Review of Fitting Kinetic Data

Transcription

1 L6-1 Review of Fitting Kinetic Data True or false: The goal of fitting kinetic data is to find the true rate expression. What are the two general methods used to fit kinetic data?

2 L6-2 Advantages and Drawbacks

3 L6-3 Review of Fitting Kinetic Data Cin = 15 mg/l Cin = 20 mg/l Cin = 25 mg/l CST Data as Obtained Concentration, mg/l Time, min

4 Review of Fitting Kinetic Data ln{(cin-c)/(cin-c0)} Transformed Transformed CST Data with CST Trendline Data (all data) Cin = 15 mg/l Cin = 20 mg/l Cin = 25 mg/l time, min ln{(cin-c)/(cin-c0)} = t R 2 = What is the order of the reaction? L6-4

5 Determining k 0, E A So, now that we know k, how can we obtain k 0 and E A?? Remember k? Plotting ln k ln k ln k 0 -E/R What s What s the the slope? slope? What s What s the the intercept? 1/T L6-5

6 Kinetics of Many Processes in Nature follow Arrhenius Relationship Some Examples Cricket chirping Ant walking Tumor growth Diffusion in solids [D = D o exp (-E D /RT)] L6-6

7 L6-7 Ant walking: Does it follow Arrhenius Law? M. I. Masel,, Chemical Kinetics and Catalysis

8 The Ants are walking one by one Raw data (speed vs temperature) Linearized data (ln u vs 1/T) Running Speed (cm/s) ln (wlaking speed) Temperature (C) L /T (K -1 )

9 Summary: Determining Rate Coefficients 1. set up 2. measure 3. set-up 4. linearize 5. use 6. do this 7. linearize, 8. use linear - slope - intercept L6-9

10 Example: determining k Nitrogen dioxide is introduced into an isothermal batch reactor at a temperature of 300 C. Its decomposition to nitrogen oxide and oxygen is followed over time and the data below is obtained. Is the reaction first order or second order? Determine the reaction rate coefficient. t (s) [NO2] (mol/l) NO2 <=> 2 NO + O2 order integrated rate law linear plot for 1 2 so, so, let s let s do do it! it! L6-10

11 L6-11 Example: 1 st order kinetics y = x R 2 = st order fit ln {[NO2 / (mol/l)} t [s]

12 L6-12 Example: 2 nd order kinetics y = x R 2 = 1 1/[NO2] / (mol/l) 120 2nd order fit t [s] perfect fit indicates

13 Determining Reaction Order L6-13 Great, so if I know the reaction order n, I can determine k. What if I do not know n? 1. Try different reaction orders (i.e. search for the best fit of the Example: data to the integrated rate laws) Works only for reasonably simple reaction orders (n = -1, 0, 1, 2) The following is measured data from some reaction A -> P t (min) C A (mol/l) C A (mol/l) time (min)

14 L6-14 Reaction Order: Example cont d (1) Assume 1 st order, linearize, and replot ln(ca/ca,0) (dimensionless) time (min)

15 L6-15 Reaction Order: Example cont d (1) Assume 2 nd order, linearize, and replot 1 1 = kt C A C A, [1/CA - 1/CA,0] (L/mol) time (min)

16 Reaction Order: Example cont d (1) Assume order 1.5, linearize, and replot r = k C 1.5 A into BR design equation: with constant volume assumption: and integration yields: 1/C A 0.5-1/CA, /C 0.5 A - 1/C 0.5 A,0 = t R 2 = What is k?? L time (min)

17 Determining Reaction Order Obviously, trial-and-error fits will not help us except for very simple cases. Hence we need a generally applicable way 2. Initial Rates (differential) Method: Measure reaction rate at several different initial concentrations. From r = k [A] n => ln r = n ln[a] + ln k Hence, the logarithmic plot of initial reaction rate r 0 vs [A] 0 yields the reaction order n as the slope. ln r 0 ln [A] 0 L6-17

18 Reaction Order, cont d In practice, one rarely deals with simple reactions with only one reactant A -> products. Therefore, the reaction order needs to be determined for each reactant separately. This is done by the Isolation Method: The concentrations of all but one reactant are kept constant during an experiment, thus isolating the effect of this reactant on the reaction rate. L6-18

19 L6-19 Summary: Determining Reaction Orders Integral + Trial and Error: Differential Method Isolation Method

20 Designing a Kinetic Experiment That s it! You now know how to design an experiment to determine all relevant kinetic data for any reaction system: 1. Select 2. set up 3. measure 4. determine 5. measure 6. determine L6-20

