1. The “Speed” of the Reaction Or Reaction Rates
Chemical Kinetics
The “Speed” of the Reaction Or
Reaction Rates

2. Graphical Analysis of Rates
Reaction Kinetics
Reactants
Products
Rate of Change
Initial rates Method
Reaction Order
Average
Instantaneous
Integrated Rate
Forms of Rate Law
Rate Law
Graphical Analysis of Rates
Half-Life

3. Reactants  What happens to the reactants in a reaction?
Map
What happens to the reactants in a reaction?
If we measure the concentration of reactants as a reaction proceeds, what would the graph look like? Or
How does the concentration vary with time?
Is it linear?
Exponential?
Random?
Do all reactions only move forward?
Assume for now there is no reverse reaction
Or the reverse reaction proceeds so slowly, we are willing to ignore it
graph

4.  Products Where do the products come from?
Map
 Products
Where do the products come from?
If we measure the concentration of the products from the first second the reaction starts,
how would the concentration vary with time?
graph

5. Rate of Change [A] = the molarity of A In the reaction A  B
Map
Rate of Change
In the reaction A  B
How does the [A] vary with time? Or
What is the rate of the reaction?
Rate = [A]time2 – [A]time Time 2 – Time 1
[A] = the molarity of A

6. Rate of Change The symbol delta, , means “the change in”
So the reaction rate can be written
Rate = [A]t2 – [A]t1 or time 2 – time 1
Rate = [A] t
graph

7. Average Rate of Change The concentration varies as time goes on.
Because we use up reactants
We can calculate an average rate of product consumption
over a period of time.
Rate is like velocity of a reaction.
the rate of change of meters vs rate of change of [A]
Instead of meters per second, it is concentration per second.
To calculate speed/velocity, divide the distance traveled by the time it took Meters = m2 – m1 =  m t2 – t1 t sec
graph

8. Average Rate of Change
What is the change in concentration? (how far did it go? If you are calculating speed)
time2 – = [A]
How much time has elapsed?
time2 – = [A] t2 – t t
graph

9. Instantaneous Rate
An average rate describes what reaction rate over a time, but does not tell us the rate at any particular moment.
The rate at any moment is the instantaneous rate

10. Map
Instantaneous Rate
If we take the average rate over a period of time and continuously make the time period smaller
When the time period is infinitesimally small, you approach the instantaneous rate
Graphically, it is the slope of the tangent line at the instant.
That’s why graphing programs have that tangent line function! Rates are important in bio, physics and chem.

11. (differential) Rate Law
Map
(differential) Rate Law
Expresses how rate depends on concentration.
Rate = - [Reactant] = k [reactant]n t
k is the rate constant
The bigger the k value, the faster the reaction
The smaller the k value, ….?
n = the order of the reaction and must be determined experimentally.

12. Reaction Rates Reaction rates are considered positive
Rateinstantaneous = kinstantaneous = - slope of tangent line
Rateaverage = k average = - ([A]2 - [A]1) t2-t1
So the rate constant, k, is always negative
Rate = - [Product] = k [Product]n t
Assuming no reverse reaction!

13. Back to:
Reactant/product
Average Rate
Reaction Rate
Instantaneous Rate
Product Formation
Average Rate of Change
From 0 to 300 s =
0.01 – = M/s
300 s

14. 2NO2  2NO + O2 SPECTROSCOPY Let’s consider the above reaction
How can we measure the rate?
What data do we need?
Measure the time
Measure the concentration
We will take advantage of color in our lab
If we are measuring light, we are doing…..
SPECTROSCOPY

15. Spectrometer One frequency Of light Many frequencies of light
Source
Monochrometer,
LED Or
Filter
Sample
Detector
The sample absorbs the light. The detector determines how much.

16. Beer’s Law
Beer’s law states that the amount of light absorbed depends on:
The material
molar absorbtivity (physical property)
How much is there?
molarity
And how big the sample holder
The light spends more “time” in
contact with a longer sample

17. Spectrometer One frequency Of light Many frequencies of light
Source
Monochrometer,
LED Or
Filter
Sample
Detector
The sample absorbs the light. The detector determines how much.

