Kinetics Mechanisms (2008-rev) Review and Examples

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1 II 25 Kinetics Mechanisms (2008-rev) Review and Examples Mechanism: series of elementary steps (uni-, bimolecular) that combine to give observed rate law elementary step - reaction order lie stoichiometry Sequential steps most interesting - bottlenec A B C means 1 2 need to form B to get C Same for A + B C + D D + E E + F, etc. D formation limit E Characteristic induction period, how intermediate form Choices: (a) B form fast, build up rapid equilibrium: A B K eq = 1 / -1 =[B]/[A] (b) very little B form, immediately go off to C result: d[b]/dt ~ 0 -- steady state approx. Rapid Equilibrium Steady State Mechanism always a model needs to be tested that may mean intermediate detection or sensitivity 25

2 II 26 Chain Reaction series of steps - bottlenec A B C means 1 2 cannot get C without forming B neglect reverse: a) -d[a]/dt = 1 [A] [A] = [A 0 ] e -1t b) d[b]/dt = 1 [A] 2 [B] [B] = 1 [A 0 ]/( 2 1 ) [e - 1t e - 2t ] c) d[c]/dt = - 2 [B] [C] = [A 0 ]{1 - e - 1t [ 1 /( 2-1 )](e - 1t - e - 2t )} induction period, t = 0 no product form, just intermediate d[c]/dt ~ 1 1 ( 2 1 / 2 1 ) ~ 0 slope=0 [ t = d[c]/dt ~ 0 (e - 0) -- reaction done] Rate Determining Steps for chain assume one step slowest bottlenec A B C lie expressway example

3 II 27 a) if 2 >> 1 as soon as B forms, it goes to C 1 controls rate of overall reaction B concentration assume small steady value d[b]/dt ~ 0 = 1 [A] 2 [B] steady state approx. [B] = 1 / 2 [A] d[c]/dt = 2 [B] = 2 [ 1 / 2 [A]] = 1 [A] - loos 1 st in A but should still have an induction C 0 rearrange: d[c] = 1{ [A 0 ]e -1t } dt t 0 [C] = [A 0 ] (1 - e - 1t ) same as before but 3 rd term drops: 2 >> 1 simpler integrated form--still has induction period b) if 1 >> 2 [B] builds up and approaches equilibrium d[b]/dt 1 [A] = 1 [A 0 ]e - 1t B only grows early [B] = [A 0 ] [A] no decay d[c]/dt = 2 [B] = 2 { [A 0 ] [A] [C] } early in reaction: d[c]/dt ~ 0 [C] ~ [A 0 ] [(e - 1t 1)] later easier, if assume A K e B come to rapid equilibrium K e = B e /A e d[c]/dt = 2 [B] = 2 K eg [A] r = '[A] again loos simple 1 st order in A 27

4 II 28 both a & b mechanisms yield simple 1 st order rate laws but complex mechanisms behind this will require testing to determine complexity c) Alternate case parallel reaction (2 paths for A, independent product) A 1 B or 2A 1 B A 2 C 2A 2 C 2A bb + cc Note: b + c give ratio of product NOT stoichiometry second case: r = -1/2 d[a]/dt = ' 1 [A] 2 + ' 2 [A] 2 A = (' 1 + ' 2 ) [A] 2 again loos simple, could hide complexity How to determine complexity for above cases: c) see 2 products whose ratio is non-stoichiometric or T-dependent a) see induction period in C b) detect intermediate (B) time Bring ideas from reaction coordinate / energy diagram 1 st step slow big E a 1 = A 1 e -E a/rt 2 nd step fast small E a 2 = A 2 e -e a/rt in this case reverse steps slow (big E a ) lie steady state case (a) 28

5 See example, Atins&de Paula, Fig pre-equilibrium can have 2 types of process (A=R, P=C) depending on barrier from B C (B intermediate) A 1 2 B 1 = A 1 e -E 1/RT C 2 = A 2 e -E -1/RT probable: -1 < 1 since E 1 < E -1 (assume A similar) favors B e : K e = B e /A e = 1 / -1 II 29 but as increase t : -1 will increase faster than 1 since this controls B and thus C could slow rate effective activation: E 1 - E -1 + E p can be negative! 29

