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Elementary Reactions Starting out with some A and B, we observe - PDF document

CEE 680 Lecture #5 1/29/2020 Print version Updated: 29 January 2020 Lecture #5 Kinetics and Thermodynamics: Fundamentals of Kinetics and Analysis of Kinetic Data (Stumm & Morgan, Chapt.2 ) (pp.16 20; 69 81) (Benjamin, 1.6) David


  1. CEE 680 Lecture #5 1/29/2020 Print version Updated: 29 January 2020 Lecture #5 Kinetics and Thermodynamics: Fundamentals of Kinetics and Analysis of Kinetic Data (Stumm & Morgan, Chapt.2 ) (pp.16 ‐ 20; 69 ‐ 81) (Benjamin, 1.6) David Reckhow CEE 680 #5 1 Elementary Reactions Starting out with some A and B, we observe that E  When reactant and F are the end products molecules collide with    A B C D the right orientation and slow energy level to form new  2 C E fast bonds     Many “observable” A D C F fast reactions are really just combinations of elementary reactions    2 A B E F David Reckhow CEE 680 #5 2 1

  2. CEE 680 Lecture #5 1/29/2020 S&M: Fig. 2.11 Pg. 72 Cont.  Elementary reactions  A single step in a reaction sequence  Involves 1 or 2 reactants and 1 or 2 products  Can be described by classical chemical kinetics David Reckhow CEE 680 #5 3 Kinetics  Examples  Fe +2 oxidation by O 2  almost instantaneous at high pH  quite slow at low pH  high D.O. may help  Oxidation of organic material  Formation of solid phases  Aluminum hydroxide  Quartz sand David Reckhow CEE 680 #5 4 2

  3. CEE 680 Lecture #5 1/29/2020 Kinetics  Base Hydrolysis of dichloromethane (DCM)  Forms chloromethanol (CM) and chloride  Classic second order reaction (molecularity of 2)     d [ DCM ] d [ OH ] d [ CM ] d [ Cl ]       Rate k [ DCM ][ OH ] dt dt dt dt  First order in each reactant, second order overall David Reckhow CEE 680 #5 5 Reaction Kinetics  Irreversible reaction  is one in which the reactant(s) proceed to product(s), but there is no significant backward reaction,  In generalized for, irreversible reactions can be represented as: aA + bB  Products  i.e., the products do not recombine or change to form reactants in any appreciable amount. An example of an irreversible reaction is hydrogen and oxygen combining to form water in a combustion reaction. We do not observe water spontaneously separating into hydrogen and oxygen. David Reckhow CEE 680 #5 6 3

  4. CEE 680 Lecture #5 1/29/2020 Reaction Kinetics: Reversibility  A reversible reaction  is one in which the reactant(s) proceed to product(s), but the product(s) react at an appreciable rate to reform reactant(s).  aA + bB  pP + qQ  Most reactions must be considered reversible An example of a reversible biological reaction is the formation of adenosine triphosphate (ATP) and adenosine diphosphate (ADP). All living organisms use ATP (or a similar compound) to store energy. As the ATP is used it is converted to ADP, the organism then uses food to reconvert the ADP to ATP. David Reckhow CEE 680 #5 7 Kinetic principles  Law of Mass Action  For elementary reactions    k aA bB products rate  a b kC A C B where, C A = concentration of reactant species A, [moles/liter] C B = concentration of reactant species B, [moles/liter] a = stoichiometric coefficient of species A b = stoichiometric coefficient of species B k = rate constant, [units are dependent on a and b] David Reckhow CEE 680 #5 8 4

  5. CEE 680 Lecture #5 1/29/2020 Reaction Kinetics (cont.)  Reactions of order dc   “n” in reactant “c” kc n dt  When n=0, we have a 90 simple zero ‐ order Concentration 80  o  reaction c c kt 70 60 50 40 30 Slope 20  10 k mg l / / min dc 10   k 0 Time (min) 0 20 40 60 80 dt David Reckhow CEE 680 #5 9 Reaction Kinetics (cont.) dc  When n=1, we   1 kc have a simple dt first ‐ order 90 80 reaction 70   kt Concentration c c e  This results in 60 o an “exponential 50 decay” 40 30 20 k   10 1 0 032 . min 0 0 20 40 60 80 Time (min) David Reckhow CEE 680 #5 10 5

  6. CEE 680 Lecture #5 1/29/2020 Reaction Kinetics (cont.) dc   1 kc  This equation can dt be linearized 100 Concentration (log scale)  good for  ln o  ln c c kt assessment of “k” from data Slope  k  1 0 032 . min 10 0 20 40 60 80 Time (min) David Reckhow CEE 680 #5 11 dc   2 kc Reaction Kinetics (cont.) dt  When n=2, we have a simple second-order reaction 90 1 80  This results in  c c 70 an especially Concentration  o 60 1 kc t wide range in o 50 rates 40  0 0015 k . L mg / / min  More typical to 30 have 2 nd order 20 in each of two 10 different 0 reactants 0 20 40 60 80 Time (min) David Reckhow CEE 680 #5 12 6

  7. CEE 680 Lecture #5 1/29/2020 Reaction Kinetics (cont.) dc   2 kc  Again, the equation can be linearized dt to estimate “k” from data Time (min) 0 20 40 60 80 0.12 1  1  kt 0.1 1/Concentration c c o 0.08 0.06 Slope 0.04  0 0015 k . L mg / / min 0.02 0 David Reckhow CEE 680 #5 13 Comparison of Reaction Orders  Curvature as order changes: 2 nd >1 st >zero 90 80 Zero Order 70 Concentration First Order 60 50 Second Order 40 30 20 10 0 0 20 40 60 80 Time (min) David Reckhow CEE 680 #5 14 7

  8. CEE 680 Lecture #5 1/29/2020 dc   1 1 kc 1 c Reaction Kinetics (cont.) 2 dt  For most reactions, n=1 for each of two different reactants, thus a second ‐ order overall reaction c  Many of these will have 2 one reactant in great excess  These become “pseudo ‐     5 1 1 k 3 . 9 x 10 Lmg min 1 st order in the limiting reactant, as the reactant in excess really doesn’t c change in concentration 1 David Reckhow CEE 680 #5 15 dc   1 1 kc 1 c 2 dt Reaction Kinetics (cont.)   k t c c e obs  Since C 2 changes little 1 1 o 90 from its initial 820 mg/L, 80 it is more interesting to    5 k obs kc 3 . 9 x 10 ( 820 ) 70 2 focus on C 1 Concentration 60   1 0 . 032 min  C 1 exhibits simple 1 st 50 order decay, called 40 pseudo ‐ 1 st order 30  The pseudo ‐ 1 st order 20 rate constant is just the 10 “observed rate” or k obs 0 0 20 40 60 80 Time (min) David Reckhow CEE 680 #5 16 8

  9. CEE 680 Lecture #5 1/29/2020 Variable Kinetic Order  Any reaction order, except n=1 dc   n kc dt 1 1   kt    n 1   n 1 n 1 c c o 1  c c    o   1     n 1  1 n 1 kc t n 1 o David Reckhow CEE 680 #5 17 Half ‐ lives  Time required for initial concentration to drop to half, i.e.., c=0.5c o  For a zero order reaction: 0 . 5 c 1  t o   c c kt   0 . 5 c c kt k o o o 1 2 2  For a first order reaction: ln( 2 )  t  kt   kt  1 1 c c e 0 . 5 c c e k 2 2 o o o 0 . 693  k David Reckhow CEE 680 #5 18 9

  10. CEE 680 Lecture #5 1/29/2020 Kinetic problem  If the half ‐ life of bromide in the presence of excess chlorine is 13 seconds (pseudo ‐ 1 st order reaction,  k       HOCl Br HOBr Cl  What is the pseudo ‐ 1 st order rate constant  how long does it take for 99% of the bromide to be oxidized? David Reckhow CEE 680 #5 19 Reactions in Series   k   k   k A B C D 1 2 3 k 1 =k 2 =k 3 =0.1 day -1 Fig. 2.9 Pg. 68 David Reckhow CEE 680 #5 20 10

  11. CEE 680 Lecture #5 1/29/2020 Reversible reaction kinetics k For a general reversible reaction: f  aA + bB pP + qQ k b And the rate law must consider both forward and reverse reactions: a b p q = k C C - k C C r A f b A B P Q where, k f = forward rate constant, [units depend on a and b] k b = backward rate constant, [units depend on a and b] C P = concentration of product species P, [moles/liter] C Q = concentration of product species Q, [moles/liter] p = stoichiometric coefficient of species P q = stoichiometric coefficient of species Q David Reckhow CEE 680 #5 21 Reversible 1 st order reactions  Kinetic law dB Fig. 2.10   k [ A ] k [ B ] 1 2 dt Pg. 69  Eventually the reaction slows and,  Reactant concentrations approach the equilibrium values dB    0 k [ A ] k [ B ] 1 2 dt [ B ] k   1 K eq [ A ] k 2 David Reckhow CEE 680 #5 22 11

  12. CEE 680 Lecture #5 1/29/2020 Analysis of Rate Data  Integral Method  Least squares regression of linearized form  Differential Method  estimate instantaneous rate at known time and reactant concentration  Initial rate Method  more rigorous, but slow  Method of Excess  only when 2 or more reactants are involved David Reckhow CEE 680 #5 23 Kinetic model for equilibrium  Consider a reaction as  The rates are: b  follows: r k { C }{ D } f  r k { A }{ B } b f  And at equilibrium the two A + B = C + D are equal, r f =r b  Since all reactions are  k { A }{ B } k { C }{ D } f b reversible, we have two  We then define an possibilities equilibrium constant (K eq ) k     A B C D f k { C }{ D }  f      K A B C D eq k { A }{ B } k b b David Reckhow CEE 680 #5 24 12

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