18. Spectroscopy
Assume the concentration is directly proportional to absorbance of light
The more stuff there is that absorbs the light
the less light that goes through …. or
More light is absorbed
Beer’s Law
a = e l c = k M
a = absorbtivity
e = molar absorbtivity (physical property)
l = length of light path
c = molarity or the solution
Molarity and absorbtivity
Are directly proportional

19. in the first 50 secs and the last 50 secs
2NO2  2NO + O2
Compare the [NO2]
in the first 50 secs and the last 50 secs
Time
[NO2]
[NO]
[O2]
0.0100 50
0.0079
0.021
0.0011
100
0.0065
0.0035
0.0018
150
0.0055
0.0045
0.0023
200
0.0048
0.0052
0.0026
250
0.0043
0.0057
0.0029
300
0.0038
0.0062
0.0031
350
0.0034
0.0066
0.0033
400
0.0069
Why does the rate slow down?

20. Formation of Products 2NO2  2NO + O2 For every two NO2 consumed
Rate of Consumption NO2 = Rate Formation NO
Rate = k[NO2 ] = - k[NO]
Because
For every two NO2 consumed
two NO formed

21. Formation of Products 2NO2  2NO + O2 For every two NO2 consumed
Rate of Consumption NO2 = 2 x Rate Formation O2
Rate = k[NO2 ] = k/2 [O2]
Because
For every two NO2 consumed
one O2 formed

22. Compare the Instantaneous Rates
At any moment in time
[NO2] = - [NO] = [O2] t t t Or
k [NO2] = - k [NO] = - k/2 [O2]
graph

23. Form of the Rate Law For aA + bB  cC +dD Rate = k [A]n [B]m
Where k is the rate constant
n = order of reactant A
m = order of reactant B
n and m must be determined experimentally
n +m = order of the reaction

24. Experimental Order the order in the integrated rate law
Rate = - [Reactant] = k [Reactant]n t
n = 0, zero order
n = 1, first order
n = 2, second order
Determine order

25. Order of Reaction A + B → C Rate = k[A]n [B]m
(n + m) = order of the reaction
= 1 unimolecular
=2 bimolecular
=3 trimolecular
This means how many particles are involved in the rate determining step

26. Method of Initial Rates
A series of experiments are run to determine the order of a reactant.
The reaction rate at the beginning of the reaction and the concentration are measured
These are evaluated to determine the order of each reactant and the overall reaction order

27. If you plot the concentration versus time of [N2O5],
you can determine the rate at 0.90M and 0.45M.
What is the rate law for this reaction?
Rate = k [N2O5]n n = the order. It is determined experimentally.

28. 2N2O5(soln)  4NO2(soln) + O2(g)
At 45C, O2 bubbles out of solution, so only the forward reaction occurs.
Data
[N2O5] Rate ( mol/l • s) M 5.4 x M 2.7 x 10-4
The concentration is halved, so the rate is halved

29. 2N2O5(soln)  4NO2(soln) + O2(g)
Rate = k [N2O5]n
5.4 x 10-4 = k [0.90]n x = k [0 45]n after algebra
2 = (2)n
n = 1 which is determined by the experiment
Rate = k [N2O5]1

30. Method of Initial Rates
Map
Method of Initial Rates
Measure the rate of reaction as close to t = 0 as you can get.
This is the initial rate.
Vary the concentration
Compare the initial rates.