6 II 30 Recall penicillin example basic chemistry, open ring N O R + H 2 O O O + NH 2 R We saw observed rate law: 1 st order: r = -d[r]/dt = [R] Here R=Lactam, previous use P, confuse with Prod How might this happen? --mechanism must sense ph N O R Lactam ring + OH - slow - O OH N R N - O R OH + HOH fast O OH NH R + OH - O OH NH R fast to equilibrium O _ O N + H 2 R Idea: 1 st reaction is slow rate controlling Once intermediate forms immediately go to product this is steady state model: d[int]/dt ~ 0 = slow [Lac][OH] fast [Int][H 2 O] [Int] = slow / fast [Lac][OH]/ [H 2 O] - but [H 2 O] ~constant~55 M ignore (part of ) d[prod]/dt = fast [Int] ([H 2 O]) ~ ( fast slow / fast )[Lac][OH] d[prod]/dt = ( slow )[Lac][OH] ~ slow [Lac] in buffer buffers mae ph ~ const., [OH] part of slow senses ph 30

7 II 31 Rate determining step is 1 st one: r ~ eff [Lac] since [ - OH] constant set by ph (~1 st order in Lac) Test: Mechanism always is a model, show consistent with data have change ph / see affect on rate Examples: Mechanisms are combination of parallel, opposed and chain steps ex. H 2 + I 2 2HI observe: 1/2 d[hi]/dt = [H 2 ][I 2 ] devise consistent mechanism: a. (old idea) assume simple bimolecular: H 2 + I 2 1 2HI ½ d[hi]/dt = 1 [H 2 ][I 2 ] consistent b. Fast equilibrium idea (subsequently detected intermediate) K eg Mech. I. I 2 2I 2I + H 2 3 2HI ½ d[hi]/dt = 3 [I] 2 [H 2 ] K eq = [I] 2 /[I 2 ] or K eq [I 2 ] = [I] 2 substitute in rate: consistent: r = 3 K eq [I 2 ][H 2 ] must be slow, 3 small (termolecular) Mech. II. I 2 K eg 1 K eg 2 2I I + H 2 H 2 I eliminate termolecular step I + H 2 I 2 2HI ½ d[hi]/dt = 2 [I][H 2 I] = 2 K 2 eq [I] 2 [H 2 ] r = 2 K 2 eq K 1 eq [I 2 ] [H 2 ] Mech. II also consistent, more flexible rate law (K eq s), Test by detection of H 2 I radical intermediate 31

8 c. Steady state example (Chain propagation) [H 2 ][Br] 1/2 1 + ' [HBr] / [Br 2 ] H 2 + Br 2 2HBr experiment: r = t=0 [HBr]=0 r~[br] 1/2 [H 2 ] but t= order = 5/2 mechanism: apparent order change with time Br 2 1 2Br initiate reaction (create radicals) Br + H 2 2 HBr + H propagate (conserve H + Br 2 3 HBr + Br cycle radicals) HBr + H 2 Br + H2 inhibit reverse 2 nd step Br + Br 1 Br2 terminate reverse 1 st step Steady state on radicals very reactive, never build up a. d[h]/dt ~ 0 = 2 [Br][H 2 ] 3 [H][Br 2 ] -2 [HBr][H] [H] = 2 [Br][H 2 ]/( 3 [Br 2 ] + -2 [HBr]) -- source of denom. b. d[br]/dt ~ 0 = 2 1 [Br 2 ] 2 [Br][H 2 ] + 3 [H][Br 2 ] + -2 [HBr][H] -1 [Br] 2 Subst. [H] result into d[br]/dt eqn., 3 rd and 4 th terms, 2 nd, 3 rd and 4 th terms will cancel sum to 0: 0 = 2 1 [Br 2 ] -1 [Br] 2 [Br] = [2 1 / 1 (Br 2 )] 1/2 substitute [Br] into [H] equation (eliminates all radicals): [H] = 2 [H 2 ] (2 1 / -1 ) 1/2 /( 3 [Br 2 ] + -2 [HBr]) rate of product formation depends on [H] and [Br]: d[hbr]/dt = 2 [Br][H 2 ] + 3 [H][Br 2 ] -2 [HBr][H] II 32 32

