elementary steps have reaction order like stoichiometry Unimolecular: A k 1 P 1 st order -d[a]/dt = k 1 [A] --> ln [A]/[A 0 ] = -k 1 t

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Arline Sparks
5 years ago
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1 B. Mechanism 009 rearrange -- Engel Ch 5.4,0,8 Series of elementary steps (uni-, bimolecular) that when combined give overall reaction and observed rate law elementary steps have reaction order lie stoichiometry Unimolecular: A - P st order -d[a]/dt = [A] --> ln [A]/[A 0 ] = - t as approach equilibrium - both forward and reverse steps must be in mech.: -d[a]/dt = [A] - [P] equilibrium: d[a]/dt = 0 = [A e ] - [P e ] II - 5 K eq = [P e ]/[A e ] = / - see if product really favored: K eq >>, then >> - -->rate law mostly lie r ~ [A] same nd order: if elementary step A + B - then: r = [A] [B] - [C] [D] r = 0 at equlibrium K eq = [C] [D]/ [A] [B] = / - C + D 5

2 reconsider st order: r = [A] - [P] [P] = [A 0 ] [A] r = [A] - - {[A 0 ]-[A]} = ( + - ) [A] - [A 0 ] r = ( + - ) {[A] [ - /( + - )] [A 0 ]} recall: K eq = [P e ]/[A e ] = (A 0 -A e )/A e = (A 0 /A e - ) = / - [A 0 ]/[A e ] = / - + = ( + - )/ - substitution: r = ( + - ) {[A] [A e ]} let A = A Ae, then -da /dt = -da/dt = r = A st order approaches an asymptote at A eq - integrate ln (A - A e )/(A 0 -A e )= - ( + - ) t c) Alternate case parallel reaction ( paths for A, forms independent products) B or A B C A C A bb + cc Note: b + c give ratio of product NOT stoichiometry II - 6 r = -/ d[a]/dt = ' [A] + ' [A] = (' + ' ) [A] A again loos simple, could hide complexity Chain Reaction series of steps find the bottlenec B C means cannot get C without forming B Rate Determining Steps for chain assume one step slowest bottlenec B C -- form C depend on or fast [this will be core issue for Mechanisms modeling!!] 6

3 Methods time dependence Engel Ch 5. Initial examples assume start [R] = A 0 + B 0 etc., [P] = 0 follow reaction forward initial rate, etc. follow [A] Monitor Absorbance, fluorescence, electrochemistry, whatever is proportional to concentration If reverse reaction or alternate steps important Need to monitor other species (intermediates) At equilibrium, rates change depend on: [R] & [P] II - 7 Equilibrium, if elementary A + B C + D r for = r rev K eq = / - = [C] e [D] e [A] e [B] e r r r f = = [A] [B] [C] [D] = Disturb equilibrium (relaxation) - change T, P, ph, how? discharge capacitor,shoc wave, Laser flash-photolysis or T-jump - system must relax to new equilibrium - reaction must go forward/reverse to new state A - B A = A e x B = B e + x x = departure from equilibrium at time = t, new T or P x = x e e -t/τ relaxation time: /τ = + - r = A - - B = (A e -x) - - (B e +x) = -da/dt = dx/dt = -( + - )x + A e - - = -( BBe + - )x Note: relaxation faster if either or - fast Since K eq = / - get both values, - from τ & K eq 7

4 Mechanism Overview: series of elementary steps (unior bimolecular) that combine to give observed rate law elementary step - reaction order lie stoichiometry Sequential steps most interesting - bottlenec B 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 forms Choices ( main models or approximations): (a) B form fast, build up rapid equilibrium: A B K eq = / - =[B]/[A] -- intermediate B balance A (b) very little B form, immediately go off to form C result: d[b]/dt ~ 0 -- steady state approx. II - 8 Rapid Equilibrium Steady State Mechanism is always a model needs to be tested -- may mean intermediate detection or sensitivity 8

5 Chain Reaction series of steps - bottlenec B C means cannot get C without forming B neglect reverse: a) -d[a]/dt = [A] [A] = [A 0 ]e - t b) d[b]/dt = [A] [B] = [A 0 ]e - t [B] integrate: [B] = [A 0 ]/( ) [e - t e - t ] c) d[c]/dt = - [B] stoichiometry: [C] = [A 0 ] [A] [B] everything from A 0 [C] = [A 0 ]{ - e - t [ /( - )](e - t - e - t )} how does this behave? t = 0 B = 0, C = 0 t = C = A 0 initially, small t C < B (early product formation slow) from expression for [C] above tae d/dt: d[c]/dt = e - t [ /( - )] (- e - t + e - t ) induction period, t = 0 no product form, just intermediate d[c]/dt ~ ( / ) ~ 0 [ t = d[c]/dt ~ 0 (e - 0) -- reaction done] II - 9 9

6 II - 0 Rate Determining Steps for chain assume one step slowest bottlenec B C lie expressway example: crash blocs all but lane, flow of cars, d(cars)/dt, limited by passing the truc a) if >> as soon as B forms, it goes to C control rate overall reaction B concentration assume small steady value d[b]/dt ~ 0 = [A] [B] steady state approx. [B] = / [A] (see graph above, right) d[c]/dt = [B] = [ / [A]] = [A] - loos st in A but should have an induction period C t rearrange: d[c] = { [A 0 ]e -t } dt 0 0 [C] = [A 0 ] ( - e - t ) same as before but 3 rd term drops: >> simpler integrated form--still has induction period b) if >> [B] builds up and approaches equilibrium d[b]/dt [A] = [A 0 ]e - t B only grows early [B] = [A 0 ] [A] no decay d[c]/dt = [B] = { [A 0 ] [A] [C] } 0

7 II - early in reaction: d[c]/dt ~ 0 [C] ~ [A 0 ] [/ (e - t )] later easier, if assume A K e B come to rapid equilibrium K e = B e /A e d[c]/dt = [B] = K eq [A] r = '[A] again loos simple st order in A time both a) & b) mechanisms yield simple st order rate law but complex mechanisms behind this will require testing to determine complexity c) Recall alternate parallel reaction ( paths for A, B or A B independent product) C A C A bb + cc Note: b & c inetic ratio of product NOT stoichiometry second case: r = -/ d[a]/dt = ' [A] + ' [A] A = (' + ' ) [A] again loos simple, could hide complexity How to determine complexity for above cases: c) see products whose ratio is non-stoichiometric or T-dependent a) see induction period in [C] b) detect intermediate [B] time

8 II - Bring ideas from reaction coordinate / energy diagram st step slow big E a = A e -E a/rt nd step fast small e a = A e -e a/rt in this case reverse steps slow (big E a ) lie steady state case (a) Example, Atins & depaula, Fig. 7.0 pre-equilibrium can have types of process (A=R, P=C) depending on barrier from B C (B intermediate) A B = A e -E /RT C = A e -E -/RT probable: - < since E < E - (assume A similar) favors B e : K e = B e /A e = / - but as increase T : - will increase faster than since this controls B and thus C could slow rate effective activation: E - E - + E p can be negative! Application: case study 7. Protein Folding in Atins To study rates of helix and sheet formation, we can observe changes in IR spectra and for tertiary (fold) formation can study fluorescence after change in T

Kinetics Mechanisms (2008-rev) Review and Examples

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