 2.0 60 28 43 149 2.3Theory of Arrhenius 12 Consider the reaction of B +C ---> D + E dB/dt = -k [B] [C] to react B and C have to collide, and the rate should depend 1. on the frequency of encounters of B and C which is proportional to the conc. of B and C and how fast B and C move (diffuse) toward each other.2. the orientation of B and C3. the fraction of the collisions that will have sufficient energy to break the bonds of B and C.The fraction of reaction species with an energy greater than the activation energy is given by the Botlzmann distribution of energies e-Ea/RT hence in rate = A e-Ea/RT[B] [C] the coef A must include frequency and orientation factors13 Activated Complex or transition state theory the reaction of B + C ---> on its way to products goes though an activated intermediate called [BC]B + C ---> [BC]‡--> [D] + [E] it is assumed that there is an equilibrium between the reactants and the activated complex [B-C]  and that [B-C] decomposes to products with a rate constant of kT/h where k = Boltzmann const 1.38 x10-23 J K-1 h= Plancks const, 1.63 x10-34 J/sec k T/h assumes that the rate constant is directly proportional to the vibrating frequency of the transition state and the energy associated with this is proportional to kT E kT  h14  = kT/h [BC] ‡--> [D] + [E] rate = kT/h [BC]‡ since we said --> ‡ B + C <-- [B-C][BC] K   [B][C] substituting for [BC]* in: rate = kT/h [BC]‡ rate = kT/h K‡ [B] [C]The equilibrium const. is related to the free energy of activation byK‡= e-G‡/RT15 and G‡ = H‡-TS‡16 kT   rate    eS /R eH /RT[B][[C]  h kT   k    eS /R eH /RT  h ‡ for a bimolecular reaction H = Ea - RTkT  k    eS /R e1 eEa /RT  h  this looks like k= A e-Ea / RT the entropy term in A --> orientation probability and temp in A may be related to the frequency of encounters; # collisions ~ (RT)1/217 Composite ReactionsSimultaneous ReactionsA--> YA---> ZCompetition reactionsA + B --> YA + C --> ZOpposing ReactionsA + B ZConsecutive reactionsA--> X--> Y-->ZFeedback A--> X--> Y---> Z18 Consecutive Reactions k1 k2 A ---> X---> Z-k1t -d[A]/dt = k1 [A] [A] = [A]o e+d[X]/dt = k1[A] d[X]/dt = +k1[A] - k2[X]-k1t d[X]/dt = +k1[A]oe +- k2[X] solution-k1t -k2t [X] =[A]ok1/(k2-k1) (e - e )-k1t -k2t [Z] =[A]o/(k2-k1) [k1(1-e ) -k1(1- e )These types of expressions are cumbersome19 (Steady- state approximation)“The rate of change of the concentration of an intermediate, to a good approximation, can be set equal to zero whenever the intermediate is formed slowly and disappears rapidly” -D.L. Chapman and L.K. Underhill, 1913Use of the Steady-State Assumption(Consecutive Reactions with an Opposing Reaction as the 1st step)Consider the reaction of OH radicals in the atmosphere with SO2 to form sulfuric acid particles. k1 ‡ OH + SO2 HOSO2 k-1 k3 ‡ + HOSO2 M ---> HOSO2 + M‡ for the rate of formation of HOSO2 , we would write20 ‡ d [HOSO2] /dt = +k1[OH][SO2] for the loss we have ‡ -k-1[HOSO2] and ‡ -k3[HOSO2] so the total rate expression is‡ d [HOSO2] /dt = +k1[OH][SO2] ‡ ‡ -k-1 [HOSO2] -k3[HOSO2]‡ at steady state d[HOSO2] /dt = zero‡ ‡ 0=+k1[OH][SO2] -k-1[HOSO2] -k3[HOSO2] k1[OH][SO2] [HOSO2 ]  k1  k3 for the formation of product HOSO221 ‡ + HOSO2 M ---> HOSO2 + M‡ d[HOSO2]/dt = +k3[HOSO2]‡ Substituting [HOSO2][HOSO ] k k [OH][SO ] 2  1 3 2 dt k1  k322 In the atmosphere the formation of ozone can be represented as hNO2 ------> NO + O. k1= 0.4xTSRM O. + O2 ----> O3 k2 = fast,fastO3 + NO -----> NO2 k3 a. derive a steady state relationship for O3 as a function of NO2, NO, k1 and k3; assume that dNO2/dt is at ss.. b. calculate the equilibrium ozone for the following +03 conditions; assume k3 = 2.64310 exp(-1370/ T) time 6:00 7:30 9:00 12:00 15:00NO (ppm) 0.06 0.11 0.08 0.03 0.005 NO2 (ppm) 0.03 0.07 0.14 0.15 0.10 Temp (oC) 25 26 28 35 34 TSR(cal cm-2min-1) 0.05 0.12 0.25 0.5 0.3 *TSR= total solar radiation flux in cal /cm2/min plot your results for NO, NO2, and O3 and if this was a real atmosphere explain.23 Rate determining steps k1 k3 A + B X Z k-1[HOSO ] k k [OH][SO ] 2  1 3 2 dt k1  k3[Z] k k [A][B]  1 3 dt k1  k3 if k3 >> k1[Z]  k [A] [B] dt 124 The Hammett Equation and rates constantsIn 1940 Hammett recognized for substituted benzoic acids the effects of substituent groups on the dissociation of the acid groupCOOH COO-+H+R R o o o G = G H +  G iEffect on the free energy change from dissociation could be represented as the sum of the free energy change by the unsubstituted benzoic acid and the contributions from the various R groups.-RTln Ka = -RTln KaH - RTln Ka,iEffect on the free energy change from dissociation could be represented as the sum of the free energy 25 change by the unsubstituted benzoic acid and the contributions from the various R groups.Hammett then says that the effects of the substituent group i, on the free energy can be represented o as:G i = -2.303 RTi; where i =ln Ka,i we know that o G = -2.303 RT log Ka and o G H = -2.303 RT log KaH o G i = -2.303 RTi so-2.303 RT log Ka=-2.303 RT log KaH +-2.303 RTi going back to:-RTln Ka = -RTln Ka,H - RTln Ka,i log Ka= log KaH +i so, log (Ka / KaH )= I and pKa = pKaH - I 26 27 Phenols and Hammet Constants anilines28 Hammet also observes that when considering other compound classes, like phenyl acetic acids the  values developed for benzoic acid can be used log (Ka / KaH) = i log (Ka / KaH )= i when considering other compound classes, like phenyl acetic acids the  values developed for benzoic acid can be used log (Ka / KaH) = iFigure 8.7 page 17429 Rate constants and Hammet sigmas if an aromatic reaction is going through a transition stateB + C ---> [BC]‡--> [D] + [E] we said that the rate rate = (kT/h) K‡ [B] [C] the rate constant is (kT/h) K‡‡ log krate= log (kT/h)+ log K since -2.303 RT log K‡= G‡oUsing the Hammett argument that‡o ‡o ‡o G = G H +  G i show that log(krate) = log krateH +  m,p30 or log(krate/krateH) =  m,pDifferent rates are obtained in different solvents, so reaction rates are not directly applicable to water if in another solvent31

Chapter 12 Environmental Chemical Reactions and Transformations

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