1. 22.5 The temperature dependence of reaction rates Arrhenius equation: A is the pre-exponential factor; E a is the activation energy. The two quantities, A and E a, are called Arrhenius parameters. In an alternative expression lnk = lnA - one can see that the plot of lnk against 1/T gives a straight line.

2. Example: Determining the Arrhenius parameters from the following data: T/K 300350400450500 k(L mol -1 s -1 )7.9x10 6 3.0x10 7 7.9x10 7 1.7x10 8 3.2x10 8 Solution: 1/T (K -1 )0.003330.002860.00250.002220.002 lnk (L mol -1 s -1 )15.8817.2218.1918.9519.58 The slope of the above plotted straight line is –E a /R, so Ea = 23 kJ mol -1. The intersection of the straight line with y-axis is lnA, so A = 8x10 10 L mol -1 s -1 0 5 10 15 20 25 00.00050.0010.00150.0020.00250.0030.0035 Series1

3. The interpretation of the Arrhenius parameters Reaction coordinate: the collection of motions such as changes in interatomic distance, bond angles, etc. Activated complex Transition state For bimolecular reactions, the activation energy is the minimum kinetic energy that reactants must have in order to form products.

4. Applications of the Arrhenius principle Temperature jump-relaxation method: consider a simple first order reaction: A ↔ B at equilibrium: After the temperature jump the system has a new equilibrium state. Assuming the distance between the current state and the new equilibrium state is x, one gets [A] = [A] eq + x; [B] = [B] eq - x;

5. 22.6 Elementary reactions Elementary reactions: reactions which involves only a small number of molecules or ions. A typical example: H + Br 2 → HBr + Br Molecularity: the number of molecules coming together to react in an elementary reaction. Molecularity and the reaction order are different !!! Reaction order is an empirical quantity, and obtained from the experimental rate law; Molecularity refers to an elementary reaction proposed as an individual step in a mechanism. It must be an integral.

6. An elementary bimolecular reaction has a second-order rate law: A + B → P If a reaction is an elementary bimolecular process, then it has second-order reaction kinetics; However, if the kinetics are second- order, then the reaction might be complex.

7. 22.7 Consecutive elementary reactions An example: 239 U → 239 Np → 239 Pu Consecutive unimolecular reaction A → B → C The rate of decomposition of A is: The intermediate B is formed from A, but also decays to C. The net formation rate of B is therefore: The reagent C is produced from the unimolecular decay of B:

8. Integrated solution for the first order reaction (A) is: Then one gets a new expression for the reactant B: the integrated solution for the above equation is: when assuming [B] 0 = 0. Based on the conservation law [A] + [B] + [C] = [A] 0

10. Example. In an industrial batch process a substance A produces the desired compound B that goes on to decay to a worthless product C, each step of the reaction being first-order. At what time will B be present in the greatest concentration? Solution: At the maximum value of B Using the equation 25.7.6 and taking derivatives with respect to t: In order to satisfy = 0 t max = The maximum concentration of B can be calculated by plugging the t max into the equation.

11. Steady State Approximation Assuming that after an initial induction period, the rates of change of concentrations of all reaction intermediates are negligibly small. Substitute the above expression back to the rate law of B 0 ≈ [B] = (k 1 / k 2 )[A] Then The integrated solution of the above equation is [C] ≈ (1 - )[A] 0

12. Self-test 22.8 Derive the rate law for the decomposition of ozone in the reaction 2O 3 (g) → 3O 2 (g) on the basis of the following mechanism O 3 → O 2 + O k1 O 2 + O → O 3 k1’ O + O 3 → O 2 + O 2 k2 Solution: First write the rate law for the reactant O 3 and the intermediate product O Applying the steady state approximation to [O] Plug the above relationship back to the rate law of [O 3 ]

13. Rate determining step

14. Simplifications with the rate determining step Suppose that k 2 >> k 1, then k 2 – k 1 ≈ k 2 therefore concentration C can be reorganized as [C] ≈ (1 - )[A] 0 The above result is the same as obtained with steady state approximation

15. Kinetic and thermodynamic control of reactions Consider the following two reactions A + B → P 1 rate of formation of P 1 = k 1 [A][B] A + B → P 2 rate of formation of P 2 = k 2 [A][B] [P 1 ]/[P 2 ] = k 1 /k 2 represents the kinetic control over the proportions of products.

16. Problems 22.6 The gas phase decomposition of acetic acid at 1189 K proceeds by way of two parallel reactions: (1)CH 3 COOH → CH 4 + CO 2, k 1 = 3.74 s -1 (2)CH 3 COOH → H 2 C=C=O + H 2 O, k 2 = 4.65 s -1 What is the maximum percentage yield of the ketene CH 2 CO obtainable at this temperature.

17. Pre-equilibrium Consider the reaction: A + B ↔ I → P when k1’ >> k2, the intermediate product, I, could reach an equilibrium with the reactants A and B. Knowing that A, B, and I are in equilibrium, one gets: and When expressing the rate of formation of the product P in terms of the reactants, we get

18. Self-test 22.9: Show that the pre-equilibrium mechanism in which 2A ↔ I followed by I + B → P results in an overall third-order reaction. Solution: write the rate law for the product P Because I, and A are in pre-equilibrium so [I] = K [A] 2 Therefore, the overall reaction order is 3.

19. Kinetic isotope effect Kinetic isotope effect: the decrease in the rate of a chemical reaction upon replacement of one atom in a reactant by a heavier isotope. Primary kinetic isotope effect: the kinetic isotope effect observed when the rate-determining step requires the scission of a bond involving the isotope: with Secondary kinetic isotope effect: the variation in reaction rate even though the bond involving the isotope is not broken to form product: with

20. Kinetic isotope effect

21. 22.8 Unimolecular reactions The Lindemann-Hinshelwood mechanism A reactant molecule A becomes energetically excited by collision with another A molecule: A + A → A* + A The energized molecule may lose its excess energy by collision with another molecule: A + A* → A + A The excited molecule might shake itself apart to form products P A* → P The net rate of the formation of A* is

22. If the reaction step, A + A → A* + A, is slow enough to be the rate determining step, one can apply the steady-state approximation to A*, so [A*] can be calculated as Then The rate law for the formation of P could be reformulated as Further simplification could be obtained if the deactivation of A* is much faster than A*  P, i.e., then in case

23. The equation can be reorganized into Using the effective rate constant k to represent Then one has

24. The Rice-Ramsperger-Kassel (RRK) model Reactions will occur only when enough of required energy has migrated into a particular location in the molecule. s is the number of modes of motion over which the energy may be dissipated, k b corresponds to k 2

25. The activation energy of combined reactions Consider that each of the rate constants of the following reactions A + A → A* + A A + A* → A + A A* → P has an Arrhenius-like temperature dependence, one gets Thus the composite rate constant also has an Arrhenius-like form with activation energy, E = E1 + E2 – E1’ Whether or not the composite rate constant will increase with temperature depends on the value of E, if E > 0, k will increase with the increase of temperature

26. Combined activation energy

27. Theoretical problem 22.20 The reaction mechanism A 2 ↔ A + A (fast) A + B → P (slow) Involves an intermediate A. Deduce the rate law for the reaction. Solution: