2015, Fall Semester Physical Chemistry II (Class 001 / )

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1 2015, Fall Semester Physical Chemistry II (Class 001 / ) Professor 김대형 : , , dkim98@snu.ac.kr Classroom : Class time : Tuesday, Thursday 9:30 ~ 10:45 Teaching Assistant 도경식 : , ksdo573@snu.ac.kr 김동찬 : , dchan1106@snu.ac.kr Textbook Atkins, Physical Chemistry, 9th ed Oxford Univ. Press [Chapter 20 ~ 23] Atkins, Physical Chemistry, 10th ed Oxford Univ. Press [Chapter 19 ~ 22]

2 Week Lecture Chapter Dates 1 Orientation, Molecular motion in gases 20 9/1, 9/3 2 Molecular motion in liquids 20 9/8, 9/10 3 Diffusion 20 9/15, 9/17 4 Diffusion, Empirical chemical kinetics 20, 21 9/22, 9/24 5 No Class ( 추석 ), Accounting for the rate laws, Ch 20 problems 보강 (10/2 6:00) 21 9/29, 10/1 6 Accounting for the rate laws 21 10/6, 10/8 7 Examples of reaction mechanisms, 개교기념일 ( 수업 ) 21 10/13, 10/15 8 Reactive encounters, Ch 21 problems 보강 (10/23 6:00) 22 10/20, 10/22 9 Activated complex theory 22 10/27, 10/29 10 Dynamics of molecular collisions, Midterm (11/5, 8:00~11:00) 22 11/3, 11/5 11 Homogeneous catalysis 23 11/10, 11/12 12 Ch 22 Summary (practice problems) 23 11/17, 11/19 13 No Class, Homogeneous catalysis 23 11/24, 11/26 14 Ch 23 Summary (practice problems), No Class 23 12/1, 12/3 15 Supplementary class, Final (12/10, 8:00~11:00) - 12/8, 12/10

3 Overview of Chapter 21 The rates of chemical reactions 1. Empirical Chemical Kinetics (1) Experimental techniques (2) Rates of reactions (3) Integrated rate laws (4) Reactions approaching equilibrium (5) Temperature dependence of reaction rates 2. Accounting for the rate laws (1) Elementary reactions (2) Consecutive elementary reactions 3. Examples of reaction mechanisms (1) Unimolecular reactions (2) Polymerization kinetics (3) Photochemistry

4 Chapter 21. The rates of chemical reactions Goal: 1) Understanding principles of chemical kinetics through mathematically expressed rates of chemical reactions and 2) determining rate parameters (figuring out unknown parameters) by using experimental data. Reaction rates depend on the concentration of reactants and products, temperature, pressure, presence of catalyst. Reaction rates can be expressed by the differential equations (called rate laws). Using solutions of reaction rates(by solving differential equation), we can predict concentrations of each species at any time after the reaction starts. Reaction rate laws provide insights for elementary steps. To conclude the reaction rate law, we can construct rate laws from proposed mechanism and compare with experimental results. Simple elementary steps have simple rate laws, which can be deduced from reaction mechanism and stoichiometry. To find out the reaction rate order and parameters, we often use approximations to simplify the reaction rate law (then it is easier to handle or compare with experimental results) and figure out unknowns. Approximation: the rate determining stage, the steady state concentration of a reaction intermediate, the existence of a pre-equilibrium

5 Empirical chemical kinetics * The steps of the kinetic analysis of reaction is to first, establish the stoichiometry of the reaction and identify any side reactions And then collect concentration of reactants and products with time through experiments Then plot experimental data and compare the graph with the established rate equation * Reactions are sensitive to the temperature. Therefore, conventionally experiments are carried out under the constant temperature condition.

6 21.1 Experimental techniques Methods to monitor concentrations depend on which species are involved in reactions and how rapidly the concentration changes. (a) Monitoring the progress of a reaction with GAS components measure pressure of gas components Reactions including at least one gas component pressure changes under constant volume system monitor pressure variations with time (then convert Pres. to Conc.) ex) 21.1 Monitoring the variation in pressure (at constant volume/temperature) Gas phase decomposition reaction: 2 N 2 O 5 (g) 4 NO 2 (g) + O 2 (g) Each mole of N 2 O 5 consumption gives rise to 5/2 mol of gas 5/2 times pressure up. If initial pressure is p 0, initial amount of N 2 O 5 is n, α fraction of N 2 O 5 decomposed, then each component has following amount: 2N 2 O 5 4NO 2 + O 2 Total Amount n (1 - α) 2 αn ½ αn n (1 + 3/2α) Assume perfect gas at constant volume/temp., then total pressure p = (1 + 3/2α)p 0 When α = 0, the pressure is p 0. When α = 1, reaction completes 5/2 times pressure up, the pressure is 5/2p 0.

7 Other Reaction Monitoring Methods (other than gas pressure) Spectrophotometry: measurement of absorption intensity change of light by a particular reaction component in a particular spectral region (This is useful when one substance in reaction mixture has a strong characteristic absorption, for example, UV-Vis Spectrometer) ex) H 2 (g) + Br 2 (g) 2HBr(g) this can be followed by measuring visible light absorption of Br 2 Conductivity measurement: Reaction having changes in number/species of ions in solution can be followed by monitoring conductivity change ex) (CH 3 ) 3 CCl(aq) + H 2 O(l) (CH 3 ) 3 COH(aq) + H + (aq) + Cl - (aq) increased conductivity ph measurement: Reaction that is involved with hydrogen ions (H + ), which changes solution ph Others: emission spectroscopy, mass spectroscopy, gas spectroscopy, nuclear magnetic resonance, electron paramagnetic resonance, etc..

8 (b) Applications of the techniques 1. Real-time analysis examples: monitoring reaction in progress (i) Continuous flow method - Mix solution of reactants in chamber flow to outlet tube observation at multiple, different times, under continuous flow - Disadvantages: large volume needed, difficult for fast reaction (ii) Stopped-flow technique - Stopping syringe (flow stops!) and then measurement, observation of time dependent concentration change - Advantage: Small amount sample required (bio, protein, enzyme) (iii) Photolysis (or flash photolysis) - Samples are exposed to flashing light (e.g., pulse laser) reaction initiated by flashing light and then reactions are monitored using time-resolved spectroscopy (electronic absorption/emission, IR, Raman scattering) - Advantage: very fast reaction can be monitored (fs or ns time scale), since the reaction is controlled (initiated) by light ex) Photolysis: light-induced dissociation of Cl 2 (g) to Cl atom Cl 2 + hν (flashing light) Cl + Cl Cl + HBr HCl* + Br (HCl*: a vibrationally excited HCl molecule) HCl* + M HCl + M (M: body (unreactive molecule that receives energy))

9 2. Quenching (time-stopping) analysis (non-real-time analysis) - Stop (quench) reaction, after the reaction proceeds for certain amount of time. - This can work only for slow reactions, for the case that quenching time (quenching is a chemical reaction that requires some time) does not affect the overall reaction. (i) Chemical quench flow method - Quenched by another chemical reagent (e.g., acid/base solution) - Advantage over stopped flow method (real time analysis method: the flow stops but the reaction continues) : fast analysis method, such as spectroscopy (ex, Raman spectroscopy), is not needed to measure concentrations of reactants or products (i.e. slow analysis method, such as chromatography, mass spectrometry, etc., is good enough to analyze the reaction.) (ii) Freeze quench method - Quench reaction by cooling within mili-seconds and then measure by spectroscopy