9 II 33 Algebra substitute in: d[hbr]/dt = 2 (2 1 / 1 ) 1/2 [Br 2 ] 1/2 [H 2 ] (2 1 / -1 ) 1/2 [H 2 ][Br 2 ] 3/2 /D (2 1 / -1 ) 1/2 [H 2 ][Br 2 ] 1/2 [HBr]/D where D= 3 [Br 2 ] + -2 [HBr] the denominator in [H] eqn. next put 1 st term over D, sum the numerators: d[hbr]/dt = { 2 (2 1 / 1 ) 1/2 [H 2 ][Br 2 ] 1/2 ( 3 [Br 2 ]+ -2 [HBr]+ 3 [Br 2 ] - -2 [HBr 2 ])}/D divide top and bottom by 3 [Br 2 ] - goal simplify denom.: d[hbr]/dt = 1/ [H ][Br ] 1/ [HBr] / [Br2 ] 3 fits experiment! = 2 2 (2 1 / -1 ) 1/2, = -2 / 3 gives bac experimental form: r = [H 2 ][Br] 1/2 1 + ' [HBr] / [Br 2 ] Comments: 1. reaction example of radical species propagating and enhancing rate but only exists as an intermediate 2. t = 0 rate ~ [H 2 ] [Br 2 ] 1/2 (Note: before wrong) initial rate is a clue right away to complexity, [ ] 1/2 from termination step (i.e. opposing step has a different order) 3. denominator is result of inhibitor step 33

10 II 34 d. Branching chain reaction see Fried p here just aiming for the idea, not details In above example always got a radical from radical or terminated chain Branching step in chain that generates more radicals: Ex 2H 2 + O 2 H 2 O H 2 + O 2 0 H 2O + O initiate O + H 2 2 OH + H branch low press H + O 2 1 OH + O 1 2 mechan. OH + H 2 3 H2O + H propagate H + OH H 2 O H + H H 2 termination O + O O 2 or H + wall 4 destruction Point is that branching creates high level of unstable species (radicals) reaction then driven very fast explodes i.e., mech. has denom., when = 0 branching out of control r H2O = 3 [OH][H 2 ] = 0 [H 2 ][O 2 ]2 1 [O 2 ]/{ [O 2 ]} alternate: r H2O ~ r 0 β/(δ β) r 0 - initiate, β branch, δ -destroy chain ν = δ β net distruction factor sensitive to container (wall collisions deactivate) and buffer gas and pressure (enhance termination) 34

11 II 35 ν = δ - β = 0 explosion limit H 2 O produce at infinite rate = 4 (T) 2 1 (T) (3RT) -1 P -last term use ideal gas law for conc. of O 2 T + P balance but each rate constant depends on T govern by wall termination 35

12 II 36 e. Practice problem test methods: CO +CI 2 COCl 2 Phosgene (poison gas) observe: d[cl 2 CO]/dt = [Cl 2 ] 3/2 [CO] 2.5 order propose mechanism Cl Cl source of half order - initiate Cl + CO 2-2 Cl ĊO propagate Cl ĊO + Cl Cl 2 CO + Cl propagate rate: d[cl 2 CO]/dt = 3 [Cl ĊO][Cl 2 ] -3 [Cl 2 CO][Ċl ] rate limit if 3 limit then this will be the correct form, but has intermediate a) Pre-equilibrium -- fast formation of intermediate K 1 = [Ċl] 2 /[Cl 2 ] and K 2 = [Ċl CO]/[Ċl] [CO] combine: [Ċl CO] = K 2 [Ċl] [CO] = K 2 (K 1 [Cl 2 ]) 1/2 [CO] plug in: d[cl 2 CO]/dt = 3 {K 2 (K 1 [Cl 2 ]) 1/2 [CO]} [Cl 2 ] *(assume -3 ~ 0) = [Cl 2 ] 3/2 [CO] consistent = K 2 K 1/2 1 3 note: if assume 2 rate limiting then r ~ 2 [Cl ][CO] ~ 2 K 1/2 1 [Cl 2 ] 1/2 [CO] but incomplete * equivalent to assuming Product very stable and will not go bac to reactant (mae problem easier oay for initial rate) 36