21 A Glimpse of Theory As we saw above, kinetic experiments are difficult and lengthy procedures (and hence also expensive ), so one would really LOVE to replace experiments by theoretical calculations. While we are still far away from being able to do this reliably, theoretical chemistry has made great progress in past decades and is becoming an increasingly important tool in chemical kinetics (even in industrial application!). The following will give just a very brief glimpse of some of the very basics of those calculations. L6-21

22 A (very brief) Theory of Reaction Rates All chemical reactions occur by collisions between molecules (or by collision of molecules with surfaces that s catalysis, and we ll talk about that a bit later ). One very simple theory of reactions builds on this observation, and is hence called collision theory. You can t react if if you don t collide! Consider the simple bimolecular reaction: In the framework of the collision theory, the rate of this reaction is described by: L6-22 r =

23 Probability of Successful Collision The probability of a successful collision can be estimated based on the fact that a bond needs to be broken during a reactive event. This requires a certain amount of energy, E. Since the energy distribution (for thermal energy) is governed by a Boltzmann distribution, the probability that a molecule at temperature T has acquired the necessary energy is p(e) ) = exp{-e/rt} E/RT}. fraction of successful collisions (= probability distribution) L6-23 Boltzmann-Distribution p(e) ) = exp{-e/rt}

24 (Simple) Collision Theory The collision rate can be calculated from the kinetic theory of gases: The frequency of collisions of a molecule of A with B is given by: The collision cross section (assuming hard spheres) is Z AB = Therein, the average thermal velocity u is given by <u> = The reduced molecular mass is given by: μ = σ AB = The collision cross section for two molecules can be regarded as L6-24

25 (Simple) Collision Theory Putting this all together, we obtain for the collision rate: Or, with concentrations: Z AB = We can hence write for the reaction rate: r = Z AB = Z AB = Compare this to the usual rate law for bimolecular (elementary) reactions: r = k 0 exp{-e A /RT} C A C B We can therefore calculate the pre-exponential factor for a 2 nd order reaction from: k 0 = L6-25

26 (Simple) Collision Theory, cont d k 0 = {8 RT/(π π μ)} μ 1/2 π d AB This value gives in effect the maximal possible rate coefficient for bimolecular reactions. In reality, the rate coefficients are typically significantly lower than this, due to steric as well as energetic considerations. Example:. OH + H. ->. O. + H 2 AB 2 L6-26

27 (Simple) Collision Theory, cont d L6-27 This value gives in effect the maximal possible rate coefficient for bimolecular reactions. In reality, the rate coefficients are typically significantly lower than this, Example 2: k 0 = {8 RT/(π π μ)} μ 1/2 π d AB due to steric as well as energetic considerations. Steric and energetic factors can reduce the rate coefficient by many orders of magnitude. Collision theory can hence only give an upper limit to k! (Comparison with similar reactions with known steric factors occasionally allows us to estimate the steric factor and hence to calculate k 0 from collision theory. => r corrected = P Z exp{-e/rt} ) AB 2

28 Example L6-28 Estimate the rate coefficient for bimolecular gas phase reactions. Assume room temperature. Values: R = 8.3 J/mol K, T = 300 K μ = d AB Dimensions: [k 0 ] = k 0 = k 0 = {8 RT/(π μ)} μ 1/2 π d 2 AB -> to obtain the usual dimensions for a 2 nd order rate constant, we need to multiply with Avogadro s number N A = 6.02 * (atoms)/mol Compare with experimental values: K + Br 2 -> KBr + Br k 0,exp CH 3 + CH 3 -> > C 2 H 6 k 0,exp H 2 + C 2 H 4 -> > C 2 H 6 k 0,exp 0,exp 0,exp 0,exp Let s assume simple molecules: N 2 and O 2 k 0 = k 0,theo P ; P

29 Collision Theory: Summary frequency of of collision fraction of of molecules facing facing each each other other correctly k = x x fraction of of molecules with with sufficient kinetic kinetic energy energy to to break break bonds bonds Which of these terms does Collision Theory capture (more or less) correctly, and which does it not? L6-29

30 Collision Theory: Unimolecular Rctns Collision theory does offer interesting insights into unimolecular reactions: It is experimentally observed that unimolecular reactions show a surprising decrease in the measured reaction coefficient towards very low pressures (so-called fall-off regime). How can that be explained? In reality, unimolecular reactions require intermolecular collisions! A -> B, r = k C A is in reality the sum reaction for a reaction sequence! A + A <=> A * + A, r 1 = A * -> B, r 2 = Since A * is a highly reactive intermediate, let s assume it s concentration is essentially constant: High pressure limit: dc A* /dt = Hence: r B = r 2 = C A* = L6-30 Low pressure limit: