1. Chemistry 100 Chapter 14 - Chemical Kinetics

2. The Connection Between Chemical Reactions and Time  Not all chemical reaction proceed instantaneously!!! 2 A  B combination reaction  Can we quantify the length of time it takes for this (or any) chemical reaction to occur?

3. Practical examples of how long! H 2 (g) + ½ O 2 (g)  H 2 O (l)  Very Slow N 2 O(g)  N 2 (g) + ½ O 2 (g)  SLOW  combustion reactions  fast process  TNT exploding  very fast reaction  Food spoilage  Drug decomposition

4. Chemical Kinetics  Chemical kinetics is concerned with determining the speed or rate at which a reaction occurs.  How is the reaction rate affected by  temperature?  states of reactants?  amount of reactants?  catalyst?  surface area of the reacting species?

5. Example Br 2 (aq) + HCOOH (aq)  2 Br - (aq) + 2 H + (aq) + CO 2 (g)  Define the average rate

6. A Sample  Reaction Br 2 (aq) + HCOOH (aq)  2 Br – (aq) + CO 2 (g) + 2 H + (aq) t / s[Br 2 ] / M 00.0120 500.0101 1000.00846 1500.00710 2000.00596 2500.00500 3000.00420 3500.00353 4000.00296

7. Average Rate Data Avg. Rate (150 s) Avg. Rate (100 s) Avg. Rate (50 s) 3.27 x 10 -5 3.54 x 10 -5 3.80 x 10 -5 2.76 x 10 -5 3.00 x 10 -5 3.28 x 10 -5 2.31 x 10 -5 2.50 x 10 -5 2.72 x 10 -5 1.93 x 10 -5 2.10 x 10 -5 2.28 x 10 -5 1.62 x 10 -5 1.76 x 10 -5 1.92 x 10 -5 1.47 x 10 -5 1.60 x 10 -5 1.34 x 10 -5

8. Instantaneous Rates  It’s best to define an instantaneous ‘speed of reaction’

9. Instantaneous Rate Data Time (s)Rate (M/s) 04.19 x 10 -5 503.52 x 10 -5 1002.95 x 10 -5 1502.48 x 10 -5 2002.08x 10 -5 2501.75 x 10 -5 3001.47 x 10 -5 4001.03 x 10 -5

11. The Rate Constant  The rate constant relates the ‘speed’ of the chemical reaction to the instantaneous reactant concentration. k = constant for constant temperature  The rate of the reaction is dependent on reactant concentration  RATE CONSTANT IS INDEPENDENT OF THE REACTANT CONCENTRATION.

12. Reaction Rates and Reaction Stoichiometry  Look at the reaction O 3 (g) + NO(g)  NO 2 (g) + O 2 (g)

13. Another Example 2 NOCl (g)  2 NO + 1 Cl 2 (g) WHY? 2 moles of NOCl disappear for every 1 mole Cl 2 formed.

14. a A + b B  c C + d D rate = -1 [A] = -1 [B] = +1 [C] = +1 [D] a t b t c t d t  Why do we define our rate in this way?  removes ambiguity  obtain a single rate for the entire equation, not just for the change in a single reactant or product. The General Case

15. The Rate Law  Relates rate of the reaction to the reactant concentrations and rate constant  For a general reaction a A + b B + c C  d D + e E rate = k[A] x [B] y [C] z The exponents (x,y, and z) are called the reactant orders.

16. The Reaction Orders  Determine the superscripts (x, y, and z) for a non-elementary chemical reaction by experimentation.  x + y + z = reaction order e.g. x = 1; y = 1; z = 0 2nd order reaction (x + y + z = 2) x = 0; y = 0; z = 1 (1st order reaction) x = 2; y = 0; z = 0 (2nd order)

17. The Isolation Method of Obtaining Rate Laws a A + b B  c C + d D rate = k[A] x [B] y  Fix the concentration of one reactant (say reactant A).  We then perform a series of experiments to examine how changing the [B] affects the initial reaction rate? rate = (constant) [B] y

18. Isolation Method (cont’d)  Now we fix the concentration of reactant B.  We then perform another series of experiments to examine how changing the [A] affects the initial reaction rate? rate = (constant) [A] x

19. Types of Reactions  The rate law gives us information about how the concentration of the reactant varies with time  How much reactant remains after specified period of time?

20. First Order Reaction A  product rate = -[A]/t = k[A] How does the concentration of the reactant depend on time? k has units of s -1

21. The Half-Life of a First Order Reaction  For a first order reaction, the half-life t 1/2 is calculated as follows.

22. Radioactive Decay  Radioactive Samples decay according to first order kinetics.  This is the half-life of samples containing e.g. 14 C, 239 Pu, 99 Tc.  Example

23. Second Order Reaction A + B  productsRate = k[A][B] A  productsRate = k[A] 2  Reaction 1 is 1 st order in A and B and 2 nd order overall  Reaction 2 is 2 nd order in A

24. The Dependence of Concentration on Time  For a second order process where rate = k[A] 2

25. Half-life for a Second Order Reaction.  [A] at t = t½ = ½ [A] 0

26. A Pseudo-First Order Reaction  Example hydration of methyl iodide CH 3 I(aq) + H 2 O(l)  CH 3 OH(aq) + H + (aq) + I - (aq) Rate = k [CH 3 I] [H 2 O]  Since we carry out the reaction in aqueous solution [H 2 O] >>>> [CH 3 I] / [H 2 O] doesn’t change by a lot

27.  Since the concentration of H 2 O is essentially constant rate = k [CH 3 I] [constant] = k`[CH 3 I] where k` = k [H 2 O]  Pseudo first order since it appears to be first order, but it is actually a second order process.