31. NH4+ + NO2-  N2 + 2H2O Rate = k[NH4+1]n [NO2-1]m
How can we determine n and m? (order)
Run a series of reactions under identical conditions. Varying only the concentration of one reactant.
Compare the results and determine the order of each reactant
Order

32. NH4+ + NO2-  N2 + 2H2O Experiment [NH4]+ Initial [NO2]- Initial Rate
Mol/L ·s 1
0.001M M
1.35 x 10-7 2
0.010 M
2.70 x 10-7 3
0.002M
0.010M
5.40 x 10-7

33. NH4+ + NO2-  N2 + 2H2O Compare one reaction to the next
1.35 x 10-7 = k(.001)n(0.050)m
2.70 x 10-7 = k (0.001)n(0.010)m
Exp
[NH4]+
Initial
[NO2]-
Initial Rate
Mol/L ·s 1
0.001M M
1.35 x 10-7 2
0.010 M
2.70 x 10-7 3
0.002M
0.010M
5.40 x 10-7
Form

34. 1.35 x = k(0.001)n(0.0050)m
2.70 x k (0.001)n(0.010)m
1.35 x = (0.0050)m
2.70 x (0.010)m
1/ = (1/2)m
m = 1
In order to find n, we can do the same type of math
with the second set of reactions

35. NH4+ + NO2-  N2 + 2H2O Compare one reaction to the next
2.70 x 10-7 = k (0.001)n(0.010)m
5.40 x 10-7 = k(.002)n(0.010)m
Exp
[NH4]+
Initial
[NO2]-
Initial Rate
Mol/L ·s 1
0.001M M
1.35 x 10-7 2
0.010 M
2.70 x 10-7 3
0.002M
0.010M
5.40 x 10-7

36. 2.70 x = k (0.001)n(0.010)m
5.40 x k(.002)n(0.010)m
= (0.5)n
n = 1
n + m = order of the reaction
= 2 or second order
Form

37. Review Method of initial rate In the form Rate = k[A]n[B]m
Where k is the rate constant
n, m = the order of the reactant
The order is determined experimentally
Rate law is important so we can gain an insight into the individual steps of the reaction

38. The Integrated Rate Law
Map
The Integrated Rate Law
Expresses how concentrations depend on time
Depends on the order of the reaction
Remember
Rate = k[A]n[B]m
Order = n + m
Integrated Rate law takes the form by “integrating” the rate function. (calculus used to determine)
The value of n and m change the order of the reaction
The form of the integrated rate depends on the value of n
You get a different equation for zero, first and second order equations.

39. Reaction Order
Order of the reaction determines or affects our calculations.
Zero order indicates the use of a catalyst or enzyme. The surface area of catalyst is the rate determining factor.
First or second order is more typical (of college problems)

40. Integrated Law - Zero Order
Rate = - [A] = k t Set up the differential equation d[A] = -kt Integral of 1 with respect to A is [A]

41. Integrated Rate Law – First Order
Map
Integrated Rate Law – First Order
Rate = [A] = k [A] n t
If n = 1, this is a first order reaction. If we “integrate” this equation we get a new form.
Ln[A] = -kt + ln[A0]
where A0 is the initial concentration

42. Why? If Rate = - [A] = k [A] 1 t
Then you set up the differential equation:
d[A] = -kdt
[A]
Integral of 1/[A] with respect to [A] is the ln[A].

43. Integrated Rate Law ln[A] = -kt + ln[A]0
The equation shows the [A] depends on time
If you know k and A0, you can calculate the concentration at any time.
Is in the form y = mx +b
Y = ln[A] m = -k b = ln[A]0
Can be rewritten ln( [A]0/[A] ) = kt
This equation is only good for first order reactions!

44. First Order Reaction [N2O5] Time (s) 0.1000 0 0.0707 50 0.0500 100
Ln[N2O5]
Time (s)

45. Zero First Second Rate Law Integrated Rate Law Line Slope = Half-life
Rate = K[A]0
Rate = K[A]1
Rate = K[A]2
Integrated Rate Law
[A] = -kt + [A]0
Ln[A] = -kt +ln[A]0
1 = kt
[A] [A]0
Line
[A] vs t
ln[A] vs t
vs t
[A]
Slope =
- k k
Half-life
t1/2 = [A]0 2k
t1/2 =
T1/2 =
k[A]0
Graph
Data
Map

