1. Chemical Kinetics What do we know about chemical reactions?
The area of chemistry that concerns reaction rates and reaction mechanisms.

2. Table Concentrations of Reactant and Products as a Function of Time for the Reaction 2NO2(g) NO + O2(g)

3. Figure Starting with a Flask of Nitrogen Dioxide at 300 degrees C, Concentrations of Nitrogen Dioxide, Nitric Oxide, and Oxygen are Plotted versus Time

4. Figure 12.2 Representation of the reaction of: 2NO2 (g) 2NO(g) + O2 (g)

5. Reaction Rate
The change in concentration of a reactant or product per unit of time

6. Reaction Rate - What is the average rate at which the
concentration of NO2 changes over the first
50 seconds of the reaction????
Change in [NO2]
Δ [NO2] = -
Time elapsed
Δ t
[.0079]t=50 – [.0100]t-0
[NO2]t=50 – [NO2]t-0 = =
50 s – 0 s
50 s – 0 s
= 4.2 x 10-5 mol/L x s

7. Table 12.2 Average Rate (in mol/L-s) of Decomposition of Nitrogen Dioxide as a Function of Time

8. Reaction Rates: 1. Can measure disappearance of reactants
2NO2(g)  2NO(g) + O2(g)
Reaction Rates:
1. Can measure
disappearance of
reactants
2. Can measure
appearance of
products
3. Are proportional
stoichiometrically

9. Reaction Rates: 4. Are equal to the slope tangent to that point
2NO2(g)  2NO(g) + O2(g)
Reaction Rates:
4. Are equal to the
slope tangent to
that point
5. Change as the
reaction proceeds,
if the rate is
dependent upon
concentration
[NO2] t

10. Rate Laws
Differential rate laws express (reveal) the relationship between the concentration of reactants and the rate of the reaction.
The differential rate law is usually just called “the rate law.”
Integrated rate laws express (reveal) the relationship between concentration of reactants and time

11. Writing a (differential) Rate Law
Problem - Write the rate law, determine the value of the rate constant, k, and the overall order for the following reaction:
2 NO(g) + Cl2(g)  2 NOCl(g)
Experiment
[NO]
(mol/L)
[Cl2]
Rate
Mol/L·s 1
0.250
1.43 x 10-6 2
0.500
5.72 x 10-6 3
2.86 x 10-6 4
11.4 x 10-6

12. Writing a Rate Law
Part 1 – Determine the values for the exponents in the rate law:
R = k[NO]x[Cl2]y
Experiment
[NO]
(mol/L)
[Cl2]
Rate
Mol/L·s 1
0.250
1.43 x 10-6 2
0.500
5.72 x 10-6 3
2.86 x 10-6 4
1.14 x 10-5
In experiment 1 and 2, [Cl2] is constant while [NO] doubles.
The rate quadruples, so the reaction is second order with respect to [NO]
 R = k[NO]2[Cl2]y

13. Writing a Rate Law
Part 1 – Determine the values for the exponents in the rate law:
R = k[NO]2[Cl2]y
Experiment
[NO]
(mol/L)
[Cl2]
Rate
Mol/L·s 1
0.250
1.43 x 10-6 2
0.500
5.72 x 10-6 3
2.86 x 10-6 4
1.14 x 10-5
In experiment 2 and 4, [NO] is constant while [Cl2] doubles.
The rate doubles, so the reaction is first order with respect to [Cl2]
 R = k[NO]2[Cl2]

14. Writing a Rate Law
Part 2 – Determine the value for k, the rate constant, by using any set of experimental data:
R = k[NO]2[Cl2]
Experiment
[NO]
(mol/L)
[Cl2]
Rate
Mol/L·s 1
0.250
1.43 x 10-6

15. Writing a Rate Law
Part 3 – Determine the overall order for the reaction.
R = k[NO]2[Cl2] 2 + 1
= 3
 The reaction is 3rd order
Overall order is the sum of the exponents, or orders, of the reactants

16. Order =3. Units are L2mol-2s-1 Order =5. Units are L4 mol-4s-1
What is the overall order of each of the following rate
laws, and what are the units of the rate constant, K,
in each of these rate laws? (Time is in seconds)
Rate= k[A] [B] [C]
Rate= k[X]2 [Y]3
Rate= k[M]2 [N]
Rate= k
(e)Rate= k [R]
Order =3. Units are L2mol-2s-1
Order =5. Units are L4 mol-4s-1
Order =3. Units are L2 mol-2s-1
Order =0 Units are mol L-1s-1
Order =1 Units are s-1

17. Analyze: We are given three boxes containing
Solution:
Analyze: We are given three boxes containing
different numbers of spheres representing mixtures containing different reactant concentrations. We are asked to use the given rate law and the compositions of the boxes to rank the mixtures in order of increasing reaction rates.
Plan: Because all three boxes have the same
volume, we can put the number of spheres of each kind into the rate law and calculate the rate of each box.
Solve: next slide

