1. Chemical Kinetics © 2009, Prentice-Hall, Inc. Reaction Mechanisms Reactions may occur all at once or through several discrete steps. Each of these processes is known as an elementary reaction or elementary process.

2. Chemical Kinetics © 2009, Prentice-Hall, Inc. Reaction Mechanisms The molecularity of a process tells how many molecules are involved in the process.

3. Chemical Kinetics © 2009, Prentice-Hall, Inc. Multistep Mechanisms In a multistep process, one of the steps will be slower than all others. The overall reaction cannot occur faster than this slowest, rate-determining step.

4. Chemical Kinetics © 2009, Prentice-Hall, Inc. Slow Initial Step The rate law for this reaction is found experimentally to be Rate = k [NO 2 ] 2 CO is necessary for this reaction to occur, but the rate of the reaction does not depend on its concentration. This suggests the reaction occurs in two steps. NO 2 (g) + CO (g)  NO (g) + CO 2 (g)

5. Chemical Kinetics © 2009, Prentice-Hall, Inc. Slow Initial Step A proposed mechanism for this reaction is Step 1: NO 2 + NO 2  NO 3 + NO (slow) Step 2: NO 3 + CO  NO 2 + CO 2 (fast) The NO 3 intermediate is consumed in the second step. As CO is not involved in the slow, rate-determining step, it does not appear in the rate law.

6. Chemical Kinetics © 2009, Prentice-Hall, Inc. Fast Initial Step The rate law for this reaction is found to be Rate = k [NO] 2 [Br 2 ] Because termolecular processes are rare, this rate law suggests a two-step mechanism. 2 NO (g) + Br 2 (g)  2 NOBr (g)

7. Chemical Kinetics © 2009, Prentice-Hall, Inc. Fast Initial Step A proposed mechanism is Step 2: NOBr 2 + NO  2 NOBr (slow) Step 1 includes the forward and reverse reactions. Step 1: NO + Br 2 NOBr 2 (fast)

8. Chemical Kinetics © 2009, Prentice-Hall, Inc. Fast Initial Step The rate of the overall reaction depends upon the rate of the slow step. The rate law for that step would be Rate = k 2 [NOBr 2 ] [NO] But how can we find [NOBr 2 ]?

9. Chemical Kinetics © 2009, Prentice-Hall, Inc. Fast Initial Step NOBr 2 can react two ways: –With NO to form NOBr –By decomposition to reform NO and Br 2 The reactants and products of the first step are in equilibrium with each other. Therefore, Rate f = Rate r

10. Chemical Kinetics © 2009, Prentice-Hall, Inc. Fast Initial Step Because Rate f = Rate r, k 1 [NO] [Br 2 ] = k −1 [NOBr 2 ] Solving for [NOBr 2 ] gives us k1k−1k1k−1 [NO] [Br 2 ] = [NOBr 2 ]

11. Chemical Kinetics © 2009, Prentice-Hall, Inc. Fast Initial Step Substituting this expression for [NOBr 2 ] in the rate law for the rate-determining step gives k2k1k−1k2k1k−1 Rate =[NO] [Br 2 ] [NO] = k [NO] 2 [Br 2 ]

12. Chemical Kinetics © 2009, Prentice-Hall, Inc. Catalysts Catalysts increase the rate of a reaction by decreasing the activation energy of the reaction. Catalysts change the mechanism by which the process occurs.

13. Chemical Kinetics © 2009, Prentice-Hall, Inc. Catalysts One way a catalyst can speed up a reaction is by holding the reactants together and helping bonds to break.

14. Chemical Kinetics © 2009, Prentice-Hall, Inc. Enzymes Enzymes are catalysts in biological systems. The substrate fits into the active site of the enzyme much like a key fits into a lock.

15. Chemical Kinetics © 2009, Prentice-Hall, Inc. * The rate constant of first order reaction is 3.46x10 -2 s -1 at 298 K. What is the rate constant at 350 K if the activation energy for the reaction is 50.2 kJ/mol : ln k 2 / k 1 = E / R [ (1/T 1 ) – ( 1/T 2 ) ] Ln k 2 / 3.46x10 -2 = 50.2x10 3 /8.314 [ (1/298) – (1/350) ] = 3.01 k 2 = 0.702 s -1 * The decomposition of hydrogen peroxide is facilitated by iodide ions, (I - ), the over all reaction is : 2H 2 O 2(aq) → 2H 2 O (l) + O 2(g), the experimental rate law is : Rate = k [ H 2 O 2 ] [ I - ], we can assume the reaction takes place in two separate elementary steps : 1- H 2 O 2 + I - → H 2 O + IO -, 2- H 2 O 2 + IO - → H 2 O + O 2 + I - a) which one of the two steps is rate determining step ? - the first step is the rate determining (slow step) b) what is the intermediate ? - IO - is the intermediate C) is there any catalyst(s) ? - I - is the catalyst