lecture 3 chemical kinetics Textbooks/References Finlayson-Pitts and Pitts, Chemistry of the Upper and Lower Atmosphere: Theory, Experiments, and Applications, Academic Press, Seinfeld and Pandis, Atmospheric Chemistry and Physics, From Air Pollution to Climate Change, Wiley and Sons, Jacob, Introduction to Atmospheric Chemistry, Princeton, Wayne, Chemistry of Atmospheres, 2nd ed., Oxford, Warneck, Chemistry of the Natural Atmosphere, Academic, Brasseur, Orlando, and Tyndall, eds., Chemistry and Global Change, Oxford, 1999.

lecture 3 chemical kinetics A  P reaction rate R: R{P}=d[A]/dt=-k 1 [A]  d[A]/[A]=- k 1 dt [A]=[A] 0 exp{-k 1 t} If A participates in more than one reaction, this is the rate of change of A with respect to this reaction. Overall d[A]/dt will include other terms. How to measure k 1 ? Plot ln([A]/[A] 0 ) versus t. What is the shape? What is the slope? How is it possible to measure small reaction constants? If [A] is very large, R will be reasonable even if k 1 is small and we will see change in [A] over time of experiment. First Order Chemical Reaction

lecture 3 chemical kinetics Lets define t=  when [A]=[A] 0 /e Then: [A] 0 /e=[A] 0 exp{-k 1  } 1=k 1   ….   =1/k 1 The lifetime (or half life) is the time it takes for the concentration to become 1/e its initial value. Not every molecule of A will last exactly this amount of time. Calculation of Lifetime in First Order Chemical Reaction

lecture 3 chemical kinetics A+B  P d[A]/dt=-d[P]/dt=-k 2 [A][B] Assume: [A]=[A] 0 -X, B=[B] 0 -X d[A]/dt=-k 2 ([A] 0 -X)([B] 0 -X) Solution: k 2 t ={1/([A] 0 -[B] 0 )} ln{([A] 0 -X)[B] 0 /([B] 0 -X)[A] 0 } If A+A  P: –lose A twice as fast as gain P, rate of loss of A is 1/2 rate of gain of P Second Order Chemical Reaction

lecture 3 chemical kinetics Chemical Reactions of General Order  : order of reaction with respect to A  : order of reaction with respect to B  +  : overall order of reaction

lecture 3 chemical kinetics If [B] 0 -X~[B] 0 then: d[A]/dt=-k 2 [A][B] 0 =-’k’[A] d[A]/[A] =-’k’t  [A]=[A] 0 e -’k’t Or: [A]=[A] 0 exp{-k 2 [B] 0 t} Conditions under which get pseudo first order reaction: 1.Concentration of one reactant (B) is so high that it is approximately constant during reaction. (Changes in A and B are small compared to [B].) 2.Concentration of one reactant is low but relatively constant over the time scale of changes in other reactants. Pseudo First Order Chemical Reaction

lecture 3 chemical kinetics Atmospheric half life of CO CO + HO.  CO 2 +H. k 2 =4x cm 3 molec -1 sec -1 HO. =10 7 molec/cm 3  = 1/’k’=1/k 2 [HO. ]=1/(4 x x10 7 )=2,500,000 sec Or  =30 days This calculation is equivalent to: Note that this method of calculating  may be simpler (e.g., in consecutive reactions). An Example of a Pseudo First Order Reaction

lecture 3 chemical kinetics A+B  C C+D  P C is involved in 2 reactions, so its overall rate of change has 2 terms: (1) gain of C from the first reaction (positive sign): +k 1 [A][B] (2) loss of C from the second reaction (negative sign): -k 2 [C][D] d[C]/dt=k 1 [A][B]-k 2 [C][D] In steady state (SS): d[C]/dt=0 If true, we can calculate C even if its levels are below any detection limit: [C]=k 1 [A][B]/k 2 [D] Also, we can use [C] to calculate concentrations of other species in the reactions: Consecutive Reactions and the Steady State Assumption

