How fast can you measure? 2 Assume: Conventional Quench and - PDF document

12/8/2011 Updated: 8 December 2011 CEE 670 Kinetics Lecture #4 1 Print version CEE 670 TRANSPORT PROCESSES IN ENVIRONMENTAL AND WATER RESOURCES ENGINEERING Kinetics Lecture #4 Rate Expressions III: More Examples Introduction David A. Reckhow How fast can you measure? 2 Assume:  Conventional Quench and measure MDL ~ 10 -7 M for [A]  k obs < 0.01 s -1 or k 2 < 10 3 M -1 s -1 ( Ɛ = 30,000 M -1 cm -1 ; abs ≥ 0.003 cm -1 ) Pseudo-1 st order kinetics say [B] ≥ 100*[A] 0 ,  Limited by reagent addition Significant drop in A during tests  Say 5 measurements at 20 second each say [A] 0 ≥ 10*[A] final then [B] ≥ 10 -5 M  100 sec total  Conventional Spectrophotometry Recall: For pseudo 1 st order in A  k obs < 0.1 s -1 or k 2 < 10 4 M -1 s -1 A + B = C k obs = k 2 [B]  Limited by shutter speed  Say 5 measurements at 2 seconds each  10 sec total  Stopped Flow Spectrophotometry  k obs < 1000 s -1 or k 2 < 10 8 M -1 s -1  k  1  Limited by mixing speed (k m )  obs  k     obs k obs k  For Durrum instrument: k m = 1700 s -1 m CEE690K Lecture #8 David A. Reckhow 1

12/8/2011 Do we care about fast reactions? 3  Competitive Kinetics  Selective contaminant destruction  Ethinylestradiol vs phenol O 3  Direction for Byproduct Formation CEE 670 Kinetics Lecture #2 David A. Reckhow Alternative #1 for very fast reactions 4  Make use of known pH speciation of reactants  Pick conditions where “formal” concentrations are easily measured, but reactive forms are orders of magnitude lower  e.g., chlorination of phenol     k HOCl C H O products 2 6 5 dc    k [ HOCl ][ C H O ] 0 2 6 5 dt  [ H ] K -2  k [ chlorine ] [ phenol ] a 2     2 tot tot K [ H ] K [ H ] -4 a 1 a 2 Log  -6  [ H ] K  a 2 -8 k k [ chlorine ] [ phenol ]     obs 2 tot tot K [ H ] K [ H ] alpha-HOCl a 1 a 2 -10 alpha-phenate product of alphas -12 CEE 670 Kinetics Lecture #5 David A. Reckhow 0 2 4 6 8 10 12 14 pH 2

12/8/2011 Alpha values 5  Background on alphas            Phenate H   OCl H       K K     a 1 a 2 HOCl Phenol          OCl HOCl OCl     K K    a 1 1 a 1        HOCl     H  HOCl H     HOCl 1         K  HOCl OCl  a 1 1    H   Phenate K    a 2    phenate  Phenol Phenate K [ H ] HOCl [ H ]    a 2     HOCl HOCl OCl K [ H ] a 1 CEE 670 Kinetics Lecture #3 David A. Reckhow Alt #1 example: Bromine + phenol 6  Flow injection analysis  Allowed 10-150 ms reaction time  Gallard et al., 2003  Wat. Res. 379:2883 0 -2 -4 Log  -6 -8 alpha-HOBr -10 alpha-phenate product of alphas -12 0 2 4 6 8 10 12 14 CEE 670 Kinetics Lecture #5 David A. Reckhow pH 3

12/8/2011 Alternative #2 for very fast reactions 7  Use competitive kinetics d [ A ] k [ A ]  A d [ B ] k [ B ] B    k d [ A ] A C P   A k [ A ][ C ] A d [ A ] k d [ B ] dt    A    k d [ B ] B C Q   B k [ B ][ C ] [ A ] k [ B ] B dt B      Time drops out [ A ] k [ B ]      ln A ln         [ A ] k [ B ]  just wait until “C” is gone 0 B 0  Must know one of the two rate constants (say k B ), and concentrations of both A & B (initial and final)  So the accessible range for k A is roughly 0.2*k B to 5*k B CEE 670 Kinetics Lecture #5 David A. Reckhow Alt #2 Example: bromine & phenols 8  HPLC determination of residual phenols  Acero et al., 2005  Wat. Res. 39:2979 CEE 670 Kinetics Lecture #5 David A. Reckhow 4