10 21.2 The rates of reactions The instantaneous rate of a reaction is the slope of the tangent of molar concentration against time. (a) The definition of rates We need molar concentration of J ([J]) ex) For a reaction: A + 2B 3C + D Rate of consumption, -d[r]/dt, R: A or B Rate of formation, d[p]/dt, P: C or D Rates follow the stoichiometry (& tangent of [J]): d[d]/dt = (1/3)d[C]/dt = -d[a]/dt = (-1/2)d[B]/dt Rate of reaction (v) can be expressed by different components ([A], [B], [C] or [D]). To prevent confusion, let s use the extent of reaction ξ [sai] *Definition: ξ = (n J n J,0 )/ν J reaction rate v = 1/V dξ/dt = 1/(Vν J ) dn J /dt (ν J is stoichiometric number of species j, negative for reactants and positive for products.) - For homogeneous reaction (at constant V), use [J]=n J /V, v = (1/ν J )(d[j]/dt) (mol L -1 s -1 ) - For heterogeneous reaction (at constant A), use σ J =n J /A, v = (1/ν J )(dσ J /dt) (mol m -2 s -1 ) (for example, surface area of catalyst occupied by the species) ex) 2 NOBr(g) 2 NO(g) + Br 2 (g), if NO formation rate is 0.16 mmoll -1 s -1 ν NO = +2 and since rate of formation of NO is 0.16 reaction rate, v = mmoll -1 s -1 ν NOBr = -2 rate of consumption of NOBr = d[nobr]/dt = mmoll -1 s -1

11 (b) Rate laws and rate constant Basically, rate law of the reaction should be determined experimentally. For example, the rate of reaction can be proportional to molar concentration of reactants (A, B). In this case, rate law: v = k[a][b], but more generally, v = f([a], [B], ) k: rate constant (~ independent of concentration but depends on temperature) For homogeneous gas-phase reaction, often the rate law is expressed by partial pressures, p J = RT[J] (PV=nRT, P=RT(n/V)=RT[J]) v = f(p A, p B, ) The rate law of reaction cannot be inferred from stoichiometry of chemical equation. Rather it should be determined by experiments (except for the elementary reaction). ex) H 2 (g) + Br 2 (g) 2HBr(g): v = k[h 2 ][Br 2 ] 3/2 /([Br 2 ] + k [HBr]) Brief Illustration) Unit conversion (refer to p788) 1cm = 10-2 m = dm = 10-1 dm 1mol = molecules, therefore 1 molecule = 1mol/ rate constant k r = cm 3 molecule -1 s -1 can be converted into dm 3 mol -1 s -1

12 (c) Reaction order If a reaction rate law is v = k[a] a [B] b.. Reaction order: a, b, : a-order in A, b-order in B overall order: a+b+.. (Overall order of a reaction is the sum of individual order for each species) ex) order of reaction law v = k[a][b], first-order in A and first-order in B, second-order overall v = k[a] 1/2 [B], half-order in A, first-order in B, 3/2-order overall v = k, zero-order (independent of concentration of the reactant) exception) H 2 (g) + Br 2 (g) 2HBr(g) v = k[h 2 ][Br 2 ] 3/2 /([Br 2 ] + k [HBr]) first order in H 2, but the reaction order of both Br 2 and HBr is not defined, also the overall order is not defined..

13 (d) The determination of the rate law (1) Isolation method - The determination of a rate law is simplified by the isolation method : - Use concentrations of all reactants except one in large excess - Then the change of concentration of reactant in large excess can be ignorable. - For v = k[a][b], if B is in large excess [B] goes to [B] 0 (constant) v = k[a][b] becomes v = k [A], k = k[b] 0 - Since 2 nd order reaction becomes 1 st order reaction by assuming that [B] in large excess is constant, it is called as a pseudo first-order rate law. - The dependence of the rate on the concentration of each species can be found by isolating each species in turn. (2) Method of initial rates The rate is measured at the beginning of the reaction for several different initial concentrations of reactants. The product concentration is very small and we assume that it does not affect to rates. A reaction with A isolated, v = k[a] a (k include other species in large excess) initial rate v 0 = k[a] 0a log v 0 = log k + a log [A] 0 For different initial concentrations, measure initial rates, and then plot them. log plot will provide straight line with slope a and intercept log k.

14 ex) Using the method of initial rates 2I(g) + Ar(g) I 2 (g) + Ar(g) (recombination of I atoms in Ar gas) Ar concentration (a) 1.0 mmoll -1, (b) 5.0 mmoll -1, (c) 10 mmoll -1 Initial conc [I] 0 /10-5 mol L v 0 /moll -1 s -1 (a) 8.70 x x x x 10-2 (b) 4.35 x x x x 10-1 (c) 8.69 x x x x 10-1 Reaction orders for I and Ar? Rate constant? sol) log plot: log v 0 vs. log [I] 0 for given Ar concentration log v 0 vs. log [Ar] 0 for given I concentration slopes: reaction orders, intercept: k log v 0 = log k + a log [I] 0 + b log [Ar] 0 The slopes are 2 and 1 for [I] 0 and [Ar] 0 v 0 = k[i] 02 [Ar] 0 second-order in [I], first-order in [Ar], third-order overall logv 0 =log k+2log[i] 0 +log[ar] 0 & from intercept: k= mol -2 L 2 s -1 cf) Method of initial rates is not universal: once the product is generated, the product can participate in the reaction and affect the rates fit using a proposed rate law throughout the reaction and compare with measured data, confirm whether the addition of products affects the reaction rates

15 Exercise (#1) 1-1) The rate of formation of C in the reaction 2A+B 2C+3D is 2.7mol dm -3 s -1. State the reaction rate, and the rates of formation or consumption of A, C, and D. v = 1 d[j] v J dt [21.3b] = 1 d[c] 2 dt = 1 2 (2.7mol dm 3 s 1 ) = 1.35mol dm 3 s 1 Rate of formation of D = 3v = 4.05 mol dm -3 s -1 Rate of consumption of A = -2v = -2.7 mol dm -3 s -1 Rate of consumption of B = -v = mol dm -3 s ) The rate law for the reaction in above problem (1) was reported as d[c]/dt=k r [A][B][C]. Express the rate law in terms of the reaction rate; what are the units for k r in each case? Given d[c] dt = k r[a][b][c], the rate of reaction is [21.3b] The units of k r, [k r ], must satisfy Therefore, [k r ]=dm 6 mol -2 s -1 v = 1 d[j] = 1 d[c] v J dt 2 dt = 1 2 k r[a][b][c] mol dm -3 s -1 =[k r ] (mol dm -3 ) (mol dm -3 ) (mol dm -3 )

16 Exercise (#1) 1-3) If the rate laws are expressed with (a) concentrations in moles per decimeter cube, (b) pressures in kilopascals, what are the units of the secondorder and third-order rate constants? (a) For a second-order reaction, denoting the units of k r by [k r ] mol dm -3 s -1 =[k r ] (mol dm -3 ) 2 ;therefore [k r ]=dm 3 mol -1 s -1 For a third-order reaction mol dm -3 s -1 =[k r ] (mol dm -3 ) 3 ;therefore [k r ]=dm 6 mol -2 s -1 (b) For a second-order reaction kpa s -1 =[k r ] kpa 2 ; therefore [k r ]=kpa -1 s -1 For a third-order reaction kpa s -1 =[k r ] kpa 3 ; therefore [k r ]=kpa -2 s -1

17 21.3 Integrated rate laws * Rate laws are differential equations. they should be integrated to find the solution of differential equation, which is concentrations of reactants and products as a function of time. * Let s find these solutions for different reaction orders. (a) First-order reaction of A, A P, v=k[a] integrated form of first order reaction rate law: [A] = [A] 0 e -kt Justification d[a]/dt = -k[a] d[a]/[a] = -kdt ([A] 0 at t=0) d[a]/[a] =-k dt ln([a]/[a] 0 ) = -kt [A] = [A] 0 e -kt For 1 st -order reaction law: If we plot ln([a]/[a] 0 ) vs t, it results in straight line with slope -k (ln([a]/[a] 0 ) = -kt) Concentration of the reactant decreases exponentially with time, and the extent of decrease is related with the rate constant k (right top graph)

