1. 1 Chapter 12 – Chemical Kinetics 1.Second order Rate Law 2.Zero Order Rate Law 3.Reaction Mechanism 4.Model for Chemical Kinetics 5.Collision 6.Catalysis 7.Heterogeneous Catalysis 8.Homogeneous Catalysis

2. 2 Second-Order Rate Law For aA  products in a second-order reaction, Integrated rate law is Plot of 1/[A] vs t will produce a straight line: slope = k

3. 3 Half-Life of a 2nd-Order Reaction t 1/2 = half-life of the reaction k = rate constant A o = initial concentration of A The half-life is dependent upon the initial concentration.

4. 4 Second-Order Rate Law – 12.5 Butadiene reacts to form its dimer 2C 4 H 6 (g)  C 8 H 12 (g) Data: [C 4 H 6 ]Time 0.010000 0.006251000 0.004761800 0.003702800 0.003133600 0.002704400 0.002415200 0.002086200 a)Reaction order? b)Value of k? c)Half-life? 1/[C 4 H 6 ]ln[C 4 H 6 ] 100-4.605 160-5.075 210-5.348 270-5.599 320-5.767 370-5.915 415-6.028 481-6.175

5. 5 Second-Order Rate Law – 12.5 Butadiene reacts to form its dimer 2C 4 H 6 (g)  C 8 H 12 (g) Data: [C 4 H 6 ]Time 0.010000 0.006251000 0.004761800 0.003702800 0.003133600 0.002704400 0.002415200 0.002086200 a)Reaction order? Rate = k[C 4 H 6 ] 2 1/[C 4 H 6 ]ln[C 4 H 6 ] 100-4.605 160-5.075 210-5.348 270-5.599 320-5.767 370-5.915 415-6.028 481-6.175

6. 6 Second-Order Rate Law – 12.5 Butadiene reacts to form its dimer 2C 4 H 6 (g)  C 8 H 12 (g) Data: [C 4 H 6 ]Time 0.010000 0.006251000 0.004761800 0.003702800 0.003133600 0.002704400 0.002415200 0.002086200 b) Value of k? k = slope 1/[C 4 H 6 ]ln[C 4 H 6 ] 100-4.605 160-5.075 210-5.348 270-5.599 320-5.767 370-5.915 415-6.028 481-6.175

7. 7 Second-Order Rate Law – 12.5 Butadiene reacts to form its dimer 2C 4 H 6 (g)  C 8 H 12 (g) k = 6.14x10 -2 L/mol*s [A] 0 = 1.000x10 -2 mol/L c) Half-life?

8. 8 Zero-Order Rate Law Zero-order reaction the rate is constant. Rate does not change with respect to concentration Rate = k[A] o = k(1) = k [A] = -kt + [A] o t 1/2 = [A] o /2k

9. 9 Reaction Mechanism 4 The series of steps by which a chemical reaction occurs. 4 A chemical equation does not tell us how reactants become products - it is a summary of the overall process.

10. 10 Often Used Terms Intermediate: formed in one step and used up in a subsequent step and so is never seen as a product. Molecularity: the number of species that must collide to produce the reaction indicated by that step. Elementary Step: A reaction whose rate law can be written from its molecularity. uni, bi and termolecular

11. 11 Reaction Mechanism We can define a reaction mechanism. It is a series of elementary steps that must satisfy two requirements: 1. The sum of the elementary steps must give the overall balanced equation for the reactions. 2. The mechanism must agree with the experimentally determined rate law.

