1.  The study of reaction rates and steps.  Spontaneous reactions are reactions that will happen - but we can’t tell how fast.  Diamond will spontaneously turn to graphite – eventually.  Reaction mechanism- the steps by which a reaction takes place.

2.  Rate = Conc. of A at t 2 -Conc. of A at t 1 t 2 - t 1  Rate =  [A] D t  Change in concentration per unit time  For this reaction  N 2 + 3H 2 2NH 3

3.  As the reaction progresses the concentration H 2 goes down ConcentrationConcentration Time [H 2 ]

4.  As the reaction progresses the concentration N 2 goes down 1/3 as fast ConcentrationConcentration Time [H 2 ] [N 2 ]

5.  As the reaction progresses the concentration NH 3 goes up. ConcentrationConcentration Time [H 2 ] [N 2 ] [NH 3 ]

6.  Average rates are taken over long intervals  Instantaneous rates are determined by finding the slope of a line tangent to the curve at any given point because the rate can change over time  Derivative.

7.  Average slope method ConcentrationConcentration Time D[H 2 ] DtDt

8.  Instantaneous slope method. ConcentrationConcentration Time  [H 2 ] D t

9.  We can define rate in terms of the disappearance of the reactant or in terms of the rate of appearance of the product.  In our example N 2 + 3H 2 2NH 3 -  [N 2 ] = -3  [H 2 ] = 2  [NH 3 ]  t  t  t

10.  Reactions are reversible.  As products accumulate they can begin to turn back into reactants.  Early on the rate will depend on only the amount of reactants present.  We want to measure the reactants as soon as they are mixed.  This is called the Initial Rate Method.

11.  Two key points 1. The concentration of the products do not appear in the rate law because this is an initial rate. 2. The order must be determined experimentally, and cannot be obtained from the equation Rate Laws

12.  You will find that the rate will only depend on the concentration of the reactants.  Rate = k[NO 2 ] n  This is called a rate law expression.  k is called the rate constant.  n is the order of the reactant -usually a positive integer.

13.  The rate of appearance of O 2 can be said to be Rate' = D [O 2 ] = k'[NO 2 ] D t  Because there are 2 NO 2 for each O 2  Rate = 2 x Rate'  So k[NO 2 ] n = 2 x k'[NO 2 ] n  So k = 2 x k'

14.  Differential Rate law - describes how rate depends on concentration.  Integrated Rate Law - describes how concentration depends on time.  For each type of differential rate law there is an integrated rate law and vice versa.  Rate laws can help us better understand reaction mechanisms.

15.  The first step is to determine the form of the rate law (especially its order).  Must be determined from experimental data.  For this reaction 2 N 2 O 5 (aq) 4NO 2 (aq) + O 2 (g) The reverse reaction won’t play a role

16. [N 2 O 5 ] (mol/L) Time (s) 1.000 0.88200 0.78400 0.69600 0.61800 0.541000 0.481200 0.431400 0.381600 0.341800 0.302000

17.  To find rate we have to find the slope at two points  We will use the tangent method. Then compare the rate at those two concentration values.

18.  Since the rate at twice the concentration is twice as fast the rate law must be Rate = - D [N 2 O 5 ] = k[N 2 O 5 ] 1 = k[N 2 O 5 ] D t  We say this reaction is first order in N 2 O 5  The only way to determine order is to run the experiment.

19.  This method requires that a reaction be run several times.  The initial concentrations of the reactants are varied.  The reaction rate is measured just after the reactants are mixed.  Eliminates the effect of the reverse reaction.