18. Box 1 contains 5 red spheres and 5 purple
spheres, giving the following rate:
Box 1: Rate = k(5)(5)2 = 125k
Box 2 contains 7 red spheres and 3 purple
Spheres:
Box 2: Rate = k(7)(3)2 = 63k
Box 3 contains 3 red spheres and 7 purple spheres:
Box 3: Rate = k(3)(7)2 = 147k

19. The slowest rate is 63k (box 2), and the
Highest is 147k (box 3).
Thus, the rates vary in the order 2<1<3
How about if the rate = k[A][B]
What would the rank of the mixtures in
order of increasing rank be now? Eh?
2=3<1

20. Order =3. Units are L2mol-2s-1 Order =5. Units are L4 mol-4s-1
What is the overall order of each of the following rate
laws, and what are the units of the rate constant, K,
in each of these rate laws? (Time is in seconds)
Rate= k[A] [B] [C]
Rate= k[X]2 [Y]3
Rate= k[M]2 [N]
Rate= k
(e)Rate= k [R]
Order =3. Units are L2mol-2s-1
Order =5. Units are L4 mol-4s-1
Order =3. Units are L2 mol-2s-1
Order =0 Units are mol L-1s-1
Order =1 Units are s-1

21. Rate Laws
Differential rate laws express (reveal) the relationship between the concentration of reactants and the rate of the reaction.
The differential rate law is usually just called “the rate law.”
Integrated rate laws express (reveal) the relationship between concentration of reactants and time

22. Determining Order with Concentration vs. Time data
(the Integrated Rate Law)
Zero Order:
First Order:
Second Order:

23. Solving an Integrated Rate Law
Time (s)
[H2O2] (mol/L)
1.00
120
0.91
300
0.78
600
0.59
1200
0.37
1800
0.22
2400
0.13
3000
0.082
3600
0.050
Problem: Find the integrated rate law and the value for the rate constant, k
A graphing calculator with linear regression analysis greatly simplifies this process!!

24. Time vs. [H2O2] Regression results: y = ax + b a = -2.64 x 10-4
Time (s)
[H2O2]
1.00
120
0.91
300
0.78
600
0.59
1200
0.37
1800
0.22
2400
0.13
3000
0.082
3600
0.050
Regression results:
y = ax + b
a = x 10-4
b = 0.841
r2 =
r =

25. Time vs. ln[H2O2] Regression results: y = ax + b a = -8.35 x 10-4
Time (s)
ln[H2O2]
120
300
600
1200
1800
-1.514
2400
-2.04
3000
-2.501
3600
-2.996
Regression results:
y = ax + b
a = x 10-4
b = -.005
r2 =
r =

26. Time vs. 1/[H2O2] Regression results: y = ax + b a = 0.00460
Time (s)
1/[H2O2]
1.00
120
1.0989
300
1.2821
600
1.6949
1200
2.7027
1800
4.5455
2400
7.6923
3000
12.195
3600
20.000
Regression results:
y = ax + b
a =
b =
r2 =
r =

27. And the winner is… Time vs. ln[H2O2]
1. As a result, the reaction is 1st order
2. The (differential) rate law is:
3. The integrated rate law is:
4. But…what is the rate constant, k ?

28. Finding the Rate Constant, k
Method #1: Calculate the slope from the
Time vs. ln[H2O2] table.
Time (s)
ln[H2O2]
120
300
600
1200
1800
-1.514
2400
-2.04
3000
-2.501
3600
-2.996
Now remember:
 k = -slope
k = 8.32 x 10-4s-1

29. Finding the Rate Constant, k
Method #2: Obtain k from the linear regresssion analysis.
Regression results:
y = ax + b
a = x 10-4
b = -.005
r2 =
r =
Now remember:
 k = -slope
k = 8.35 x 10-4s-1

30. Half Life t1/2
Half Life: is defined as the time required for a reactant to reach half its original concentration.
0.693
t1/2 = k

31. Half Life t1/2 Try this: A certain first-order reaction
has a half-life of 20.0 minutes.
A. Calculate the rate constant for this
reaction.
K = 3.47 x 10-2 min-1
What am I NOT dependent on???

32. Rate Laws Summary Rate = k Rate = k[A] Rate = k[A]2 [A] = -kt + [A]0
Zero Order
First Order
Second Order
Rate Law
Rate = k
Rate = k[A]
Rate = k[A]2
Integrated Rate Law
[A] = -kt + [A]0
ln[A] = -kt + ln[A]0
Plot the produces a straight line
[A] versus t
ln[A] versus t
Relationship of rate constant to slope of straight line
Slope = -k
Slope = k
Half-Life

33. Reaction Mechanism
The reaction mechanism is the series of elementary steps by which a chemical reaction occurs.
The sum of the elementary steps must give the overall balanced equation for the reaction
The mechanism must agree with the experimentally determined rate law

34. Rate-Determining Step
In a multi-step reaction, the slowest step is the rate-determining step. It therefore determines the rate of the reaction.
The experimental rate law must agree with the rate-determining step

35. Identifying the Rate-Determining Step
For the reaction:
2H2(g) + 2NO(g)  N2(g) + 2H2O(g)
The experimental rate law is:
R = k[NO]2[H2]
Which step in the reaction mechanism is the rate-determining (slowest) step?
Step #1 H2(g) + 2NO(g)  N2O(g) + H2O(g)
Step #2 N2O(g) + H2(g)  N2(g) + H2O(g)
Step #1 agrees with the experimental rate law

36. Table 12.7 Examples of Elementary Steps
The molecularity of a process tells how many molecules are involved in the process.