28. Collision Theory of Kinetics  With few exceptions, the reaction rate increases with increasing temperature.  Chemical reactions take place due to collisions between reactant molecules i.e. rate  number of collisions / unit time A 2 + B 2  product rate = k[A 2 ][B 2 ]

29. The Reaction Profile  How does the energy of the reactants vary during the reaction sequence?

30. The Activation Energy  The minimum amount of energy need for initiation of a chemical reaction is the activation energy (E a ).  Colliding reactant molecules possess kinetic energy > the activation energy or E a.

31. The Activated Complex  The species temporarily formed by the reactant molecules – the activated complex.  A small fraction of molecules usually have the required kinetic energy to get to the transition state The concentration of the activated complex is extremely small.

32. The Arrhenius Equation Arrhenius showed how the rate constant depended on temperature. A is called the frequency factor – an estimate of the number of reactive collisions in the system E a is the activation energy

33.  Reaction possesses a large activation energy  small rate constant Slow reaction!!  Measure k at several different temperatures Activation Energies and the Arrhenius Equation R = 8.314 J/(K mole) T in Kelvin units!!!

34. Arrhenius Equation (cont’d)  The Arrhenius equation is best suited for studying reactions between simple species (atoms, diatomic molecules).  The orientation of the reactants (how they collide) becomes very important when the species get bigger.

35. Catalysts  So far, we have considered one way of speeding up a reaction increasing T usually increases k.  Another way is by the use of a catalyst.  A catalyst - a substance that speeds up the rate of the reaction without being consumed in the overall reaction.

36.  look at the following two reactions A+B  Crate constant k A+B  C rate constant with catalyst is k c  NOTE: RATE WITH CATALYST > RATE WITHOUT CATALYST

37. Types of Catalyst  We will briefly discuss three types of catalysts. The type of catalyst depends on the phase of the catalyst and the reacting species. Homogeneous Heterogeneous Enzyme

38. Homogeneous Catalysis  The catalyst and the reactants are in the same phase  e.g. Oxidation of SO 2 (g) to SO 3 (g) 2 SO 2 (g) + O 2 (g)  2 SO 3 (g)SLOW  Presence of NO (g), the following occurs. NO (g) + O 2 (g)  NO 2 (g) NO 2 (g) + 2 SO 2 (g)  2 SO 3 (g) + NO (g) FAST

39.  SO 3 (g) is a potent acid rain gas H 2 O (l) + SO 3 (g)  H 2 SO 4 (aq)  Note the rate of NO 2 (g) oxidizing SO 2 (g) to SO 3 (g) is faster than the direct oxidation.  NO x (g) are produced from burning fossil fuels such as gasoline, coal, oil!!

40. Heterogeneous Catalysis  The catalyst and the reactants are in different phases adsorption the binding of molecules to the surface to a surface.  Adsorption on the surface occurs on active sites.  An active site is a place where reacting molecules are adsorbed and physically bond to the metal surface.

41.  The hydrogenation of ethene (C 2 H 4 (g)) to ethane C 2 H 4 (g) + H 2 (g)  C 2 H 6 (g)  Reaction is energetically favourable  r H = -136.98 kJ/mole of ethane.  With a finely divided metal such as Ni (s), Pt (s), or Pd(s), the reaction goes very quickly.

42. Common Heterogeneous Catalysts  Two other important heterogeneous catalysis processes petroleum cracking (refining crude oil) catalytic converters  very efficient in reducing exhaust emission when hot; cold is another story!

43. Enzyme Catalysis  Enzymes - proteins (M > 10000 g/mol)  High degree of specificity (i.e., they will react with one substance and one substance primarily  Living cell > 3000 different enzymes

44. The Lock and Key Hypothesis  Enzymes are folded into fixed configurations.  According to Fischer, active site is rigid.  The substrate’s molecular structure exactly fits the “lock” (hence, the “key”).

45. Simplified Model for Enzyme Catalysis  E  enzyme; S  substrate; P  product E + S  ES ESP + E rate = k [ES]  The reaction rate depends directly on the concentration of the substrate.

46. Rate Laws for Multistep Processes  Chemical reactions generally proceed via a large number of elementary steps - the reaction mechanism  The experimentally established rate law must reflect the reaction rate of the slowest elementary step  the rate determining step (rds)

47. What do we mean by an rds?  A commuter goes through a two step process to get to work in Halifax. (1)highway  MacKay Bridge Toll booth (2)toll booth  downtown

48. MacKay Bridge Toll Booth  Overall reaction highway  downtown  Situation 1.  highway clogged, toll booth is fast.  Situation 2.  fast highway, clogged toll booth.

49. The Rate Determining Step (rds)  Situation 1 - clogged highway is the slowest step in the commuting process (rds).  Situation 2 - the clogged toll-booth is the slowest step in the commuting process (the rds).  Speed of overall process (highway  downtown) depends which step is slowest! Which is the rate-determining step.

50. Elementary steps and the Molecularity  Any chemical reaction occurs via a sequence of elementary steps.  Kinetics of the elementary step only depends on the number of reactant molecules in that step! Molecularity  the number of reactant molecules that participate in elementary steps

51. The Kinetics of Elementary Steps  Classes of elementary steps A bimolecular step A unimolecular step

52.  For the step A termolecular (three molecule) step.