13 II 37 b) Alternative Steady State d[ċl CO]/dt = 2 [Ċl][CO] -2 [Ċl CO] 3 [Ċl CO][Cl 2 ] = 0 [Ċl CO] = 2 [Ċl][CO]/ [Cl 2 ] *(again neglect -3 ) i) assume fast equilibrium from first step K 1 = [Cl] 2 /[Cl 2 ] [Ċl CO] = 2 {K 1 [Cl 2 ]} 1/2 [CO/ [Cl 2 ]] rate: d[cl 2 CO]/dt = 3 K 1/ 2 1 [Cl ] 1/2 2 [CO] + [Cl ] 2 [Cl 2 ] 2 cases: a. -2 >> 3 [Cl 2 ] r = [Cl 2 ] 3/2 [CO] same as before = 3 2 K 1 1/2 / -2 = 3 K 2 K 1 1/2 i.e. this wors b. 3 [Cl 2 ] >> -2 r = ' [Cl 2 ] 1/2 [CO] does not fit observed rate law -2 >> 3 [Cl 2 ] test by vary [Cl 2 ] observed law should deviate high 37

14 II 38 Microscopic Reversibility Once get to elementary steps the reaction can go forward and bac on same path can not get reversible reaction by cyclic mechanism B 1 2 A 4 3 C Product dash lines must be included - complete However reverse steps may be fast/slow equilibria maes rate constants interdependent K f = P/A = 1 2 / -1-2 K r = A/P = 3 4 / o but K f = K r 1 2 / -1-2 = -3-4 / 3 4 rate limiting idea may favor solid line path e But if -1 = 0 then K f = or -2 = 0 = 1 2 / -1-2 clearly then -3, -4 0 or K e r = impossible! Summary: r f = r r at equilibrium { detailed balance Exponential behavior often analyze [conc] vs. t by fit to exponential function 1 st order: -da/dt = A A = A 0 e -t ex. Protein folding vary conditions /protein fold on own should be 1 st order exponential -- if simple if fit to multiple exponential multi step process 38

15 II 39 Bio. Mechanism. Tinoco pp Renature DNA Duplex DNA - complementary strand A + A' AA' High T unfold and separate Cool refold speed depends on alignment, 2 nd order Sonicate DNA brea into small segments, e.g. ~20 bp uniqueness depends on the repeat pattern of DNA if melt strand separate and mix, recombine slow if unique, faster if repeat Complexity: Recombination rate vary heterogeneous 1. Brea to ~400 bp 2. Denature to ss 3. Cool, renature Different curves for sequences with increasing complexity, left to right: polyupolya, Mouse, MS-2, T4, e.coli, calf r C 0 conclusion complexity (arrows) See that simple sequences fold faster because they can find a mate (less complex) for segment More repeat sequences, less complex 39

16 II 40 N = number of bp in smallest repeat sequence Initial concentration [A 0 ] ~ C 0 /N C 0 concentration all ss N poly A poly U = 1 = 1 smallest repeat each bp E. coli almost no repeat N = number of bp So point is r = -d[a]/dt = [A][A'] since A ~ C/N more complex big N more complex A small / rate slow Complementarity [A] = [A'] from how strands broen -d[a]/dt = [A] 2 = (C/N) 2 Thus more complex, slows the reaction: t 1/2 ~N/C 0 or half-life varies inversely with complexity 40

17 II 41 Blood Clotting example rapid equilibrium Tinoco, p response to wound is very complex but ey is fibrogen fibrin which forms clot To turn this on need Thrombin (proteolytic enzyme) To get this need activate prothrombin by protolysis Prothrombin Ca +2 Thrombin Antithrombin Inactive Thrombin Mechanism: Fibrogen Fibrin (clot) Thrombin is then an intermediate / build fast equilibrium decay 41

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