Kinetics Mechanisms (2012) Examples Atkins Ch 7 Tinoco Ch.7 (p ), Engel Ch , Ch

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1 II 3 Kinetics Mechanisms (01) Examples Atins Ch 7 Tinoco Ch.7 (p ), Engel Ch , Ch Recall penicillin example basic chemistry, open ring N O R + H O O O We saw observed rate law: 1 st order: r = -d[r]/dt = [R] Here R = Lactam, previous used P, confuse with Prod + NH R How might this happen? --mechanism must sense ph R= N O R Lactam ring + OH - slow N - O OH R =Int N - O R OH + HOH fast O OH NH R + OH - O OH NH R fast to equilibrium O _ O N + H R =P 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 O] [Int] = slow / fast [Lac][OH]/ [H O] note [Int] small, f >> s - but [H O] ~constant ~55 M ignore (part of ) d[prod]/dt = fast [Int] ([H O]) ~ ( fast slow / fast )[Lac][OH] [H O] cancels d[prod]/dt = ( slow )[Lac][OH] ~ slow [Lac] in buffer 1 st order buffers mae ph ~ const., [OH] part of slow senses ph 3

2 II 4 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 change ph / see affect on rate Examples: Mechanisms are combination of parallel, opposed and chain steps ex. H + I HI observe: 1/ d[hi]/dt = [H ][I ] devise consistent mechanism: a. (old idea) assume simple bimolecular: H + I 1 HI ½ d[hi]/dt = 1 [H ][I ] consistent, not proven b. Fast equilibrium idea --(subsequently intermediate was detected) Mech. I. I I + H K eg I assume rapid equilibrium 3 HI ½ d[hi]/dt = 3 [I] [H ] K eq = [I] /[I ] or K eq [I ] = [I] substitute in rate: consistent: r = 3 K eq [I ][H ] but must be slow, due 3 -termolecular Mech. II. I I + H I + H I K eg 1 K eg I H I HI eliminate termolecular step (wor out) ½ d[hi]/dt = [I][H I] = K eq [I] [H ] r = K eq K 1 eq [I ] [H ] Mech. II also consistent, more flexible rate law (K eq s), Test by detection of H I radical intermediate 4

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

4 Algebra substitute in: d[hbr]/dt = ( 1 / 1 ) 1/ [Br ] 1/ [H ] + 3 ( 1 / -1 ) 1/ [H ][Br ] 3/ /D - - ( 1 / -1 ) 1/ [H ][Br ] 1/ [HBr]/D where D = { 3 [Br ] + - [HBr]} the denominator in [H] eqn. II 6 next put 1 st term over D, sum the numerators: d[hbr]/dt = { ( 1 / 1 ) 1/ [H ][Br ] 1/ ( 3 [Br ]+ - [HBr]+ 3 [Br ] - - [HBr])}/D divide top and bottom by 3 [Br ] - simplify denom.(to 1+ form): d[hbr]/dt = 1/ 1 [H ][Br ] 1/ 1 1+ [HBr] / [Br ] 3 Let -- = ( 1 / -1 ) 1/, = - / 3 fits experiment! [H ][Br ] r = gives bac experimental form: 1+ ' [HBr] / [Br ] 1/ Comments: 1. reaction is an example of radical species propagating and enhancing rate but radical only exists as an intermediate. t = 0 rate ~ [H ] [Br ] 1/ initial rate is a clue right away to complexity, [ ] 1/ from termination step signal for radical formation (i.e. opposing step has a different order) 3. denominator is result of inhibitor step, {1 + [conc]} form lets you test limiting conditions, here initial [HBr] = 0 6

5 II 7 d. Branching chain reaction see Fried p here just aiming for the idea, not details (cultural!) In chain rxn. example always got a radical from radical (1 1) or terminated chain (1 0 or 0) Branching steps in chain that generate more radicals: Ex H + O H O H + O 0 H O + O initiate O + H OH + H branch low press H + O 1 OH + O 1 mechanism OH + H 3 H O + H propagation H + OH H O H + H H termination, also high press. O + O O 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 Solve by steady state on all radicals: H, O and OH - read Combining them with significant algebra gives OH yielding: r HO = 3 [ OH][H ] = ( 0 [H ][O ])( 1 [O ])/{ 4-1 [O ]} alternate: r HO ~ r 0 β/(δ β) r 0 - initiate, β branch factor, δ -destroy chain ν = δ β net distruction factor 7