18 ex) Analysing 1 st -order reaction Partial pressure of azomethane was measured with time at 600 K Is this decomposition first-order in azomethane? What is the rate constant at 600 K? CH 3 N 2 CH 3 (g) CH 3 CH 3 (g) + N 2 (g) t/s p/(10-2 Torr) Sol) First, let s assume that it is a first order reaction (ln([a]/[a] 0 ) = -kt). Since partial pressure, p, is proportional to concentration, [A], in the gas phase reaction, let s plot ln(p/p 0 ) vs. t to see if it is straight line. (If it is first order reaction, ln([a]/[a] 0 ) = -kt) The plot is straight (right top corner graph). This reaction is a first-order reaction!! From the graph, slope = s -1, and therefore k = s -1

19 (b) Half-life and time constant Half-life (t 1/2 ): time taken for the reactant concentration to fall to half of initial value From 1 st order rate law, ln([a]/[a] 0 ) = -kt kt 1/2 = -ln(½ [A] 0 /[A] 0 ) = -ln½ = ln2 t 1/2 = ln2/k (ln 2 ~ 0.693) * Half-life of 1 st order reaction is independent of its initial concentration Time constant (τ): time for the reactant concentration to fall to 1/e of initial value From 1 st order rate law, ln([a]/[a] 0 ) = -kt kτ = -ln{([a] 0 /e)/[a] 0 } = -ln(1/e) = 1 τ = 1/k * Time constant of 1 st order reaction is independent of its initial concentration

20 (c) Second-order reactions (1) 2 nd -order reaction of A, A P (A+A P), v=k[a] 2 d[a]/dt = -k[a] 2 d[a]/[a] 2 = -k dt ([A] 0 at t = 0) d[a]/[a] 2 = k dt 1/[A] 1/[A] 0 = kt or [A] = [A] 0 /(1 + kt[a] 0 ) For 2 nd -order reaction law: If we plot 1/[A] vs. t and if it is straight line, then it is 2 nd -order reaction with slope, k, and intercept, 1/[A] 0. [A] approaches zero more slowly (fractional decrease) than 1 st -order reaction (exponential decrease) with same [A] 0 *Half life t 1/2 at [A] = ½[A] 0, t 1/2 = 1/k[A] 0 : Half-life varies with [A] 0 : longer at low [A] 0 *In general, for n th order reaction (n>1) of A P The half life varies, which is related to k and [A] 0 as t 1/2 = (2 n-1-1)/(n-1)k[a] 0 n-1 1 st 2 nd

21 (2) Another type of 2 nd -order reaction A+B P v=k[a][b] d[a]/dt = -k[a][b] if [A] is consumed by the amount x, then [A] = [A] 0 -x, [B] = [B] 0 -x d[a]/dt = -k([a] 0 x)([b] 0 x) and by differentiating [A]=[A] 0 -x with t, then d[a]/dt = -dx/dt dx/dt = -d[a]/dt = k([a] 0 x)([b] 0 x) with boundary condition of x = 0 at t = 0 dx/[([a] 0 x)([b] 0 x)] = k dt, [A] = [A] 0 and [B] = [B] 0 (or x = 0) at t = 0 Using standard integration table, dx/[([a] 0 x)([b] 0 x)] = {1/([B] 0 [A] 0 )}{ln[[a] 0 /([A] 0 x)] - ln[[b] 0 /([B] 0 x)]} = {1/([B] 0 [A] 0 )}{ln[[a] 0 / [A]] - ln[[b] 0 / [B]]} = kt ln[([b]/[b] 0 )/([A]/[A] 0 )] = ([B] 0 [A] 0 )kt Plot ln[([b]/[b] 0 )/([A]/[A] 0 )] vs. t straight line: from the slope, k can be obtained If [A] 0 = [B] 0, then [A] = [A] 0 x = [B] 0 x = [B], same with previous case

22

23 Exercise (#2) A second-order reaction of the type A+B P was carried out in a solution that was initially 0.075mol dm -3 in A and 0.050mol dm -3 in B. After 1.0h, the concentration of B had fallen to 0.020mol dm -3. (a) Calculate the rate constant. The integrated rate law of 2 nd order reaction is 1 k t t = ln( [B]/[B] 0 )[21.19] [B] 0 [A] 0 [A]/[A] 0 (a) The stoichiometry of the reaction shows that after 1 hour [A] = mol dm -3, since [B] =( )mol dm -3 = mol dm -3 Thus, [A]=0.075mol dm mol dm -3 =0.045mol dm -3 when [B]=0.020mol dm -3. Therefore, k r t= 1 ( )mol dm 3 ln(0.020/0.050 ) 0.045/0.075 k r 1.0h=16.2dm 3 mol -1 so, k r =16.2dm 3 mol -1 h -1 ( 1h 3600s ) = dm 3 mol -1 s -1

24 Exercise (#2) A second-order reaction of the type A+B P was carried out in a solution that was initially 0.075mol dm -3 in A and 0.050mol dm -3 in B. After 1.0h the concentration of B had fallen to 0.020mol dm -3. (b)what is the half-life of the reactants? (b) The half-life with respect to A is the time required for [A] to fall to mol dm -3 (and then [B] to =0.0125mol dm -3 ). We can solve eqn , 1 k t t = ln( [B]/[B] 0 )[21.19] for half life with k [B] 0 [A] 0 [A]/[A] r =16.2dm 3 mol -1 h -1, 0 1 t 1/2 (A)=( ) (16.2dm 3 mol 1 h 1 ) ( 0.025mol dm 3 ) ln(0.0125/0.050 )=1.71h= s 0.50 Similarly, the half-life with respect to B is the time required for [B] to fall to mol dm -3 (and then [A] to 0.050mol dm -3 ). 1 t 1/2 (B)=( ) ln( 0.50 (16.2dm 3 mol 1 h 1 ) ( 0.025mol dm 3 ) 0.050/0.075 )=0.71h= s

25 21.4 Reactions approaching equilibrium - So far all rate laws did not consider the reverse reaction by products. For example, we assumed initial rates. - But when the reaction is close to equilibrium, product concentration is too high to ignore reverse reactions. - In practice, therefore, most kinetic study is done far from equilibrium, like the case of initial rates, where reverse reaction is not important and ignorable. (a) First-order reaction close to equilibrium A B B A v = k[a] : forward reaction v = k [B] : reverse reaction Net rate of A = consumption of A + generation of A, d[a]/dt = -k[a] + k [B] If initial condition of A is [A] 0 and [B] 0 =0 at t=0, always [A] + [B] = [A] 0, since one A is converted into one B d[a]/dt = -k[a]+k ([A] 0 -[A]) = -(k+k )[A]+k [A] 0 Integrate the above differential equation with [A] 0 at t=0 [A] = {[k + ke -(k+k )t ]/(k + k)}[a] 0, [B]=[A] 0 -[A] (Time dependent decrease of [A] and increase of [B] )

26 As t, the concentrations reach their equilibrium values, From [A] = {[k + ke -(k+k )t ]/(k + k)}[a] 0, [B]=[A] 0 -[A] [A] eq = k [A] 0 /(k + k ), [B] eq = [A] 0 [A] = k[a] 0 /(k + k ) From physical chemistry 1 (chemical equilibrium): For reaction: equilibrium constant: Equilibrium constant of reaction A B K = [B] eq /[A] eq = k[a] 0 /(k + k )/k [A] 0 /(k + k ) = k/k (from above results) The same conclusion (equilibrium constant) can be derived by following relation: k[a] eq = k [B] eq (at equilibrium, forward and reverse rate same, by definition) K = [B] eq /[A] eq = k/k Thermodynamically derived quantity, equilibrium constant, K, can be related to the kinetically derived quantity, rate constants (rate constants, k and k ). More generalized overall equilibrium constant, K, for general reactions, can be expressed with reaction rate constants, k a, k b, k a, k b,, as follows: K = k a /k a x k b /k b x.. (k: rate constant for forward steps, k : reverse steps)