46. Given the Reaction 2C4H6  C8H12
And the data
[C4H6] mol/L Time (± 1 s)

47. Equations
2C4H6  C8H12

48. Graphical Analysis
Map
Data
___1___
[C4H6]
Ln [C4H6]

49. Experimental Derivation of Reaction Order
Arrange data in the form
1/[A] or ln [A] or [A]
Plot the data vs time
Choose the straight line
y = mx b
Determine the k value from the slope
Graphical rate laws
1/[A] = kt + b → 2nd
ln[A] = kt + b → 1st
[A] = kt + b → zero

50. Half-life The time it takes 1/2 of the reactant to be consumed
This can be determined
Graphically
Calculate from the integrated rate law

51. Half-Life Graphical Determination

52. Half-Life Algebraic Determination
Map
Zero
First
Second
Half-life
t1/2 = [A]0 2k
t1/2 = k
T1/2 =
k[A]0
Equations are derived from the Integrated Rate Laws.

53. As temperature increases, the rate increases.
Temperature and Rate
Most reactions speed up as temperature increases (EX. food spoils faster when not refrigerated.)
As temperature increases, the rate increases.
Typically the rate doubles for every 10ºC increase in temperature.

54. Why? Temperature and Rate
Since the rate law has no temperature term in it, the rate constant must depend on temperature.
Why?
The temperature effect is quite dramatic.

55. Temperature and Rate
The Collision Model
Observations: rates of reactions are affected by concentration and temperature.
In order for a reaction to occur the reactant molecules must collide in the correct orientation and with enough energy to form products.

56. Temperature and Rate
The Collision Model
The more molecules present, the greater the probability of effective collisions and the faster the rate.
The higher the temperature, the more energy available to the molecules and the faster the rate.

57. Temperature and Rate
The Collision Model
Complication:
Not all collisions lead to products. In fact, only a small fraction of collisions lead to product.

58. Temperature and Rate
Activation Energy

59. Temperature and Rate
Activation Energy
Arrhenius: molecules must possess a minimum amount of energy to react.
In order to form products, bonds must be broken in the reactants.
Bond breakage requires energy.
Activation energy, Ea, is the minimum energy required to initiate a chemical reaction.

60. Temperature and Rate
Activation Energy

61. Temperature and Rate Activation Energy
Consider the rearrangement of acetonitrile:
In H3C-NC, the C-NC bond bends until the C-N bond breaks and the NC portion is perpendicular to the H3C portion. This structure is called the activated complex or transition state.
The energy required for the twisting and breaking of the reactant bonds is the activation energy, Ea.
Once the C-N bond is broken, the NC portion can continue to rotate forming a C-CN bond.

62. Temperature and Rate
Activation Energy

63. Temperature and Rate Key Points for Activation Energy
The change in energy, ∆E, for the reaction is the difference in energy between reactants (CH3NC) and products (CH3CN).
The activation energy is the difference in energy between the reactants and the transition state.
The reaction rate depends on Ea. The lower the Ea, the faster the reaction.
Notice that if a forward reaction is exothermic (CH3NC  CH3CN), then the reverse reaction is endothermic (CH3CN  CH3NC).

64. Temperature and Rate Activation Energy
Re-consider the reaction between Cl and NOCl:
If the Cl effectively collides with the Cl end of NOCl, then the products are Cl2 and NO.
If the Cl collided with the O of NOCl, then no products are formed.
We need to quantify this effect.

65. Temperature and Rate
As the temperature increases, more molecules have sufficient energy to have effective collisions and react.
Lower temp
Minimum energy needed for reaction, Ea
Higher temp
Fraction of molecules
Kinetic energy

66. Temperature and Rate The Arrhenius Equation
Arrhenius discovered most reaction rate data obeyed the equation:
k is the rate constant, Ea is the activation energy, R is the gas constant (8.314 J/K-mol) and T is the temperature in K.
A is called the frequency factor. It is a measure of the probability of a favorable collision.
Both A and Ea are specific to a given reaction.

67. Temperature and Rate The Arrhenius Equation
If you have a lot of data, you can determine Ea and A graphically by rearranging the Arrhenius equation:
If you have only two data points, then you can use

68. Catalysis A catalyst changes the rate of a chemical reaction.
There are two types of catalysts: homogeneous and heterogeneous.
Homogeneous Catalysis
The catalyst and reaction are in one phase.
Hydrogen peroxide decomposes very slowly: 2H2O2(aq)  2H2O(l) + O2(g).

