Chapter 14

1. Chapter 14 Chemical Kinetics reaction rate- the speed at which a chemical reaction occurs reaction mechanism- step-by-step molecular level view of the pathway from reactants to products

2. Factors That Affect Reaction Rates • Physical state of the reactants -if reactants are in different phases, the reaction is limited by the area of contact -inc. surface area of solid to inc rate 2. Reactant concentrations -inc. conc. = inc. collisions = inc. rate 3. Reaction temp -inc. temp = inc. KE/collisions = inc. rate • Presence of a catalyst

3. rate of appearance of product = ∆ [product] ∆ time [ ] = concentration rate units = M/s -the conc. of reactants dec. with time while the conc. of product inc. -rate of disappearance of reactants = rate of appearance of product rate of disappearance = -∆ [reactant] ∆ time

4. Sample Exercise 14.1 page 560 Using data in figure 14.3 pg 559, calculate the rate at which A disappears over the time interval from 20s to 40s. rate of disappearance = -∆ [reactant] ∆ time rate = -(0.30M-0.54M) (40s-20s) = 0.012M/s

5. Practice Exercise rate of product = (0.70M - 0M) (40s - 0s) = 0.018M/s

6. -it is typical for rates to dec. over time WHY? -the conc. of reactants dec. page 561 Table 14.1 and Fig 14.4

7. instantaneous rate- rate at a particular instant during a reaction -determined from the slope of the curve at a particular point in time -tangent lines are drawn to find the rate initial rate- instantaneous rate at t=0 Page 561 figure 14.4 *Problems page 562

8. Rate = -∆ [reactant] = ∆ [product] ∆ t ∆ t aA+ bB→ cC + dD -a,b,c,d are the coefficients from the balanced equation rate = -1∆[A] = -1∆[B] = +1∆[C] = +1∆[D] a ∆t b ∆t c ∆t d ∆t

9. Rate Law -relationship between the rate of the reaction and the conc. of the reactants -consider this reaction: aA+ bB→ cC + dD -the rate law would be: rate = k[A]m[B]n k = rate constant m and n= reaction order (small whole numbers) *the value of m and n determines how the rate depends on the conc. of the reactant- can only be found experimentally

10. Example: look at page 563 Table 14.2 rate = k[A]m[B]n -m and n are both first order based on data 5.4x10-7M/s = k(0.0100M)(0.200M) k = 2.7x10-4/M∙s *can use rate law and k to calculate rate for any set on conc. rate = 2.7x10-4/M∙s(0.100M)(0.100M) = 2.7x10-6M/s

11. -a large value of k (≥ 109) means a fast reaction -a small value of k (10 or lower) means a slow reaction -units of k change depending on overall order of reaction zero order = Ms-1 1st order = s-1 2nd order = M-1s-1 -overall order of reaction found by adding together orders of each reactant

12. First Order Reactions -rate is directly proportional to a single reactant rate = -∆ [A] = k[A] ∆t Integrated Rate Law -relationship between conc of reactant and time ln[A]t = -kt + ln[A]0 y = mx + b *has form of general equation for a straight line *if when t vs. ln[A]t is graphed and a straight line is produced then it is first order

13. Second-Order Reactions -rate is proportional to the square of the [A] or to reactants each raised to first power rate = -∆ [A] = k[A] 2 ∆t -more sensitive to conc. of the reactants than first order Integrated Rate Law 1/[A]t= kt+ 1/[A]0 [A]t = 1/ (kt + 1/[A]0) *if when t vs. 1/[A]tis graphed and a straight line is produced then it is second order

14. Zero-Order Reactions -rate of reaction is independent of the [A] rate = -∆ [A] = k ∆t -rate is the same at any concentration Integrated Rate Law [A]t = -kt + [A]0 *if when t vs. [A]tis graphed and a straight line is produced then it is zeroorder

15. Half-life -time required for the conc. of a reactant to fall to one half of its initial value Zero-Order t1/2 = [A]0 /2k First-Order t1/2 = 0.693/k Second-Order t1/2 = 1/ k[A]0

16. Temperature and Rate -as temp inc. rate inc. Ex- Glow sticks collision model- molecules must collide in order to react -the greater the # of collisions/sec the greater the reaction rate -inc temp and conc of reactants inc the collisions which inc the rate

17. -for a reaction to occur more is required than just collisions -molecules must be oriented in a certain way to make sure they are positioned to form new bonds -page 576 figure 14.15

18. -molecules must also possess a certain amount of energy to react -this energy comes from collisions -KE is used to stretch, bend and break bonds activation energy/energy barrier- minimum amount of energy required to initiate a chemical reaction (Ea) -as Eainc. reaction rate dec.

19. activated complex/transition state- a high energy intermediate step between reactants and products -page 577 figure 14.17 -for reverse reaction Ea = ∆E + Ea *must change sign on ∆E because going from right -Ea will dec. with inc. temp -page 578 figure 14.18 -the fraction of molecules that have an energy ≥ Eais given by: f = e-Ea /RT

20. Arrhenius found that most reaction data obeyed an equation based on: a) the fraction of molecules possessing energy Eaor greater b) # of collisions per second c) the fraction of collisions that have the appropriate orientation Arrhenius Equation: k = Ae-Ea/RT A= frequency factor related to frequency of collisions and the probability that the collisions are favorably oriented

21. Sample Exercise 14.10 page 579 Rank slowest to fastest 2 < 3 < 1 Practice Exercise Reverse from slowest to fastest 2 < 1 < 3

22. Reaction Mechanisms -the steps by which a reaction occurs Elementary Reactions -when a reaction occurs in a single event or step molecularity- defined by the # of reactants in an elementary reaction -if a single molecule is involved, the reaction is unimolecular bimolecular- collision of two reactants termolecular- collision between three molecules (not as probable as uni or bi)

23. Multistep Mechanisms -sequence of elementary equations ex- page 582 -the elementary reactions in a multistep mechanism must always add to give equation of overall process intermediate- not a reactant or a product, formed in one elementary reaction and consumed in the next

24. Sample Problem 14.12 page 582 • Molecularity = uni and bi b) Write equation for overall reaction 2O3(g)  3O2(g) • Identify intermediates O(g)

25. Practice Exercise • yes it is consistent b/c both equations add to give overall reaction • Molecularity = uni and bi c) Intermediates = Mo(CO)5

26. Rate Laws for Elementary Reactions -rate law is based on molecularity Table 14.3 page 584 Sample Problem 14.13 page 584

27. Multistep Mechanisms -each step in a mechanism has its own rate constant and Ea -often one step is much slower than the others -the overall rate cannot exceed the rate of the slowest elementary step rate determining step/rate limiting step- limits overall reaction rate -determines the rate law

28. Mechanisms with slow initial step Page 585 -since step 1 is slow it is rate-determining step -rate of overall reaction depends on rate of step 1 -rate law equals rate law of step 1 -step 1 is bimolecular so using page 584 it is second order rate = k1[NO2]2

29. Mechanisms with fast initial step Page 587 -step 1 has forward and reverse (k-1) reaction -step 2 is slow and determines overall rate rate = k2[NOBr2][NO] -NOBr2 is an intermediate (unstable, low conc.) -rate law depends on unknown conc. -not desirable -want to express rate law in terms of reactants and products