27 (b) Relaxation methods Relaxation: the return of a system to equilibrium after a sudden shift of equilibrium condition to new one by external influence (ex. temperature or pressure jump) Temperature jump in A B equilibrium (1 st -order) At initial temperature (before temperature jump) d[a]/dt = -k a [A]+k b [B] (at equilibrium, d[a]/dt = 0, by def.) k a [A] eq = k b [B] eq (at initial temperature) When temperature increased suddenly, rate constants change to k a, k b (rate constant is temp. dependent, previous ones: k a, k b ) and reach new equilibrium (k a, k b ), where d[a]/dt = 0 again. k a [A] eq = k b [B] eq (at new temperature) * Concentration of A and B relax into new equilibrium with new rate constants Deviation of [A] from equilibrium value, as amount of x, [A]=[A] eq +x, [B]=[B] eq -x From d[a]/dt = -k a [A]+k b [B] and [A] = [A] eq +x, ( d[a]/dt = dx/dt) d[a]/dt = -k a ([A] eq +x)+k b ([B] eq -x) = -(k a +k b )x, (since k a [A] eq = k b [B] eq by definition) dx/dt = -(k a +k b )x integration: x = x 0 e -t/τ, 1/τ = k a +k b, where τ is the relaxation time (x and x 0 : deviation from equilibrium at the new temperature and immediately after the temperature jump, respectively)

28 ex) 21.4 Analysing a temperature-jump experiment H 2 O(l) H + (aq)+oh - (aq) reaction relaxes to equilibrium after a small external perturbation with 37μs relaxation time at 298 K and ph ~ 7 and K w = [H + ] eq [OH - ] eq = Forward reaction: 1 st -order, reverse reaction: 2 nd -order (1 st for each). Calculate rate constants for forward and reverse reactions (k 1 and k 2 ). solution) Rate equation: d[h 2 O]/dt = -k 1 [H 2 O] + k 2 [H + ][OH - ] (k 1 : forward, k 2 : reverse) Deviation from equilibrium: [H 2 O] = [H 2 O] eq +x, [H + ] = [H + ] eq x, [OH - ] = [OH - ] eq -x d[h 2 O]/dt = dx/dt = -{k 1 +k 2 ([H + ] eq +[OH - ] eq )}x k 1 [H 2 O] eq + k 2 [H + ] eq [OH - ] eq + k 2 x 2 Small deviation: x>>x 2 neglect x 2, Also at equilibrium k 1 [H 2 O] eq = k 2 [H + ] eq [OH - ] eq dx/dt -{k 1 +k 2 ([H + ] eq +[OH - ] eq )}x x = x 0 e -t/τ, where 1/τ= k 1 +k 2 ([H + ] eq +[OH - ] eq ) = k 2 {K + [H + ] eq + [OH - ] eq } (where K=k 1 /k 2 ) Since τ is given above, we will find out K first. Then k 2 and then k 1. K = k 1 /k 2 = [H + ] eq [OH - ] eq /[H 2 O] eq = K w /[H 2 O] eq = K w /55.6 (moll -1 ) (since, molar concentration of pure water, [H 2 O] eq : 55.6 moll -1 ) K = K w /55.6 = /55.6= , Also at ph ~ 7, [H + ] eq =[OH - ] eq =K w ½ 1/τ = k 2 {K+[H + ] eq +[OH - ] eq } = k 2 {K+K w½ +K w½ } = ( ) k 2 moll -1 = 1/( ) k 2 = 1/[( s) ( moll -1 )] = Lmol -1 s -1 k 1 = k 2 K = s -1

29 Exercise (#3) The equilibrium NH 3 (aq)+h 2 O(l) NH 4+ (aq)+oh - (aq) at 25 C is subjected to a temperature jump that slightly increases the concentration of NH 4+ (aq) and OH - (aq). The measured relaxation time is 7.61ns. The equilibrium constant for the system is at 25 C, and the equilibrium concentration of NH 3 (aq) is 0.15mol dm -3. Calculate the rate constant. (Use a condition of pseudo-first-order protonation of NH 3 in excess water, i.e. water concentration does not change and the forward reaction can be considered as a first order reaction.) The reactions : NH 3 (aq)+h 2 O(l) kr k rnh 4+ (aq)+oh - (aq) The rate constants are related by: K b = k r k = [NH 4 + ][OH ] r [NH 3 ] = mol dm -3 where the concentrations are equilibrium concentrations. (We assign units to K b, which technically is a pure number, to help us keep track of units in the rate constants. Keeping track of the units makes us realize that k r is a pseudo-first-order protonation of NH 3 in excess water, for which water does not appear in the above expression.) We need one more relationship between the constants, obtained as in Example 21.4: Rate equation: d[nh 3 ]/dt = -k r [NH 3 ] + k r [NH 4+ ][OH - ] Deviation from equilibrium: [NH 3 ]= [NH 3 ] eq +x, [NH 4+ ] = [NH 4+ ] eq x, [OH - ] = [OH - ] eq -x d[nh 3 ]/dt = dx/dt = -{k r + k r ([NH + ] eq + [OH - ] eq )}x - k r [NH 3 ] eq + k r [NH 4+ ] eq [OH - ] eq + k r x 2 Small deviation: x>>x 2 neglect x 2, also at equilibrium k r [NH 3 ] eq = k r [NH 4+ ] eq [OH - ] eq dx/dt -{ k r + k r ([NH 4+ ] eq + [OH - ] eq )}x By solving differential equation, x = x 0 e -t/τ, where 1 τ =k r+k r ([NH 4+ ]+[OH - ]) Substitute k r and [NH 4+ ]&[OH - ] into this expression with k r =K b k r and [NH 4+ ]=[OH - ]=(K b [NH 3 ]) 1/2 hence, 1 τ =K bk r +2 k r (K b [NH 3 ]) 1/2 = k r {K b +2(K b [NH 3 ]) 1/2 }, where we know, K b, [NH 3 ]

30 (Cont d) So the reverse rate constant can be calculated 1 k r = τ{k b +2(K b [NH 3 ]) 1/2 } 1 = s{ mol dm 3 +2( ) 1/2 mol dm 3= dm -3 mol s -1 And the forward constant is k r =K b k r = mol dm dm 3 mol -1 s -1 = s -1 recall that k r is the rate constant in the pseudo-first-order rate law (where we assume that [H 2 O] does not change due to its excess amount) d[nh 3] dt =k r [NH 3 ] (from pseudo-first-order) Let us call k as the rate constant in the bimolecular rate law d[nh 3] dt =k[nh 3 ][H 2 O] (from real bimolecular reaction) Setting these two expressions equal to each other yields the following (Molar concentration of water [H 2 O]=55.6mol/L) k= k r = s 1 =1.28 [H 2 O] (1000g dm 3 )/(18.02g mol 1 ) 104 dm 3 mol -1 s -1

31 21.5 The temperature dependence of reaction rate (a) Arrhenius parameters The rate constant of most reactions increase as T increases. Experimentally found: ln k vs. 1/T straight line Arrhenius equation: ln k = ln A E a /RT or k= A exp (-E a /RT) Arrhenius parameters A and E a A: pre-exponential factor or frequency factor E a : activation energy ln A: intercept at 1/T = 0 (infinite temperature) -E a /R : slope

32 ex) 21.5 Determining the Arrhenius parameters Rate of 2 nd -order decomposition of acetaldehyde (CH 3 CHO) over K. E a and A? T/K k/(lmol -1 s -1 ) Sol) ln k = ln A E a /RT plot lnk vs. 1/T and from slope and intercept Slope: , intercept: 27.7 E a = K J K -1 mol -1 = 189 kj mol -1 A = e 27.7 dm 3 mol -1 s -1 = dm 3 mol -1 s -1 E a is given by the slope of ln k vs 1/T graph: k= A exp (-E a /RT) Higher E a stronger temperature dependence of k If Zero E a rate is independent of T (k is constant) If Negative E a reaction rate as T Some reactions are not Arrhenius-like ( plot of lnk vs 1/T is not straight line). But still possible to express activation energy as temperature derivative: E a = RT 2 [d(lnk)/dt]: definition of the activation energy (Try to differentiate ln k = ln A E a /RT and rearrange.)

33 (b) The interpretation of the parameters Arrhenius equation: k = A e -Ea/RT Definitions Collision: A and B reactants come into contact Reaction coordinate: collection of motions, (ex. changes of inter-atomic distances and bond angles, which are directly involved in the formation of products from reactants) Activation complex: cluster of atoms that corresponds to the potential energy maximum Transition state: crucial configuration to make products E a : minimum kinetic energy that reactants must have, in order to form products the fraction of collisions with a kinetic energy in excess of E a is given by Boltzmann distribution as e -Ea/RT : fraction of collisions that have enough kinetic energy to lead to the reaction Pre-exponential factor (A): a measure of the rate at which collisions occur k=a e -Ea/RT measure of collision rate fraction of enough E to react the measure of rate of successful collisions and reactions Fig Potential E profile of exothermic reaction. The height of the E barrier is the activation E of the reaction.

34 <Summary of Ch 21.1 ~ Ch 21.5> ξ: extent of reaction ξ = (n J n J,0 )/ ν J, v= 1/V dξ/dt = 1/(Vν J ) dn J /dt Reaction rates: rate laws, rate constants, reaction order Determine rate law: isolation method, method of initial rates The intergrated rate laws (solution of reaction rate law) 1 st order reaction: [A] = [A] 0 e -kt, t 1/2 = ln 2/k, τ = 1/k, k(s -1 ) 2 nd order reaction: 1/[A] 1/[A] 0 = kt t 1/2 = 1/k[A], k(m -1 s -1 ) K = [B] eq /[A] eq = k/k (general: K = k a /k a x k b /k b x..) Relaxation: x = x 0 e -t/τ, 1/τ = k a +k b Arrhenius equation: ln k = ln A E a /RT or k= A exp (-E a /RT) E a = RT 2 [d(lnk)/dt]: definition of the activation energy

35 Accounting for the rate laws 21.6 Elementary reactions Elementary reaction is a chemical reaction in which one or more of the chemical species react directly to form products in a single reaction step and with a single transition state. Molecularity is the number of molecules coming together to react in an elementary reaction Unimolecular elementary reaction is a reaction, in which a single molecule shakes itself and changes its atoms into a new arrangement. ex) A P, d[a]/dt = -k[a] Bimolecular elementary reaction is a reaction, in which a pair of molecules collide and exchange energy between atoms, or groups of atoms, and form products. ex) A + B P, d[a]/dt = -k[a][b] Molecularity and Reaction Order is a different concept. Molecularity refers to number of molecules in an elementary reaction (an individual step in an entire mechanism). Reaction order is an empirical quantity, obtained from experimental rate law.

36 21.7 Consecutive elementary reactions Some reactions proceed through intermediate (I), as in the consecutive unimolecular reactions: k a k b 23.5 min 2.35 day A I P ex) 239 U 239 Np 239 Pu (a) The variation of concentrations with time Rate equations: d[a]/dt = -k a [A], d[i]/dt = k a [A] k b [I], d[p]/dt = k b [I] Suppose only A was present initially: [A]=[A] 0, [I]=[I] 0 =0, [P]=[P] 0 =0 at t=0 (initial condition) Then the integration of first differential equation using above i.c.: [A] = [A] 0 e -kat Insert [A] into second differential equation: d[i]/dt+k b [I] = k a [A] 0 e -kat, [I] 0 = 0 at t=0 Then the integration of second differential equation: [I] = {k a /(k b k a )}(e -kat e -kbt )[A] 0 During the reaction, total concentration is constant, since this is unimolecular reaction, therefore, total concentration = [A]+[I]+[P] = [A] 0 +[I] 0 +[P] 0 = [A] 0 [P] = [A] 0 -[A]-[I] = {1 + [(k a e -kbt k b e -kat )/(k b k a )]}[A] 0 If we plot three concentrations vs time [A] decays exponentially. [I] rises to a maximum, then falls to zero. [P] rises from zero toward [A] 0

37 ex) 21.6 Analyzing consecutive reactions A produces I (desired) and I goes to C (worthless), each-step is 1 st order reaction when desired product, [I], will be maximized? Sol) From the concentration equation in the previous slide: [I] = {k a /(k b k a )}(e -kat e -kbt )[A] 0 To estimate the maximum value, differentiate [I] over t and make it equal to 0: d[i]/dt = -{k a /(k b k a )}(k a e -kat k b e -kbt )[A] 0 = 0 This rate = 0 when k a e -kat = k b e -kbt t max = {1/(k a - k b )}ln(k a /k b )

38 (b) The steady-state approximation If the reaction mechanism has more than couple of elementary steps, the analytical solution is too complicated to solve. Either a numerical approach or an approximation (such as steady-state approximation ) to simplify it is required. Steady-state approximation: during major part of reaction (after initial reaction), the rate of changes of concentrations of all reaction intermediates are negligibly small, d[i]/dt 0 (we are assuming that the lifetime of I is very short) k a k b In the reaction : A I P d[a]/dt = -k a [A], d[i]/dt = k a [A] k b [I], d[p]/dt = k b [I] The intermediate is assumed not to change: d[i]/dt 0 (steady state) [A] = [A] 0 e -kat, [I] (k a /k b )[A] d[p]/dt = k b [I] k a [A] [P] = k a [A] 0 e -kat dt = (1 e -kat )[A] 0 Let s compare it with general solution w/o approximation: [P] = [A] 0 -[A]-[I] = {1 + [(k a e -kbt k b e -kat )/(k b k a )]}[A] 0 If k b >>k a (I is consumed much faster than generation), it reduces to [P] = (1 e -kat )[A] 0 (dotted line is under steady state approximation: similar)

39 ex) 21.7 using the steady-state approximation 2N 2 O 5 (g) 4NO 2 (g) + O 2 (g), Q) write rate law for N 2 O 5 decomposition.. Mechanisms: N 2 O 5 NO 2 + NO 3 k a NO 2 + NO 3 N 2 O 5 k a NO 2 + NO 3 NO 2 + O 2 + NO k b NO + N 2 O 5 NO 2 + NO 2 + NO 2 k c Intermediates are NO 3 and NO. (N 2 O 5 is reactant, NO 2 and O 2 are products) Let s assume that net rates of change of their concentration 0. d[no]/dt = k b [NO 2 ][NO 3 ] k c [NO][N 2 O 5 ] 0 d[no 3 ]/dt = k a [N 2 O 5 ] - k a [NO 2 ][NO 3 ] k b [NO 2 ][NO 3 ] 0 [NO]= k b [NO 2 ][NO 3 ] / k c [N 2 O 5 ], [NO 3 ] = k a [N 2 O 5 ] / (k a +k b )[NO 2 ] rate law of N 2 O 5 concentration: d[n 2 O 5 ]/dt = -k a [N 2 O 5 ] + k a [NO 2 ][NO 3 ] k c [NO][N 2 O 5 ] Substitute [NO] and [NO 3 ] and continue calculation, d[n 2 O 5 ]/dt = - 2k a k b /(k a + k b ) [N 2 O 5 ] Simplified into 1 st order differential equation, which can be easily solved

40 (c) The rate-determining step k a k b A I P Suppose k b >> k a, as soon as I is produced from A, I decays rapidly into P (like steady state approximation) Since e -kbt <<e -kat and k b k a k b, under above assumption, [P] = {1 + [(k a e -kbt k b e -kat )/(k b k a )]}[A] 0 reduces to [P] (1-e -kat )[A] 0 * P depends on only the smaller rate constant (k a ) of two rate constants (or slower rxn) Step A I (slower reaction) is the rate-determining step * Rate-determining step : slowest step in a mechanism and a crucial gateway for the formation of the products, which control the overall rate of the reaction. RD step RD step No RD