69. Catalysis Homogeneous Catalysis
2H2O2(aq)  2H2O(l) + O2(g).
In the presence of the bromide ion, the decomposition occurs rapidly:
2Br-(aq) + H2O2(aq) + 2H+(aq)  Br2(aq) + 2H2O(l).
Br2(aq) + H2O2(aq)  2Br-(aq) + 2H+(aq) + O2(g).
Br- is a catalyst because it can be recovered at the end of the reaction.
Generally, catalysts operate by lowering the activation energy for a reaction.

70. Catalysis
Homogeneous Catalysis

71. Catalysis Homogeneous Catalysis
Catalysts can operate by increasing the number of effective collisions.
From the Arrhenius equation: catalysts increase k by increasing A or decreasing Ea.
A catalyst may add intermediates to the reaction.
Ex: When Br- is added, Br2(aq) is generated as an intermediate in the decomposition of H2O2.

72. Catalysis Heterogeneous Catalysis
The catalyst is in a different phase than the reactants and products.
First step is adsorption (the binding of reactant molecules to the catalyst surface).
Adsorbed species (atoms or ions) are very reactive.

73. Catalysis
Heterogeneous Catalysis

74. C2H4(g) + H2(g)  C2H6(g), H = -136 kJ/mol.
Catalysis
Heterogeneous Catalysis
Consider the hydrogenation of ethylene:
C2H4(g) + H2(g)  C2H6(g), H = -136 kJ/mol.
The reaction is slow in the absence of a catalyst.
In the presence of a metal catalyst (Ni, Pt or Pd) the reaction occurs quickly at room temperature.
First the ethylene and hydrogen molecules are adsorbed onto active sites on the metal surface.
The H-H bond breaks and the H atoms migrate about the metal surface.

75. C2H4(g) + H2(g)  C2H6(g), H = -136 kJ/mol.
Catalysis
Heterogeneous Catalysis
Consider the hydrogenation of ethylene:
C2H4(g) + H2(g)  C2H6(g), H = -136 kJ/mol.
When an H atom collides with an ethylene molecule on the surface, the C-C  bond breaks and a C-H  bond forms.
When C2H6 forms it desorbs from the surface.
When ethylene and hydrogen are adsorbed onto a surface, less energy is required to break the bonds and the activation energy for the reaction is lowered.

76. Catalysis Enzymes Enzymes are biological catalysts.
Most enzymes are protein molecules with large molecular masses (10,000 to 106 amu).
Enzymes have very specific shapes.
Most enzymes catalyze very specific reactions.
Substrates undergo reaction at the active site of an enzyme.
A substrate locks into an enzyme and a fast reaction occurs.

77. Catalysis
Enzymes

78. Catalysis Enzymes The products then move away from the enzyme.
Only substrates that fit into the enzyme lock can be involved in the reaction.
If a molecule binds tightly to an enzyme so that another substrate cannot displace it, then the active site is blocked and the catalyst is inhibited (enzyme inhibitors).
The number of events (turnover number) catalyzed is large for enzymes ( per second).

79. Catalysis Nitrogen Fixation and Nitrogenase
Nitrogen gas cannot be used in the soil for plants or animals.
Nitrogen compounds, NO3, NO2-, and NO3- are used in the soil.
The conversion between N2 and NH3 is a process with a high activation energy (the NN triple bond needs to be broken).
An enzyme, nitrogenase, in bacteria which live in root nodules of legumes, clover and alfalfa, catalyses the reduction of nitrogen to ammonia.

80. Catalysis
Nitrogen Fixation and Nitrogenase

81. Catalysis Nitrogen Fixation and Nitrogenase
The fixed nitrogen (NO3, NO2-, and NO3-) is consumed by plants and then eaten by animals.
Animal waste and dead plants are attacked by bacteria that break down the fixed nitrogen and produce N2 gas for the atmosphere.