12. 12 Reaction Mechanism 4 The reaction NO 2 (g) + CO(g)  NO(g) + CO 2 (g) has many steps in the reaction mechanism. Rate = k[NO 2 ] 2 NO 2 (g) + NO 2 (g)  NO 3 (g) + NO(g) NO 3 (g) + CO(g)  NO 2 (g) + CO 2 (g) NO 2 (g) + NO 2 (g) + NO 3 (g) + CO(g)  NO 3 (g) + NO(g) + NO 2 (g) + CO 2 (g) Overall reaction: NO 2 (g) + CO(g)  NO(g) + CO 2 (g)

13. 13 Rate-Determining Step In a multistep reaction, it is the slowest step. It therefore determines the rate of reaction. – Slow (rate determing) NO 2 (g) + NO 2 (g)  NO 3 (g) + NO(g) – Fast NO 3 (g) + CO(g)  NO 2 (g) + CO 2 (g) Rate of formation of NO 3 =  [NO 3 ]/  t = K 1 [NO 2 ] 2 Overall rate =  [NO 3 ]/  t = k 1 [NO 2 ] 2

14. 14 A Summary (continued)

15. 15 Model for Chemical Kinetics Collision Model Key Idea: Molecules must collide to react. However, only a small fraction of collisions produces a reaction. Why? Arrhenius: An activation energy must be overcome.

16. 16 Reaction Progress The arrangement of atoms found at the top of the potential energy barrier is call the activated complex or transition state.

17. 17 Reaction Progress  E has no effect on the rate of reaction. Rate depends on the size of the activation energy E a.

18. 18 Reaction Progress A certain minimum energy is required for the molecules to “get over the hill” At a given temperature only a fraction of the collisions possess enough energy to be effective - Lower temperature -Effective collisions - Small - Higher temperature -Effective collisions – Increase exponentially

19. 19 Reaction Progress - Collisions Collisions with E a = (Total Collisions)e -Ea/RT Observed rate smaller than the rate of collisions with E a Molecular Orientations:

20. 20 Reaction Progress - Collisions Requirements must satisfied for reactants to collide successfully 1. Collisions must involve enough energy to produce the reaction; Collision energy must equal or exceed the activation energy 2. Relative orientation of reactants must allow for the formation of any new bonds necessary to produce products. Rate constant: k = zpe -Ea/RT z = collision frequency R = 8.3145 J/K mol p = steric factor (fraction with effective orientation)

21. 21 Catalysis Catalyst: A substance that speeds up a reaction without being consumed Enzyme: A large molecule (usually a protein) that catalyzes biological reactions. Homogeneous catalyst: Present in the same phase as the reacting molecules. Heterogeneous catalyst: Present in a different phase than the reacting molecules.

22. 22 Catalysis Increases the number of effective collisions by providing a reaction pathway with a lower activation energy

23. 23 Heterogeneous Catalysis Heterogeneous catalysis most often involves gaseous reactant being absorbed on the surface of a solid surface. Hydrogenation is an example:Changes C=C into saturated H-C-C-H 1.Adsorption to addition of a substance to the surface of another. 2.Migration of absorbed reactants on the surface 3.Reaction of absorbed substances 4.Desorption of products

24. 24 Heterogeneous catalytic ethylene hydrogenation: C 2 H 4 + H 2 → C 2 H 6

25. 25 Homogeneous Catalysis Reactants and Catalysis are in the same phase. Gas-Gas or Liquid-Liquid N 2 (g) + O 2 (g)  2NO(g) Product of high-temperature combustion when N 2 is present. However catalytic in production of ozone 2NO(g) + O 2 (g)  2NO 2 (g) NO 2 (g)  NO(g) + O(g) (light) O 2 (g) + O(g)  O 3 (g) 3 / 2 O 2 (g)  O 3 (g)

26. 26 Homogeneous Catalysis In the upper atmosphere, NO has opposite effect. 2NO(g) + O 3 (g)  NO 2 (g) + O 2 (g) O + NO 2 (g)  NO(g) + O 2 (g) O 3 (g) + O(g)  2O 2 (g) Nitric Oxide is catalytic in production of O 2. O 3 required in upper atmosphere to block uv radiation.

27. 27 Homogeneous Catalysis Freon - ChloroFluoroCarbons CCl 2 F 2 (g)  CClF 2 (g) + Cl(g) Light Cl + O 3 (g)  ClO(g) + O 2 (g) O(g) + ClO(g)  Cl(g) + O 2 (g) O 3 (g) + O(g)  2O 2 (g) Cl(g) destroyer of ozone – catalytic in destruction of O 3.