lecture 3 chemical kinetics d[C]/dt=k 1 [A][B]-k 2 [C][D] 1.If k 2 >> k 1, C is short-lived and d[C]/dt is low with respect to [A], [B], and [P]. 2.If production and loss do not vary significantly over lifetime of C. Example: In photochemistry when the solar radiation intensity (SRI) is at max (high noon) or min value, it does not vary much over lifetime of photochemically produced oxygen atom: If [O]  SRI and d[SRI]/dt=0 then d[O]/dt=0 O 3 +h  O+O 2 rate constant: J 1 {O 3 } O+O 2  O 3 rate constant: k 2 d[O]/dt=J 1 {O 3 }[O 3 ]-k 2 [O][O 2 ]=0 [O]= J 1 {O 3 }[O 3 ]/k 2 [O 2 ] For longer-lived species, the SS assumption is valid when the lifetime is much longer than seasonal changes. When is the SS Assumption Valid?

lecture 3 chemical kinetics NO 2 +h  NO+ O( 3 P  J  O( 3 P)+O 2  O 3 k 2 d[O( 3 P  /dt=J 1 [NO 2 ]-k 2 [O( 3 P  ][O 2 ] In SS: d[O( 3 P  /dt=0 [O( 3 P  =J 1 [NO 2 ]/k 2 [O 2 ] [O( 3 P  =5x10 -3 x5.4x10 12 /5x x5.4x10 18 =10 3 molecules cm -3 =3.7x ppm 1ppm=10 -6 x6.02x10 23 /22400(273/T)=2.69x10 13 molecules cm -3 Another Example of the Steady State Assumption

lecture 3 chemical kinetics X  P k 1 X+A  Pk 2 X+A+B  Pk 3 R r =-d[X]/dt=k 1 [X]+k 2 [A][X]+k 3 [A][B][X]= R r = [X]{k 1 +k 2 [A]+k 3 [A][B]}=[X]/  Or  =[X]/R r Note that this is the same form as what we wrote for the single reaction: CO + HO.  CO 2 +H. It is correct based on the fact that first order lifetimes are additive: X  P a ; X  P b d[X]/dt= k a [X]+k b [X]= [X](k a +k b ) After integrationln[X]/[X] 0 =(k a +k b )t Species Lifetime Calculation for Consecutive Reactions

lecture 3 chemical kinetics A+B+C  P This is not a simultaneous three body collision, rather A+B  AB* AB*+C  P Example O+O 2 +M  O 3 +M d[O 3 ]/dt=R{O 3 }=k 3 [O][O 2 ][M]=‘k 3 ’[O]pseudo 1st order (Why?) For normal atmospheric conditions M=1atm, O 2 =0.2atm, or 10 6 and 2x10 5 ppm, or 2.7x10 19 and 5.4x10 18 molecules cm -3 respectively Third Order Chemical Reaction

lecture 3 chemical kinetics First order rate constant: sec -1 or min -1 Second order rate constant: R{P}=k 2 [A][B] k 2 =R{P}/[A][B]={molec cm -3 sec -1 } /{molec cm -3 } 2 = cm 3 molec -1 sec -1 ={ppm min -1 } /{ppm} 2 =ppm -1 min -1 Third Order: cm 6 molec -2 sec -1 or ppm -2 min -1 Rate Constant Units

lecture 3 chemical kinetics Reverse Reactions and Chemical Equilibrium Reverse reactions: At steady state (equilibrium): Equilibrium constant:

lecture 3 chemical kinetics Pressure Dependence of Equilibrium If products are in the vicinity of other molecules (bath gas), reaction rate depends on pressure: What kind of reaction is last reaction? Quenching. High pressure limit (high [M]) Which is the rate determining step? The slowest reaction: In this limit, the rate does not depend on pressure. Low pressure limit (low [M]) Which is the rate determining step? In this limit, the rate does depend on pressure (on [M]).

lecture 3 chemical kinetics Steady State Assumption and Pressure Dependence Assume AB* is short-lived (k 2 and k -1 >> k 1 ): Solve for [AB*]: Use [AB*] to solve for production rate of [AB]: What order is the rate constant? 3

lecture 3 chemical kinetics Steady State Assumption and Pressure Dependence cont. High pressure limit (k 2 [M] >> k -1 ): Low pressure limit (k -1 >> k 2 [M]): What will the rate constant look like as a function of pressure ([M])?