37. Figure 12.9 A Molecular Representation of the Elementary Steps in the Reaction of NO2 and CO

38. NO2 (g) + CO (g)  NO (g) + CO2 (g)
The rate law for this reaction is found experimentally to be
Rate = k [NO2]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.

39. A proposed mechanism for this reaction is
Rate = k [NO2]2
A proposed mechanism for this reaction is
Step 1: NO2 + NO2  NO3 + NO (slow)
Step 2: NO3 + CO  NO2 + CO2 (fast)
NO2 (g) + CO (g)  NO (g) + CO2 (g)

40. Identifying Intermediates
For the reaction:
2H2(g) + 2NO(g)  N2(g) + 2H2O(g)
Which species in the reaction mechanism are intermediates (do not show up in the final, balanced equation?)
Step #1 H2(g) + 2NO(g)  N2O(g) + H2O(g)
Step #2 N2O(g) + H2(g)  N2(g) + H2O(g)
2H2(g) + 2NO(g)  N2(g) + 2H2O(g)
 N2O(g) is an intermediate

41. 2NO (g) + Br2(g)  2 NOBr(g)
Challenge Time!
2NO (g) + Br2(g)  2 NOBr(g)
Rate = k [NO]2 [Br2]
A proposed mechanism for this reaction is:
Step 1: NO(g) + Br2(g) NOBr2(g) (fast)
Step 2: NOBr2(g) + NO(g) 2NOBr(g) (slow) k1
k-1 k2
dddddddddd
dddddddddd
dddddddddd

42. Intermediates are intrinsically UNSTABLE
Rate = k [NO]2 [Br2] k1
Step 1: NO(g) + Br2(g) NOBr2(g) (fast)
Step 2: NOBr2(g) + NO(g) 2NOBr(g) (slow)
k-1 k2
Dynamic equilibrium gets reached. So….
K1[NO][Br2] = k-1[NOBr2]
FINALLY

43. Dynamic equilibrium gets reached. So…. K1[NO][Br2] = k-1[NOBr2]
Rate = k [NO]2 [Br2]
Dynamic equilibrium gets reached. So….
K1[NO][Br2] = k-1[NOBr2]
Solving for [NOBr2] k1
[NOBr2] = [NO][Br2]
k-1
Sub into RDS
Step 2: NOBr2(g) + NO(g) 2NOBr(g) (slow)
Step 2: [NO][Br2] + NO(g) 2NOBr(g) (slow)

44. Let’s check our understanding of Integrated Rate Laws & Half-life and Reaction Mechanisms
Please begin (at your workstations) the
following Problems: pg. 571: 49-52

45. A Model For Chemical Kinetics

46. Collision Model Key Idea: Molecules must collide to react.
However, only a small fraction of collisions produces a reaction. Why?

47. Collision Model
Collisions must have sufficient energy to produce the reaction (must equal or exceed the activation energy). 1.
Colliding particles must be correctly oriented to one another in order to produce a reaction. 2.

48. Factors Affecting Rate
Increasing temperature always increases the rate of a reaction.
Particles collide more frequently
Particles collide more energetically
Double the temp…effective collisions
increase exponentially!
Increasing surface area increases the rate of a reaction
Increasing Concentration USUALLY increases the rate of a reaction
Presence of Catalysts, which lower the activation energy by providing alternate pathways

49. Endothermic Reactions

50. Exothermic Reactions

51. The Arrhenius Equation
k = rate constant at temperature T
A = frequency factor
Ea = activation energy
R = Gas constant, J/K·mol

52. Catalysis
Catalyst: A substance that speeds up a reaction without being consumed
Enzyme: A large molecule (usually a protein) that catalyzes biological reactions.
Homogeneous catalyst: Present in the same phase as the reacting molecules.
Heterogeneous catalyst: Present in a different phase than the reacting molecules.

53. Lowering of Activation Energy by a Catalyst

54. Catalysts Increase the Number of Effective Collisions

55. Heterogeneous Catalysis
Carbon monoxide and nitrogen monoxide adsorbed on a platinum surface
Step #1: Adsorption and activation of the reactants.

56. Heterogeneous Catalysis
Carbon monoxide and nitrogen monoxide arranged prior to reacting
Step #2:
Migration of the adsorbed reactants on the surface.

57. Heterogeneous Catalysis
Carbon dioxide and nitrogen form from previous molecules
Step #3:
Reaction of the adsorbed substances.

58. Heterogeneous Catalysis
Carbon dioxide and nitrogen gases escape (desorb) from the platinum surface
Step #4:
Escape, or desorption, of the products.