6 sensitive to container (wall collisions deactivate) and buffer gas and pressure (enhance termination) II 8 ν = δ - β = 0 explosion limit H O prod. infinite rate (denom.) = 4 (T) 1 (T) (3RT) -1 P -last term use ideal gas law for conc. of O T + P balance but each rate constant depends on T govern by wall termination depend on size and shape nd limit higher pressure, need additional steps (rxns) include wall reactions and H O formation/destruction pressure then needs more temp to explode (other path) 3 rd limit self heating speeds up the reactions, lower explosion limit results 8

7 II 9 e. Practice problem test methods: CO +CI COCl Phosgene (poison gas) observe: d[cl CO]/dt = [Cl ] 3/ [CO].5 order propose mechanism Cl 1-1 Cl source of half order - initiate Cl + CO - Cl ĊO propagate Cl ĊO + Cl 3-3 Cl CO + Cl propagate rate: d[cl CO]/dt = 3 [Cl ĊO][Cl ] -3[Cl CO][Ċl ] rate limit if 3 limit then this will be the correct form, but this has intermediate - need eliminate NOTE due to propagation steps, Cl conc. can be signif. * simplify, assume neglect reverse -3 step (see if OK) a) Pre-equilibrium -- assume fast form intermediates K 1 = [Ċl] /[Cl ] and K = [Ċl CO]/[Ċl] [CO] combine: [Ċl CO] = K [Ċl] [CO] = K (K 1 [Cl ]) 1/ [CO] plug in: d[cl CO]/dt = 3 {K (K 1 [Cl ]) 1/ [CO]} [Cl ] r = [Cl ] 3/ [CO] consistent *(recall - assumed -3 ~ 0) = K K 1/ 1 3 note: if assume rate limiting get wrong rate law then r ~ [Cl ][CO] ~ K 1/ 1 [Cl ] 1/ [CO] * equivalent to assuming Product very stable and not reform reactant (maes problem easier oay for initial rate) 9

8 II 30 b) Alternative Steady State on [Ċl CO] d[ċl CO]/dt = [Ċl][CO] - [Ċl CO] 3 [Ċl CO][Cl ] = 0 [Ċl CO] = [Ċl][CO]/( [Cl ]) *(again neglect -3 ) i) assume fast equilibrium from first step K 1 = [Cl] /[Cl ] [Ċl CO] = {K 1 [Cl ]} 1/ [CO]/( [Cl ]) rate: d[cl CO]/dt = 3 1/ K 1 1/ [Cl ] [CO] + [Cl ] [Cl ] complex-- cases: a. - >> 3 [Cl ] r = [Cl ] 3/ [CO] same as before = 3 K 1 1/ / - = 3 K K 1 1/ i.e. this wors also tend justify neglect -3 note -if - fast, then K fast equil. lie above b. 3 [Cl ] >> - r = ' [Cl ] 1/ [CO] does not fit observed rate law - >> 3 [Cl ] test by vary [Cl ] observed law should deviate at high [Cl ] conc. Note equivalent to rate limiting not wor

9 II 31 Microscopic Reversibility Once get to elementary steps the reaction can go forward and bac on same path cannot get reversible reaction by a cyclic mechanism A B Product C Solid line path not enough dash lines must be included complete (but can be slow, and neglect in solve mechanism) However reverse steps may be fast or slow equilibria maes rate constants interdependent K f = P e /A e = 1 / -1 - K r = A e /P e = 3 4 / but K f = K r 1 / -1 - = -3-4 / 3 4 rate limiting idea may favor solid line path e But if -1 = 0 then K f = or - = 0 = 1 / -1 - 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 equiv to fit ln[a] vs. t 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 31

10 II 3 (extras) Bio. Mechanism. Tinoco pp Renature DNA Duplex DNA - complementary strand A + A' AA' High T unfold and separate Cool refold speed depends on alignment, nd order Sonicate DNA brea into small segments, e.g. ~0 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. Denature to ss 3. Cool, renature strands pair up, first find right alignment Different curves for sequences with increasing complexity, left to right: polyupolya, Mouse, MS-, 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 3

11 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] = (C/N) II 33 Thus more complex, slows the reaction: t 1/ ~N/C 0 or half-life varies inversely with complexity 33

12 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 II 34 Prothrombin Ca + Thrombin Antithrombin Inactive Thrombin Mechanism: Fibrogen Fibrin (clot) Thrombin is then an intermediate / build fast equilibrium decay 34