41 (d) Pre-equilibria Let s see more complicated mechanism: Pre-equilibrium: Intermediate I is in equilibrium with reactants, A and B. This happens when intermediate, I, goes back to reactants (A and B) much faster, rather than forms product, P (i.e. k a >> k b ). k a k b A + B I P k a Since A, B and I are in equilibrium (pre-equilibrium), the equilibrium constant K can be defined as follows (Assumption: I P is too slow to affect pre-equilibrium, I A and B is much faster) : K = [I]/([A][B]), [I] = K[A][B], K = k a /k a (by definition of equilibrium constant) d[p]/dt = k b [I] = k b K[A][B] = k a k b /k a [A][B] d[p]/dt = k[a][b] where k = k b K = k a k b /k a : 2 nd -order rate law

42 ex) 21.8 Analyzing a pre-equilibrium Repeat pre-equilibrium calculation but without the assumption. i.e. P formation is not too slow (I slowly leaking away to form P) k a k b A + B I P k a Sol) Net rates of change of P and I: d[p]/dt = k b [I] Let s use steady-state approximation for the intermediate I : d[i]/dt = k a [A][B] - k a [I] k b [I] 0 [I] = k a [A][B]/(k a + k b ) d[p]/dt = k b [I] = k[a][b],where k = k a k b /(k a + k b ) (cf. if we use the assumption (P formation is slow) k b << k a k k a k b /k a, same result with the previous slide)

43 (e) Kinetic and thermodynamic control of reaction Suppose two products, P 1 and P 2, are produced by following competing reactions: A + B P 1 rate of formation of P 1 P 1 = k 1 [A][B] A + B P 2 rate of formation of P 2 P 2 = k 2 [A][B] Relative proportion of two products (before reaching equilibrium) = ratio of two rates = ratio of two rate constants (due to the same reactants) [P 2 ]/[P 1 ] = k 2 /k 1 which means kinetic control of proportion of products in competing reactions (If the reaction reaches equilibrium, the proportion is determined not by kinetics but by thermodynamics.)

44 <Summary of Ch 21.6 ~ Ch 21.7> Definitions: Reaction coordinate, Activation complex, Transition state e -Ea/RT : fraction of collisions that have enough kinetic energy to lead to reaction A : a measure of the rate which collisions occur irrespective of their energy Definitions: Elementary reactions, Molecularity Unimolecular elementary reaction and Bimolecular elementary reaction Consecutive elementary reactions A I P The steady-state approximation d[i]/dt ~ 0 The rate-determining step Pre-equilibrium Kinetic and thermodynamic control of reaction

45 Examples of reaction mechanisms 21.8 Unimolecular reactions Many gas phase reactions are first order reaction. ex) Isomerization of cyclopropane: A P cyclo-c 3 H 6 CH 3 CH=CH 2 v = k[cyclo-c 3 H 6 ] How this 1 st -order and unimolecular reaction can happen? (Presumably a molecule gets energy usually through collision, which is a bimolecular event..)

46 (a) Lindemann-Hinshelwood mechanism L-H mechanism explains unimolecular reaction, in which the reactant is excited by collision with another reactant in a bimolecular step: k a k a Activation: A + A A* + A d[a*]/dt = k a [A] 2 Deactivation: A + A* A + A d[a*]/dt = -k a [A][A*] k Unimolecular decay: A* b P d[a*]/dt = -k b [A*] If the unimolecular step (unimolecular decay) is much slower than the bimolecular step, i.e. k b << k a & k a (k b to be the rate determining step), the overall reaction will be the first order kinetics (dominated by slower reaction, unimolecular decay). Let s use Steady-state approximation ( [A*] is constant) to find the general solution: d[a*]/dt = k a [A] 2 - k a [A][A*] k b [A*] 0 [A*] = k a [A] 2 /(k b + k a [A]) d[p]/dt = - d[a*]/dt = k b [A*] = k a k b [A] 2 /(k b + k a [A]) (1) This rate law, d[p]/dt, is not the first order. But if deactivation rate, A-A* collision, is much greater than the unimolecular decay rate (i.e. unimolecular decay is much slower than bimolecular reactions; ex. at high partial pressure of A, A* formation/deactivation is much faster than A* consumption) : k a [A*][A] >> k b [A*] or k a [A] >> k b d[p]/dt k[a], k = k a k b /k a becomes 1 st -order reaction

47 As [A] is consumed (and thus low partial pressure/concentration of A), A* formation/deactivation rate becomes slower than A* consumption rate; bimolecular << unimolecular k a [A] << k b, then The general solution, d[p]/dt = k b [A*] = k a k b [A] 2 /(k b + k a [A]), goes to d[p]/dt = k a [A] 2 turns to 2 nd order reaction (at low partial pressure of A) The change of order also can be understood physically: at high pressure, the rate determining step is the unimolecular reaction of A*, first order at low pressure, the rate determining step is the bimolecular reaction of A*, second order If we write a generalized full equation, From d[p]/dt = k b [A*] = k a k b [A] 2 /(k b + k a [A]) d[p]/dt = k[a], in which the effective rate constant k is defined as k = k a k b [A]/(k b + k a [A]). * By rearranging the effective rate constant k 1/k = k a /k a k b + 1/k a [A] Plot 1/k vs. 1/[A]: straight line? Yes, at low concentrations: linear at low P (high 1/[A]) (2 nd order rxn; RDS: bimolecular) deviation at high P (low 1/[A]) (1 st order rxn; RDS: unimolecular)

48 (b) The activation energy of a composite reaction What is the temperature dependency of composite reaction (a combination of reaction steps)? (Normally reaction rate of each step is increased as temperature increases by Arrhenius equation) Let s see high-pressure limit / high [A] of Lindemann-Hinshelwood mechanism Unimolecular decay is the rate determining step. d[p]/dt = k[a], where k = k a k b [A]/(k b + k a [A]) reduces to k = k a k b /k a (at high pressure) Let s use Arrhenius relation: k = k a k b /k a = [(A a e -Ea(a)/RT )(A b e -Ea(b)/RT )]/(A a e -Ea (a)/rt ) k = (A a A b /A a )e-{ea(a) + Ea(b) -Ea (a)}/rt Composite rate constant k has Arrhenius-like form E a = E a (a) + E a (b) - E a (a) If E a (a) + E a (b) > E a (a) positive, rate increases as T If E a (a) + E a (b) < E a (a) negative, rate decreases as T temperature dependency is determined by the relative size of each activation energy.

49 21.9 Polymerization kinetics (1) Stepwise polymerization: - Growth can start at any pair of monomers. - New chains begin to form throughout the reaction, between any species. - Monomers are removed early in the reaction. - Average molar mass grows with time. (2) Chain polymerization: - An activated monomer attacks another monomer, becomes oligomer and it attacks another monomer, and so on. - Chain grows as each chain that acquires additional monomers. - Only activated radical can proceed reaction (Amount of initiator determines the reaction site number.). - High molecular weight polymers are formed rapidly. - The yield of the polymer, not the average molar mass, is increased by allowing long reaction time.

50 (a) Stepwise polymerization * Stepwise polymerization commonly proceeds by a condensation reaction. (Small molecule (typically H 2 O) is eliminated in each step through condensation rxn.) Example 1) Polyamides production (nylon-66): H 2 N(CH 2 ) 6 NH 2 + HOOC(CH 2 ) 4 COOH H 2 N(CH 2 ) 6 NHCO(CH 2 ) 4 COOH + H 2 O (repeat n times) H [NH(CH 2 ) 6 NHCO(CH 2 ) 4 CO] n OH + nh 2 O Polyesters and polyurethanes are formed similarly Example 2) Polyester production: stepwise condensation of HO-M n -COOH + nh 2 O * The condensation reaction is overall second order reaction of two functional groups, OH and COOH group, that make a byproduct as a result of condensation, H 2 O. d[a]/dt = -k[oh][a], A = -COOH group, 2 nd -order reaction Also [A] = [-COOH] = [OH] (reaction between one OH group for each COOH group) d[a]/dt = -k[oh][a] = -k[a] 2 by integration, [A] = [A] 0 /(1 + kt[a] 0 ) * Definition of p, Fraction of Consumption of A p : fraction of COOH groups that have consumed at time t (p increases from 0 to 1) p = consumed/initial = (initial-current)/initial = ([A] 0 [A])/[A] 0 = kt[a] 0 /(1 + kt[a] 0 )

51 * Definition of <n>, Degree of Polymerization <n> : average # of monomer per polymer molecule <n> = [A] 0 /[A] (since there is only one A group per polymer molecule, i.e., polyester has only one COOH group per one polymer chain: HO-M n -COOH) Also from the definition of p (p= ([A] 0 [A])/[A] 0 ) <n> = [A] 0 /[A] = 1/(1 p) ex) initially 1000 A group, now 10 A group each polymer is 100 units on average <n> = [A] 0 /[A] = 1/(1 p) = 100 p = 1-1/100 = 99/100 If we use p = kt[a] 0 /(1 + kt[a] 0 ), <n> = 1/(1 p) = 1 + kt[a] 0 * Average chain length ( degree of polymerization) grows linearly with time. the longer the polymerization, the higher the average molar mass * From intro slide: - Monomers are removed early in the reaction. - Average molar mass grows with time.