12/8/2011 Mixed Second Order    k A B products 2 9  Two different reactants dx  k [ A ][ B ] 2  dt 1 d 1 d [ A ]       rate     k [ A ] x [ B ] x V dt dt A 2 0 0  Initial Concentrations are different; [A] 0 ≠ [B] 0  The integrated form is: Similar to 1 [ ] [ ] B A  0 ln k t equ 9.18 in  2 [ A ] [ B ] [ A ] [ B ] Clark 0 0 0  Which can be expressed as: [ A ]   [ B ]    0 log 0 . 43 k [ A ] [ B ] t log 2 0 0 [ B ] [ A ] 0 [ A ] log B [ ] [ A ] log B 0 [ ] 0 t CEE 670 Kinetics Lecture #2 David A. Reckhow Mixed Second Order    k A B products 2 10  Initial Concentrations are the same; [A] 0 =[B] 0      [ A ] [ B ] [ A ] x [ B ] x 0 0 dx  k [ A ][ A ] 2 dt       k [ A ] x [ A ] x 2 0 0 d [ A ] 1 1       k dt 2 k t  The integrated form is: A 2 2 2 [ A ] [ A ] [ A ] 0 1 1  Which can be integrated:   2 k t 2 [ A ] [ A ] 0 1 [ A ] 1 [ A ] 0 t CEE 670 Kinetics Lecture #2 David A. Reckhow 5

12/8/2011 Pseudo first order    k A B products 2 11  For most reactions, n=1 for each of two different reactants, thus a second-order overall reaction c  Many of these will have one B reactant in great excess (e.g., B)  These become “pseudo-1 st order in the limiting reactant, as the reactant in excess really doesn’t change in     5 1 1 k 3 . 9 x 10 Lmg min concentration c dc A   1 1 k c A c 2 B dt CEE 670 Kinetics Lecture #2 David A. Reckhow Pseudo-1 st order (cont.) dc   1 1 k c A c 2 B dt 12   k t c c e obs  Since C 2 changes little A Ao from its initial 820 mg/L, it 90 80 is more interesting to focus    5 k k c 3 . 9 x 10 ( 820 ) 70 on C A Concentration obs 2 B   60 1 0 . 032 min  C A exhibits simple 1 st 50 order decay, called 40 pseudo-1 st order 30  The pseudo-1 st order rate 20 constant is just the 10 “observed rate” or k obs 0 0 20 40 60 80 Time (min) CEE 670 Kinetics Lecture #2 David A. Reckhow 6

12/8/2011 Example: O 3 & Naphthalene 13  How long will it take for ozone (4.8 1 1   mg/L dose) to reduce the k t 2 [ A ] [ A ] concentration of naphthalene by 0 99%? 1 1  Used in moth balls and as a chemical   3000 t intermediate   6 4 10 10  2 nd order reaction; k 2 = 3000 M -1 s -1  Table 1 in Hoigne & Bader, 1983 [Wat.  Res. 17:2:173] 3000 t 990 , 000  Industrial WW with 0.1mM naphthalene  t 330 sec  Both reactants are at same (0.1mM)  concentration 5 . 5 min  Therefore, this reduces to a simple 2 nd order reaction CEE 670 Kinetics Lecture #2 David A. Reckhow O 3 & Naphthalene (cont.) 14  Contaminated river water (0.001 mM)  Now ozone is in great molar excess, so this is a pseudo-1 st order reaction   k [ B ] t [ A ] [ A ] e 2 0 0   [ A ]     ln k [ B ] t   [ A ] 2 0   0    8  10   4 ln 3000 10 t  6 10    4 . 605 0 . 3 t  t 15 . 4 sec CEE 670 Kinetics Lecture #2 David A. Reckhow 7

12/8/2011 Molecularity of three: 3 rd order kinetics 15  Quite improbably, but sometimes happens  Three different reactants     k A B C products 3 dx          k [ A ][ B ][ C ] k [ A ] x [ B ] x [ C ] x 3 3 0 0 0 dt  Complicated integrated form exists  Two different reactants    k 2 A B products 3 dx         2 2 k [ A ] [ B ] k [ A ] 2 x [ B ] x 3 3 0 0 dt  Integrated form:      [ ] [ ] 2 [ ] [ ] [ ] [ ]   A A B A B A    2 0 0 0 0 ln 2 [ B ] [ A ] k t 0 0 3 [ A ] [ A ] [ A ] [ B ] 0 0 CEE 670 Kinetics Lecture #2 David A. Reckhow 3 rd Order (cont.)  3  k 3 A products 16  Only one reactant or Initial Concentrations are the same dx  k [ A ][ A ][ A ] 3 dt         k [ A ] x [ A ] x [ A ] x 3 0 0 0 1 1 d [ A ]          2 k t 6 k t  The integrated form is: k dt A 3 3 2 2 [ A ] [ A ] A 3 3 [ A ] 0 1 1  Which can be integrated:   6 k t 3 2 2 [ A ] [ A ] 0 1 2 [ A ] 1 2 [ A ] 0 t CEE 670 Kinetics Lecture #2 David A. Reckhow 8