1. Chemical Kinetics Unit 11

2. Chemical Kinetics Chemical equations do not give us information on how fast a reaction goes from reactants to products. KINETICS: the study of reaction rates and their relation to the way the reaction proceeds, i.e. its mechanism We can use thermodynamics to tell if a reaction is product – or reactant – favored Only kinetics will tell us how fast the reaction happens!

3. Rate of Reaction A rate is any change per interval of time.  Example: speed (distance/time) is a rate! Reaction rate = change in concentration of a reactant or product with time

4. Expressing a Rate For the reaction A  P = Appearance of product Disappearance of reactant

5. Reaction Conditions & Rates Collision Theory of Reactants Reactions occur when molecules collide to exchange or rearrange atoms Effective collisions occur when molecules have correct energy and orientation

6. Factors Affecting Rates 1. Concentrations (and physical state of reactants and products) 2. Temperature 3. Catalysts Catalysts are substances that speed up a reaction but are unchanged by the reaction

7. Effect of Concentration on Reaction Rate To propose a reaction mechanism, we study the reaction rate and its concentration dependence.

8. Rate Laws or Rate Expressions The rate law for a chemical reaction relates the rate of reaction to the concentration of reactants. For aA + bB  cC + dD The rate law is: Rate = k[A] m [B] n The exponents in a rate law must be determined by experiment. They are NOT derived from the stoichiometry coefficients in an overall chemical equation.

9. Rate Laws & Orders of Reactions Rate Law for a reaction: Rate = k[A] m [B] n [C] p The exponents m, n, and p  Are the reaction order  Can be 0, 1, 2, or fractions (may be other whole numbers in fictional examples)  Must be determined by experiment Overall Order = sum of m, n, and p

10. Interpreting Rate Laws If m = 1 (1 st order) Rate = k [A] 1 If [A] doubles, then the rate doubles (goes up by a factor of 2) If m = 2 (2 nd order) Rate = k [A] 2 If [A] doubles, then rate quadruples (increases rate by a factor of 4) If m = 0 (zero order) Rate = k [A] 0 If [A] doubles, rate does not change! Rate = k[A] m [B] n [C] p

11. Rate Constant, k Relates rate and concentration at a given temperature. General formula for units of k: M (1- overall order) time -1 Overall OrderUnits of k 0 M time -1 1 Time -1 2 M -1 Time -1 3 M -2 Time -1

12. Rate Law Problem: The initial rate of decomposition of acetaldehyde, CH 3 CHO, was measured at a series of different concentrations and at a constant temperature. Using the data below, determine the order of the reaction – that is, determine the value of m in the equation CH 3 CHO(g)  CH 4 (g) + CO(g) Rate = k[CH 3 CHO] m CH 3 CHO (mol/L) 0.1620.1950.2730.4100.518 Rate (mol/L*min) 3.154.568.9420.235.2

13. Strategy Use the equation: Pick any two points from the given data!

15. Deriving Rate Laws Rate of rxn = k[CH 3 CHO] 2 Here the rate goes up by FOUR when the initial concentration doubles. Therefore, we say this reaction is SECOND order overall.

16. Example: Using the same set of data from the previous example, and knowing the order of the reaction, determine: b) the value of the rate constant, k (w/ units!) c) the rate of the reaction when [CH 3 CHO] = 0.452 mol/L Strategy:  Use any set of data to find k.  Solve for rate using k, rate order equation, and given concentration.

18. The data below is for the reaction of nitrogen (II) oxide with hydrogen at 800 o C. 2NO(g) + 2H 2 (g)  N 2 (g) + 2H 2 O(g) Determine the order of the reaction with respect to both reactants, calculate the value of the rate constant, and determine the rate of formation of product when [NO]=0.0024 M and [H 2 ]=0.0042 M. Strategy: Choose two experiments where concentration of one reactant is constant and other is changed; solve for m and n separately!

20. Example: The initial rate of a reaction A + B  C was measured with the results below. State the rate law, the value of the rate constant, and the rate of reaction when [A] = 0.050 M and [B] = 0.100 M. Experiment [A] (M)[B] (M)Initial Rate (M/s) 1 0.1 4.0x10 -5 20.10.24.0x10 -5 30.20.116.0x10 -5

22. Potential Energy Diagrams Molecules need a minimum amount of energy for a reaction to take place.  Activation energy (E a ) – the minimum amount of energy that the reacting species must possess to undergo a specific reaction Activated complex - a short-lived molecule formed when reactants collide; it can return to reactants or form products.  Formation depends on the activation energy & the correct geometry (orientation)

23. Potential Energy Diagram

24. Potential Energy Diagrams

26. Catalyzed Pathway Catalysts lower activation energy!!!

27. Reaction Mechanisms Mechanism – how reactants are converted to products at the molecular level Most reactions DO NOT occur in a single step! They occur as a series of elementary steps (a single step in a reaction).

28. Rate Determining Step Rate determining step – the slowest step in a reaction COCl 2 (g)  COCl (g) + Cl (g)fast Cl (g) + COCl 2 (g)  COCl (g) + Cl 2 (g) slow 2 COCl (g)  2 CO (g) + 2 Cl (g)fast 2 Cl (g)  Cl 2 (g)fast

29. Getting the Overall Reaction COCl 2 (g)  COCl (g) + Cl (g)fast Cl (g) + COCl 2 (g)  COCl (g) + Cl 2 (g) slow 2 COCl (g)  2 CO (g) + 2 Cl (g)fast 2 Cl (g)  Cl 2 (g)fast 2 COCl 2 (g)  2 Cl 2 (g) + 2 CO (g) Adding elementary steps gives the net (or overall) reaction!

30. Intermediates Intermediates are produced in one elementary step but reacted in another NO (g) + O 3 (g)  NO 2 (g) + O 2 (g) NO 2 (g) + O (g)  NO (g) + O 2 (g) O 3 (g) + O (g)  2 O 2 (g)

31. Catalysts Catalyst – a reactant in an elementary step but unchanged at the end of the reaction  A substance that speeds up the reaction but is not permanently changed by the reaction  Both an original reactant and a final product NO (g) + O 3 (g)  NO 2 (g) + O 2 (g) NO 2 (g) + O (g)  NO (g) + O 2 (g) O 3 (g) + O (g)  2 O 2 (g)

32. Example Cl 2 (g)  2 Cl (g) Fast Cl (g) + CHCl 3 (g)  CCl 3 (g) + HCl (g) Slow CCl 3 (g) + Cl (g)  CCl 4 (g)Fast Identify:  The rate determining step  The overall (net) reaction  The identity of any intermediates  The identity of any catalysts

33. Example H 2 O 2 (aq) + I 1- (aq)  H 2 O(l) + IO 1- (aq) Slow H 2 O 2 (aq) + IO 1- (aq)  H 2 O(l) + O 2 (g) + I 1- (aq) Fast Identify:  The rate determining step  The overall (net) reaction  The identity of any intermediates  The identity of any catalysts

34. Example O 3 (g) + Cl (g)  O 2 (g) + ClO (g) Slow ClO (g) + O (g)  Cl (g) + O 2 (g) Fast Identify:  The rate determining step  The overall (net) reaction  The identity of any intermediates  The identity of any catalysts