20.  For the reaction BrO 3 - + 5 Br - + 6H + 3Br 2 + 3 H 2 O  The general form of the Rate Law is Rate = k[BrO 3 - ] n [Br - ] m [H + ] p  We use experimental data to determine the values of n, m, and p

21. Initial concentrations (M) Rate (M/s) BrO 3 - Br - H+H+ 0.100.100.108.0 x 10 -4 0.200.100.101.6 x 10 -3 0.200.200.103.2 x 10 -3 0.100.100.203.2 x 10 -3 Now we have to see how the rate changes with concentration

23.  Expresses the reaction concentration as a function of time.  Form of the equation depends on the order of the rate law (differential).  Changes Rate =  [A] n  t  We will only work with n=0, 1, and 2

24.  For the reaction 2N 2 O 5 4NO 2 + O 2  We found the Rate = k[ N 2 O 5 ] 1  If concentration doubles rate doubles.  If we integrate this equation with respect to time we get the Integrated Rate Law ln[N 2 O 5 ] = - kt + ln[N 2 O 5 ] 0  [N 2 O 5 ] 0 is the initial concentration.

25.  General form Rate =  [A] /  t = k[A]  ln[A] = - kt + ln[A] 0  In the form y = mx + b y = ln[A]m = -k x = tb = ln[A] 0  A graph of ln[A] vs time is a straight line.

26.  By getting the straight line you can prove it is first order  Often expressed in a ratio

27.  ln(2) = kt 1/2 The time required to reach half the original concentration. If the reaction is first order [A] = [A] 0 /2 when t = t 1/2

28.  t 1/2 = 0.693/k  The time to reach half the original concentration does not depend on the starting concentration.  An easy way to find k

29.  Rate = -  [A] /  t = k[A] 2  integrated rate law  1/[A] = kt + 1/[A] 0 y= 1/[A] m = k x= t b = 1/[A] 0  A straight line if 1/[A] vs t is graphed  Knowing k and [A] 0 you can calculate [A] at any time t

30.  [A] = [A] 0 /2 at t = t 1/2

31.  Rate = k[A] 0 = k  Rate does not change with concentration.  Integrated [A] = -kt + [A] 0  When [A] = [A] 0 /2 t = t 1/2  t 1/2 = [A] 0 /2k

32.  Most often when reaction happens on a surface because the surface area stays constant.  Also applies to enzyme chemistry.

33. Time ConcentrationConcentration

34. ConcentrationConcentration  A]/  t tt k =  A]

35.  BrO 3 - + 5 Br - + 6H + 3Br 2 + 3 H 2 O  For this reaction we found the rate law to be Rate = k[BrO 3 - ][Br - ][H + ] 2  To investigate this reaction rate we need to control the conditions

36. the reactants are in large excess.  We set up the experiment so that two of the reactants are in large excess. [BrO 3 - ] 0 = 1.0 x 10 -3 M [Br - ] 0 = 1.0 M [H + ] 0 = 1.0 M  As the reaction proceeds [BrO 3 - ] changes noticeably  [Br - ] and [H + ] don’t

37.  This rate law can be rewritten Rate = k[BrO 3 - ][Br - ] 0 [H + ] 0 2 Rate = k[Br - ] 0 [H + ] 0 2 [BrO 3 - ] Rate = k’[BrO 3 - ]  This is called a pseudo first order rate law.  k = k’ [Br - ] 0 [H + ] 0 2

38.  The series of steps that actually occur in a chemical reaction.  Kinetics can tell us something about the mechanism  A balanced equation does not tell us how the reactants become products.

39.  2NO 2 + F 2 2NO 2 F Rate = k[NO 2 ][F 2 ]  The proposed mechanism is  NO 2 + F 2 NO 2 F + F (slow)  F + NO 2 NO 2 F(fast)  F is called an intermediate It is formed then consumed in the reaction

40.  Each of the two reactions is called an elementary step.  The rate for a reaction can be written from its molecularity.  Molecularity is the number of pieces that must come together.

41.  Unimolecular step involves one molecule - Rate is first order.  Bimolecular step - requires two molecules - Rate is second order.  Termolecular step- requires three molecules - Rate is third order.  Termolecular steps are almost never heard of because the chances of four molecules coming into contact at the same time are miniscule.

42.  A products Rate = k[A]  A+A productsRate= k[A] 2  2A productsRate= k[A] 2  A+B productsRate= k[A][B]  A+A+B products Rate= k[A] 2 [B]  2A+B products Rate= k[A] 2 [B]  A+B+C products Rate= k[A][B][C]  Only for the elementary steps

43.  Use the reactions that form them  If the reactions are fast and irreversible - the concentration of the intermediate is based on stoichiometry.  If it is formed by a reversible reaction set the rates equal to each other.

