Mechanism: series of elementary steps (uni-, bimolecular) that combine to give observed rate law elementary step - reaction order like stoichiometry

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1 II 25 Kinetics Mechanisms (2) eview 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 intereting - bottlenec A 1 B 2 C means 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 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. apid Equilibrium Steady State Mechanism always a model needs to be tested that may mean intermediate detection or sensitivity ecall penicillin example basic chemistry, open ring + H 2 + H 2 We saw observed rate law: 1 st order: r = -d[]/dt = [] Here =Lactam, previous use P, confuse with Prod How might this happen? --mechanism must sense ph Lactam ring + H - slow - H 25

2 II 26 - H + HH fast H H + H - H H fast to equilibrium _ + H 2 Idea: 1 st reaction is slow nce intermediate forms immediately go to product this is steady state model: d[int]/dt ~ 0 = slow [Lac][H] fast [Int] [Int] = slow / fast [Lac][H] -([H 2 ] ~constant) d[prod]/dt = fast [Int] ([H 2 ]) ~ ( fast slow / fast )[Lac][H] d[prod]/dt = ( slow )[Lac][H] ~ slow [Lac] in buffer buffers mae ph ~ constant ate determining step is 1 st one: r ~ eff [Lac] since [ - H] 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 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 ] b. Fast equilibrium example (since subsequently have detected intermediate) K eg Mech. I. I 2 2I 2I + H 2 3 2HI ½ d[hi]/dt = 3 [I] 2 [H 2 ] K eg = [I] 2 /[I 2 ] or K eg [I 2 ] = [I] 2 substitute in rate: consistent: r = 3 K eg [I 2 ][H 2 ] must be slow, 3 small (termol.) 26

3 II 27 Mech. II. I 2 K eg 1 2I K eg 2 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 eg [I] 2 [H 2 ] r = 2 K 2 eg K 1 eg [I 2 ] [H 2 ] Mech. II also consistent, more flexible rate law (K eq s), Test by detection of H 2 I radical intermediate c. Steady state example (Chain propagation) H 2 + Br 2 2HBr exp r = [H 2 ][Br] 1/2 1 + ' [HBr] / [Br 2 ] mechanism: 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 Br + Br 1 Br2 terminate 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 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. result of d[h]/dt eqn., 2 nd, 3 rd, and 4 th terms 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] 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 27

4 II 28 divide top and bottom by 3 [Br 2 ] - goal simplify denom.: 1/ [H ][Br ] 1/ d[hbr]/dt = 1+ 2 fits experiment! [HBr] / [Br2 ] 3 = 2 2 (2 1 / -1 ) 1/2, = -2 / 3 gives exper. 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 (ote: 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 Branching chain reaction see Fried p In above example always got a radical from radical or terminated chain Branching Variation include step in chain that generates more radicals: Ex 2H H 2 H H 2 + initiate + H 2 2 H + H branch low press H H mechan. H + H 2 3 H2 + H propagate H + H H 2 H + H H 2 termination + 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., denominator = 0 branching out of control r H2 ~ r 0 β/(δ β) r 0 - initiate, β branch, δ -destroy chain 28

5 II 29 sensitive to container (wall collisions deactivate) and buffer gas and pressure (enhance termination) δ - β = 4 (T) 2 1 (T) (3T) -1 P -last term conc. of 2 T + P balance eview C +CI 2 CCl 2 Phosgene (poison gas) observe: d[cl 2 C]/dt = [Cl 2 ] 3/2 [C] 2.5 order propose mechanism Cl 2 1 Cl + C Cl Ċ + Cl 2-1 2Cl source of half order Cl Ċ Cl 2 C + Cl rate: d[cl 2 C]/dt = -3 [Cl Ċ][Cl 2 ] -3 [Cl 2 C][Ċl ] rate limit if 3 limit then this will be the correct form, but has intermediate a) Pre-equilibrium -- fast form intermediate K 1 = [Ċl] 2 /[Cl 2 ] and K 2 = [Ċl C]/[Ċl] [C] combine: [Ċl C] = K 2 [Ċl] [C] = K 2 [K 1 [Cl 2 ]] 1/2 [C] 29

6 II 30 plug in: d[cl 2 C]/dt = 3 {K 2 (K 1 [Cl 2 ]) 1/2 } [Cl 2 ] [C] *(assume -3 ~ 0) = [Cl 2 ] 3/2 [C] consistent = 2 1 1/2 3 note: if assume 2 rate limiting then r ~ 2 [Cl ][C] ~ 2 1 1/2 [Cl 2 ] 1/2 [C] but incomplete * equivalent to assuming Product very stable and will not go bac to reactant (mae problem easier oay for initial rate) b) Alternative Steady State d[ċl C]/dt = 2 [Ċl][C] -2 [Ċl C] 3 [Ċl C][Cl 2 ] = 0 [Ċl C] = 2 [Ċl][C]/ [Cl 2 ] *(again neglect -3 ) i) assume fast equilibrium from first step K 1 = [Cl] 2 /[Cl 2 ] [Ċl C] = 2 {K 1 [Cl 2 ]} 1/2 [C/ [Cl 2 ]] rate: d[cl 2 C]/dt = 3 [Cl ] 1/2 2 [C] + [Cl ] 1/ 2 2 K1 [Cl 2 ] cases: -2 >> 3 [Cl 2 ] r = [Cl 2 ] 3/2 [C] same as before = 3 2 K 1 1/2 / -2 = 3 K 2 K 1 1/2 3 [Cl 2 ] >> -2 r = ' [Cl 2 ] 1/2 [C] does not fit observed rate law -2 >> 3 [Cl 2 ] test by vary [Cl 2 ] observed law should deviate high 30

7 II 31 Microscopic eversibility nce get to elementary steps the reaction can go forward and bac on same path B 1 2 A 4 3 Product C dash lines must be included - complete However reverse steps may be fast/slow rate limiting idea may favor solid e But if -1 = 0 then K f = or -2 = 0 = 1 2 / -1-2 clearly then -3, -4 0 or K r e = 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 31