52 (b) Chain polymerization Chain polymerization occurs by addition of monomers to a growing polymer that has an active radical. ex) polymerization of methyl methacrylate (PMMA), styrene (PS) -CH 2 CHX + CH 2 =CHX -CH 2 CHXCH 2 CHX and subsequent reactions Polymerization rate is proportional to square root of initiator concentration, v = k[i] ½ [M] Justification: The rate of chain polymerization, v = k[i] ½ [M] (a) Initiation I R + R v i = k i [I] (slow reaction) M + R M 1 (fast reaction, not a rate determining step) (where, I: initiator, R: radical, M: monomer) * Rate-determining step radical R formation from initiator initiation rate equation, v i = k i [I] (b) Propagation M + M 1 M 2 M + M 2 M 3 M + M n-1 M n propagation rate equation v p = k p [M][ M] * We can assume that the propagation rate is independent of chain length for sufficiently large chains. Then propagation rate, v p, is equal to the rate of overall polymerization, v. * Since the chain reaction propagates quickly, the total concentration of radicals is equal to the initiation step, which is a rate-determining step. If, f is defined as the fraction of radical R that successfully initiates a chain, (d[m ]/dt) production = 2fk i [I] (since one initiation reaction, [I], makes two [R ] (2 times) and only f fraction of generated radicals, f[r ], initiates the reaction.)

53 (c) Termination * There are three different reaction termination paths: M n + M m M n+m (mutual termination: combination of two chains) M n + M m M n + M m (disproportionation: hydrogen transfer from one to another) M + M n M + M n (chain transfer: new chain initiates instead of current one) if we suppose that only mutual termination occurs, termination rate, v t = k t [ M] 2 Rate of [total radical] depletion: (d[ M]/dt) depletion = -2k t [ M] 2 (since one termination reaction removes two M (2 times)) * From (a) Initiation, (b) Propagation and (c) Termination, Let s use the steady-state approximation of intermediate, [ M] d[ M]/dt = generation consumption = (d[m ]/dt) production + (d[ M]/dt) depletion = 2fk i [I] 2k t [ M] 2 = 0 [ M] = (fk i /k t ) ½ [I] ½ Since the propagation rate, v p, is equal to the rate of overall polymerization, v, v = v p = k p [ M][M] = k p (fk i /k t ) ½ [I] ½ [M] = k[i] ½ [M] (where k = k p (fk i /k t ) ½ ) End of justification

54 Kinetic chain length (v) Definition v = # of consumed monomer units / # of produced activated centers = rate of propagation of chains / rate of production of radicals (Because, as the propagation rate increases, more monomers are consumed. Also as radical production rate increases, more activated centers are generated.) With steady-state expression (constant radical concentration at steady state) production rate of radicals = termination rate of radicals Then, v = rate of propagation of chains / rate of termination v = k p [ M][M] / 2k t [M ] 2 = k p [M]/2k t [ M] Since [ M] = (fk i /k t ) ½ [I] ½ in the previous slide, v = k[m][i] -1/2, where k = ½k p (fk i k t ) -1/2 We are considering a polymer produced by mutual termination. Average # of monomers in a polymer molecule (degree of polymerization), <n>, is the sum of numbers in two combining polymer chains (two propagating chains react to form one final polymer chain in mutual termination). <n> = 2v = 2k[M][I] -1/2, where k = ½k p (fk i k t ) -1/2 The slower the initiation of the chain (the smaller the initiator concentration and the smaller the initiation rate constant), the greater the kinetic chain length, and the higher molar mass of the polymer.

55 21.10 Photochemistry Photochemical reactions are initiated by the absorption of external electromagnetic radiation, such as light (photon).

56 Kinetics of photochemical processes Photochemical processes are initiated by the absorption of radiation by at least one component of reaction mixture, which is sensitive to the external radiation. Primary process: products are formed directly from the excited state of a reactant ex) fluorescence (S* S + hν f ) cis-trans photoisomerization of retinal Secondary process: products are formed from intermediates that are formed directly from the excited state of a reactant ex) photosynthesis, photochemical chain reactions Photochemical product formation is competing with primary photophysical processes that can deactivate the excited states (Table 21.6, primary photophysical processes: fluorescence, phosphorescence, internal conversion) Primary Primary Primary

57 (a) The primary quantum yield *The rates of deactivation of excited state (through either radiative or non-radiative or chemical processes) determine the yield of product in photochemical reaction. * Definition of Primary quantum yield (φ, fai) # of photophysical or photochemical events that lead to primary products # of photons absorbed by the molecules in the same interval = rate of radiation-induced primary events rate of photon absorption * Definition of φ in terms of rates of processes φ = # of photo-events / # of photons absorbed = rate of primary process / rate(intensity) of light absorbed = v/i abs * An excited molecule either decays to ground state or reacts to from photochemical product. # of activated molecules = # of deactivated + # of reacted species # of deactivation by radiative and non-radiative processes + photochemical reaction (# of photo-events) = # of excited species produced by absorption (# of photons absorbed) φ i = # of photo-events / # of photons absorbed = (v i /I abs ) = 1 (ex. Primary photophysical process decay only: φ f +φ IC +φ p = 1; Primary photophysical and photochemical decay: φ f +φ IC +φ p +φ r = 1; where φ f : fluorescence, φ IC : internal conversion, φ p : phosphorescence, φ r : photochemical reaction) I abs = v i φ i = v/i abs = v i / v i (divide numerator and denominator with the time interval)

58 (b) Mechanism of decay of excited singlet states (S * ) Let s see the case of formation and decay of an excited singlet state in the absence of the photo-chemical reaction (i.e. Primary photophysical process decay only: φ f +φ IC +φ p = 1) : Absorption: S + hν i S* v abs = I abs Fluorescence: S* S + hν f v f = k f [S*] Internal conversion: S* S v IC = k IC [S*] Intersystem crossing: S* T* v ISC = k ISC [S*] Three photo-physical processes of deactivation (where, S: absorbing species, S*: excited singlet state, T*: excited triplet state, hν i and hν f : energies of incident and fluorescent photons) Rate of formation of [S*] = I abs Rate of decay of [S*] = - k f [S*] k IC [S*] k ISC [S*] d[s*]/dt = formation + decay = I abs -k f [S*] k IC [S*] k ISC [S*] (continue in the next slide )

59 When the light is turned off no light absorption (I abs =0) excited singlet state (S*) decays by a first-order process therefore exponentially decay (since, it is first order diff. equation) [S*]/dt= I abs -k f [S*] k IC [S*] k ISC [S*]= (k f +k IC +k ISC )[S*] (when light is turned off, I abs =0) By integrating the first order differential equation, [S*] t = [S*] 0 e -t/τ0, where the observed lifetime of excited singlet state, τ 0, of the first excited singlet state is defined as τ 0 = 1/(k f +k IC +k ISC ) If we use steady-state approximation for [S*] (If [S*] is small and constant), then d[s*]/dt = formation + decay = I abs -k f [S*] k IC [S*] k ISC [S*] = I abs (k f +k IC +k ISC )[S*] = 0 I abs = (k f +k IC +k ISC )[S*] If we calculate the Primary Quantum Yield of Fluorescence (φ f ) using the above result, From the definiton of primary quantum yield, φ i = v i /I abs = v i / v i, and above result, φ f = v f /I abs = k f [S*]/{(k f +k IC +k ISC )[S*]} = k f /(k f +k IC +k ISC ) τ 0 = 1/(k f +k IC +k ISC ) = φ f / k f : Observed lifetime of excited singlet state can be obtained by measuring the fluorescence decay. Or from observed quantum yield and lifetime, the fluorescence rate constant can be calculated.

