1. Chemistry 40S Unit 3: Chemical Kinetics Lesson 4

2. Learning Outcomes O C12-3-09 Explain the concept of a reaction mechanism. Include: rate-determining step O C12-3-10 Determine the rate law and order of a chemical reaction from experimental data. Include: zero-, first-, and second-order reactions and reaction rate versus concentration graphs

3. Reaction Mechanism O A reaction mechanism summarizes the individual steps a reaction follows O Each individual step is called an elementary step or an elementary process O The balanced equation of a reaction represents the net reaction mechanism

4. Reaction Mechanism O Example 1: 2NO(g) + O 2 (g)  2NO 2 (g) Step 1: 2NO(g)  N 2 O 2 (g) Step 2: N 2 O 2 (g) + O 2 (g)  2NO 2 (g) Net reaction: 2NO(g) + O 2 (g)  2NO 2 (g) O As the N 2 O 2 appears in the reaction mechanism but not in the overall chemical equation, it is called an intermediate.

5. Reaction Mechanism O Example 2: 2O 3 (g)  3O 2 (g) Step 1: Cl 2 (g) + O 3 (g)  ClO(g) + O 2 (g) Step 2: O 3 (g)  O 2 (g) + O(g) Step 3: ClO(g) + O(g)  Cl 2 (g) + O 2 (g) Net reaction: 2O 3 (g)  3O 2 (g) O Cl 2 (g) is a catalyst and the ClO(g) is an intermediate O Catalysts are unchanged & not in the overall reaction O Intermediates are not in the overall reaction because they are transformed

6. Rate-Determining Step O Not all steps in a mechanism have the same speed O Rate-Determining Step - slowest of the elementary processes O Determines the rate of the reaction because it has a stronger affect than the others

7. Rate-Determining Step Problem O Example 1: Given the following mechanism: P + Q  X + T (slow) X + P  Y + R (fast) Y + S  T (moderate) a) What is the net reaction? b) What are the reaction intermediates? c) Which step is the rate determining step?

8. Rate-Determining Step Problem d) What would be the effect of increasing the concentration of P? e) What would be the effect of decreasing the concentration of Q? f) What would be the effect of increasing the concentration of S?

9. Rate Law

10. O Rate of consumption is directly proportional to concentration O The faster the reactants are consumed, the lower its concentration O This can be represented by: Rate = k[A] x O k – specific rate constant O [A] – concentration of A O x – order of the reaction O The rate constant is unique to each reaction at a specific temperature (its value depends on the speed/types of molecules) O Therefore = changing T  change in the rate constant

11. Order of Reaction O Order of reaction – indicates the effect of concentration of reactants on reaction rate O First Order Reaction (x = 1) – reaction rate is directly proportional to changes in reactant concentration O i.e. doubling reactant concentration = doubling rate

12. Order of Reaction O Second Order Reaction (x =2) – reaction rate is proportional to the square of the change in concentration O i.e. doubling concentration = 2 2 = 4x increase in rate O Zero Order Reaction (x=0) – the reaction rate does not depend on the concentration of reactants

13. Rate Law O When a reaction has two reactants the rate depends on the concentration of both reactants O Each reactant affects the rate differently therefore: Rate = k[A] x [B] y O The total order of the reaction is the sum of the order of all reactants (i.e. x + y)

14. Determining the Rate Law of a Reaction O Rate law can only be determined experimentally (despite our knowledge of reaction stoichiometry) O There are 3 ways to determine rate law, but all three work to determine the order of each reactant: O Differential Rate Law (calculus) O Integrated Rate Law (time vs. concentration graphs) O Initial Rates

15. Rate Law using Initial Rates Example 1: What is the rate law for the below reaction based on the provided experimental data? H 2 O 2 + 2 HI → 2 H 2 O + I 2 Trial[H 2 O 2 ] (M)[HI] (M)Initial Rate (mol/L·s) 10.10 0.0076 20.100.200.0152 30.200.100.0152

16. Rate Law using Initial Rates Example 2: For the reaction A + B → products, the following data was collected. Determine the rate law. Trial[A] (M)[B] (M)Initial Rate (mol/L·s) 10.100.202.0 20.300.2018.0 30.200.4016.0

17. Determining the Specific Rate Constant Example 1: Determine the value of the specific rate constant, k, based on the below information. Rate = k[A] 2 [B] Trial[A] (M)[B] (M)Initial Rate (mol/L·s) 10.100.202.0 20.300.2018.0 30.200.4016.0

18. Rate Law & Stoichiometry O For single step (elementary reactions) the order of each reactant in the rate law is equal to the coefficient in the reaction's balanced equation Example 1: aA + bB → cC + dD rate = k[A] a [B] b

19. Rate Law & Reaction Mechanisms O Rate law corresponds to the stoichiometry of the rate determining step O Example 1: The mechanism for the reaction: 3 M + N → P + 2 Q O a) What is the rate law for this reaction? b) What would be the effect of tripling the [M]? c) What would be the effect of doubling the [N]?