Print version Updated: 28 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 + → + the right orientation and A B C D 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
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
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
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
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 − c c kt reaction 70 60 50 40 Slope 30 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 − = Concentration kt 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
Reaction Kinetics (cont.) dc = − 1 kc dt This equation can 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 50 o 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
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
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 1 st order in the limiting k 3 . 9 x 10 Lmg min 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
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 − = = 1 kt 1 c c e 0 . 5 c c e k 2 2 o o o 0 . 693 = k David Reckhow CEE 680 #5 18
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
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
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: f = r k { C }{ D } 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
Kinetic model with moles In terms of molar concentrations, the rates are: [ ] [ ] D [ ] [ ] B = γ γ = γ γ r k C D r k A B f f A b b C And at equilibrium the two are equal, r f =r b [ ] [ ] [ ] [ ] D γ γ = γ γ k A B k C D f A B b C And solving for the equilibrium constant (K eq ) [ ] [ ] [ ][ ] k γ γ γ γ C D C D ≡ f = = K C D C D [ ] [ ] [ ][ ] eq γ γ γ γ k A B A B b A B A B David Reckhow CEE 680 #5 25
To next lecture David Reckhow CEE 680 #5 26
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