60 (c) Quenching Quenching : shortening of the lifetime of the excited state by another species Quenching can be studied by monitoring the emission from the excited state Addition of quencher, Q, opens a new route for deactivation of excited singlet state, S* Quenching: S* + Q S + Q v Q = k Q [Q][S*] (now it becomes 2 nd order) Stern-Volmer equation (justification in the next slide) S-V equation (φ f,0 /φ f = 1 + τ 0 k Q [Q]) relates φ f and φ f,0 (fluorescence quantum yield with and without quencher) τ 0 : observed fluorescence lifetime k Q : 2 nd order rate constant of quenching [Q]: the concentration of quencher φ f : fluorescence quantum yield with quencher φ f,0 : fluorescence quantum yield without quencher φ f,0 /φ f = 1 + τ 0 k Q [Q], slope: τ 0 k Q It is a straight line to the quencher concentration

61 Justification of Stern-Volmer equation Let s apply the steady-state approximation for [S*] : We need to add another consumption term of S* by the quencher (v Q = k Q [Q][S*]) to the original differential equation of the singlet state. ([S*]/dt = I abs -k f [S*] k IC [S*] k ISC [S*]) d[s*]/dt = I abs (k f + k IC + k ISC + k Q [Q])[S*] = 0 From the definition of primary quantum yield: φ i = v i /I abs = v i / v i fluorescence quantum yield with quencher: φ f = k f /(k f + k IC + k ISC + k Q [Q]) fluorescence quantum yield without quencher: φ f,0 = k f /(k f + k IC + k ISC ) φ f,0 /φ f = 1 + k Q /(k f + k IC + k ISC )[Q] τ 0 (observed fluorescence lifetime) = 1/(k f +k IC +k ISC ), φ f,0 /φ f = 1 + k Q /(k f + k IC + k ISC )[Q] = 1+ τ 0 k Q [Q] (straight line with the slope of τ 0 k Q ) End of justification

62 Example 21.9) Determine quenching rate constant k Q for the reaction of S* + Q S + Q of 2,2 -bipyridine + Ru 2+ ion Ru(bpy) 3 2+ complex (MLCT transition at 450 nm) Quenching of excited state, *Ru(bpy) 3 2+, by quencher, Fe(H 2 O) 6 3+, is monitored by the emission lifetime at 600nm and the data set is as follows: [Fe(OH 2 ) 6 3+ ]/(10-4 mol dm -3 ) τ/(10-7 s) Sol) Let s use the S-V equation, φ f,0 /φ f = 1 + τ 0 k Q [Q]. Since we have quencher concentration vs. lifetime data, quantum yield ratio term can be replaced with lifetime ratio. i.e. τ 0 = 1/(k f +k IC +k ISC ) = φ f,0 /k f τ 0 /τ = φ f,0 /φ f ( k f constant). φ f,0 /φ f = τ 0 /τ = 1+τ 0 k Q [Q], then rearrange this. 1/τ = 1/τ 0 + k Q [Q] (τ 0 = ) Plot the quenching data above 1/τ vs [Q] From the slope(τ 0 k Q ), k Q = dm 3 mol -1 s -1

63 Exercise (#4) Consider the quenching of an organic fluorescent species with τ 0 =6.0 ns by a d-metal ion with k Q = dm 3 mol -1 s -1. Predict the concentration of quencher required to decrease the fluorescence intensity of the organic species to 50 % of the unquenched value. The fluorescence intensity is proportional to the fluorescence quantum yield. The Stern-Volmer equation(eqn ) relates the ratio of fluorescence quantum yields in the absence and presence of quenching. φ f,0 /φ f = 1 + τ 0 k Q [Q]. Since the fluorescence intensities are proportional to fluorescence quantum yields, f,0 f = 1 + τ 0 k Q [Q] = I f,0 I f Since the fluorescence intensity decreases to 50% of the unquenched value, I f,0 I f = 2, and we know τ 0 and k Q. Then solve this equation for [Q], [Q]= (I f,0/i f ) 1 τ 0 k Q = 2 1 ( s) ( dm 3 mol 1 s 1 ) =0.56 mol dm-3

64 * Three common mechanisms for bimolecular quenching of an excited singlet (S*) (or triplet, T*) state: Collision Deactivation (CD): S* + Q S + Q Resonance Energy Transfer (RET): S* + Q S + Q* Electron Transfer (ET): S* + Q S + + Q - or S - + Q + * Criteria that governs relative efficiencies of different quenching mechanisms. ~ Collisional Deactivation quenching is particularly efficient among above 3 mechanisms, when quencher is heavy species (such as iodine ion, I - ): I - receives energy from S* and decay primarily by internal conversion to the ground state. ~ Electron Transfer quenching is explained by Marcus theory: The rate of electron transfer in E.T. quenching depends on three factors. i) As the distance between donor and acceptor decreases, ET efficiency increases. ii) As the reaction becomes more exergonic (releasing energy during reaction), ET mechanism becomes more efficient. iii) As the reorganization energy between donor, acceptor and medium are more closely matched each other, the efficiency becomes higher. ~ Resonance Energy Transfer quenching is explained by Fröster theory: next slide

65 (d) Resonance energy transfer * Resonance energy transfer (RET): S* + Q S + Q* 1) The incoming electromagnetic field (ex. incoming photon) has an oscillating frequency,, and if = E s /h, where E s is the energy difference between the excited and ground state of S, then S will absorb the incoming electromagnetic field and be excited (becomes S*). 2) The oscillating dipole in S* affects electronics in Q. If the frequency of the oscillating dipole of S* is = E Q /h, it induces oscillating electric dipole moment in Q and Q will absorb the energy. 3) By absorption of electromagnetic radiation, energy is transferred (resonance condition). * Definition of Efficiency of resonance energy transfer: T = 1 (φ f /φ f,0 ) (where φ f, φ f,0 are fluorescence quantum yield with and without quencher) According to Fröster theory, resonance energy transfer is efficient when: Fig Distance between energy donor (S*) & acceptor (Q) is short (~ nm scale or less) 2. Photons emitted by the excited state of the donor (S*) can be directly absorbed by the accepter (Q) (Significant overlap between emission and absorption spectrum of donor and acceptor molecule is important for high RET efficiency Fig 21.26)

66 * 1) Dependence of T on doner-acceptor distance, R T = 1 (φ f /φ f,0 ) = R 06 /(R 06 + R 6 ) The efficiency of energy transfer, T, increases with distance, R, decreases. ( T, which is an efficiency of energy transfer in terms of the donor-acceptor distance ; R 0, characteristic parameter of each donor-acceptor pair) Table 21.7 for R 0 2) Radiation frequency overlap If the emission spectrum of S* is significantly overlapped by the absorption spectrum of Q, the rate of energy transfer can be maximized. Figure (previous slide) * Application example of dependence of T on R in FRET: T = R 06 /(R 06 + R 6 ) The above equation is a basis of fluorescence resonance energy transfer (FRET) T = 1 (φ f /φ f,0 ) = R 06 /(R 06 + R 6 ), which can be deduced from fluorescence measurement, is used to estimate distances, R, between the energy doner and acceptor in biological systems. In FRET, one biological system is covalently labeled with doner, while the other is labeled with accepter. Then the distance between each label is calculated by using above equation and known characteristic parameter of each donor-acceptor pair, R 0. FRET is useful to measure distances ranging from 1 to 9 nm.