44.  2 NO + O 2 2 NO 2  Mechanism 2 NO N 2 O 2 (fast) N 2 O 2 + O 2 2 NO 2 (slow)  Based on slow step: rate = k 2 [N 2 O 2 ][O 2 ]  Substitute for intermed: k 1 [NO] 2 = k -1 [N 2 O 2 ] rate = k 2 (k 1 / k -1 )[NO] 2 [O 2 ]=k[NO] 2 [O 2 ]

45.  2 IBr I 2 + Br 2 Mechanism  IBrI + Br (fast)  IBr + Br I + Br 2 (slow)  I + II 2 (fast)  Rate = k[IBr][Br] but [Br]= [IBr] since fast reversible reaction occurs, conc are related stoichiometrically.  Rate = k[IBr][IBr] = k[IBr] 2

46.  Molecules must collide to react.  Concentration affects rates because collisions are more likely if more molecules are present.  Must collide hard enough.  Temperature and rate are related.  Only a small number of collisions produce reactions.

47. Potential EnergyPotential Energy Reaction Coordinate Reactants Products

48. Potential EnergyPotential Energy Reaction Coordinate Reactants Products Activation Energy E a

49. Potential EnergyPotential Energy Reaction Coordinate Reactants Products Activated complex

50. Potential EnergyPotential Energy Reaction Coordinate Reactants Products EE }

51. Potential EnergyPotential Energy Reaction Coordinate 2BrNO 2NO + Br Br---NO 2 Transition State

52.  Activation energy - the minimum energy needed to make a reaction happen.  Activated Complex or Transition State - The arrangement of atoms at the top of the energy barrier.

53.  Said that the reaction rate should increase with temperature.  At high temperature more molecules have the energy required to get over the barrier.  The number of collisions with the necessary energy increases exponentially.

54.  Number of collisions with the required energy = z e -E a /RT  z = total collisions  e is Euler’s number (opposite of ln)  E a = activation energy  R = ideal gas constant  T is temperature in Kelvin

55.  Observed rate is less than the number of collisions that have the minimum energy.  Due to molecular orientation  written into equation as p the steric factor.

56. O N Br O N O N O N O N O N O N O N O N O N No Reaction

57. k = zpe -E a /RT = Ae -E a /RT  A is called the frequency factor = zp  ln k = -(E a /R)(1/T) + ln A  Another line !!!!  ln k vs t is a straight line

59.  There is an activation energy for each elementary step.  Activation energy determines k.  k = Ae - (E a /RT)  k determines rate  Slowest step (rate determining) must have the highest activation energy.

60. + This reaction takes place in three steps

61. + EaEa First step is fast Low activation energy

62. Second step is slow High activation energy + EaEa

63. + EaEa Third step is fast Low activation energy

64. Second step is rate determining

65. Intermediates are present

66. Activated Complexes or Transition States

67.  Speed up a reaction without being used up in the reaction.  Enzymes are biological catalysts.  Homogenous Catalysts are in the same phase as the reactants.  Heterogeneous Catalysts are in a different phase than the reactants.

68.  Catalysts allow reactions to proceed by a different mechanism - a new pathway.  New pathway has a lower activation energy.  More molecules will have this activation energy.  Do not change  E

69. Pt surface HHHH HHHH  Hydrogen bonds to surface of metal.  Break H-H bonds

70. Pt surface HHHH C HH C HH

71. HHHH  The double bond breaks and bonds to the catalyst. C HH C HH

72. Pt surface HHHH  The hydrogen atoms bond with the carbon C HH C HH

73. Pt surface H C HH C HH HHH

74.  Chlorofluorocarbons catalyze the decomposition of ozone.  Enzymes regulating the body processes. (Protein catalysts)

75.  Catalysts will speed up a reaction but only to a certain point.  Past a certain point adding more reactants won’t change the rate.  Zero Order

76. Concentration of reactants RateRate Rate increases until the active sites of catalyst are filled. Then rate is independent of concentration