1. Chemical Kinetics
Chapter 14
AP Chemistry

2. Chemical Kinetics
Kinetics – the area of chemistry concerned with the rate (or speed) of a reaction.
Kinetics vs. Thermodynamics
Applications: Medicine, Chemical Engineering

3. Reaction Rate Factors Physical state of reactants Concentration
Surface area
Concentration
Rate increases with concentration increase
Temperature
Rate doubles every 10oC increase
Catalyst
Increase the reaction rate w/o being used up

4. Reaction Rates
Speed is the change in a particular quantity with respect to a change in time.
In chemistry, we define the reaction rate
The change in concentration of the reactants or products over time
Units are usually M/sec
Rate =

5. C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq)

6. C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq)
A graph of concentration vs. time is often plotted.
Slope of the tangent line at any point along the curve is the instantaneous rate.
Rate decreases with time.
Reactants decrease with time

7. Average Rate
Because the rate of reaction changes with time, it is useful to consider an average rate.
Rate =
The average rate for a reaction is usually take as the early stages of a reaction.

8. Measuring Rates
To determine the progress of a reaction, we can measure two quantities:
Disappearance of the reactant
Formation of products
Reaction rate is a positive value.
Reaction rate is the same, no matter the method of measurement.

9. aA + bB  cC + dD
The rate of reaction is given by the following equalities:
A: Rate =
B: Rate =
C: Rate =
D: Rate =
Rate =

10. H2O2(g)  H2(g) + O2 (g)
Write the rate in terms of each species.

11. SO2(g) + O2(g)  SO3(g)
Write the rate in terms of each species.

12. H2(g) + O2(g)  H2O(g)
Hydrogen is burning at the rate of 0.85 M/sec. Rate of oxygen consumption? Rate of water vapor formation?

13. How?
How is it possible to measure the concentration of reactants or products?
There are a variety of methods.
One of the more common methods is spectroscopy.
Measures the ability to absorb/transmit light and converts the data to a concentration.

14. Spectrophotometer (Spec-20)

15. Beer-Lambert Law
There is a linear relationship between the concentration of a sample and its absorbance.
A = -logT
Beer’s Law: A = εbC
Standards to find slope
Convert T to A to C

16. NH3(g)  N2(g) + H2(g)
Nitrogen is forming at the rate of M/sec. Rate of ammonia consumption? Rate of hydrogen formation?

17. Rate Law General Equation: aA + bB  cC + dD Rate =
m is the order of A
n is the order of B
(m+n) is the overall reaction order
k is the rate constant
specific for a rxn, Temperature dependent

18. Rate = k[A]m[B]n The rate law must be experimentally determined.
m and n are NOT the stoichiometric coefficient
Unit of rate constant k
M-p s-1 or 1/(Mp s )
p = (m+n) – 1
Rate depends on reactant conc…k does not depend on reactant conc.

19. Rate = k[A][B] What happens to the rate if we…
Double conc of A (everything else the same)?
Double conc of B (everything else the same)?
Triple conc of A and double conc of B?
Order of A? B? Overall?

20. Rate = k[A]2[B] What happens to the rate if we…
Double conc of A (everything else the same)?
Double conc of B (everything else the same)?
Triple conc of A and double conc of B?
Order of A? B? Overall?

21. Rate = k[A]0[B]3 What happens to the rate if we…
Double conc of A (everything else the same)?
Double conc of B (everything else the same)?
Triple conc of A and double conc of B?
Order of A? B? Overall?

22. Rate = k[A]m[B]n Two ways to determine the rate law
Initial rate method
Can have many reactants
Graphical Method
Can only have one reactant
Solving a rate Law:
Need to determine the orders of the reactants
Need to determine the rate constant k

23. 2 NO(g) + 2 H2(g)  N3(g) + 2H2O(g)
Determine the rate law:
2 NO(g) H2(g)  N3(g) + 2H2O(g)
Experiment
[NO]
[H2]
Initial Rate (M/s) 1
0.10
1.23 x 10-3 2
0.20
2.46 x 10-3
4.92 x 10-3

24. a A(g) + b Bg)  c C(g) + d D(g)
Determine the rate law:
a A(g) + b Bg)  c C(g) + d D(g)
Experiment
[A]
[B]
Initial Rate (M/s) 1
0.40
0.30
1.00 x 10-4 2
0.80
4.00 x 10-4
0.60
1.60 x 10-3

25. S2O82-(aq) + 3 I1-(aq)  2 SO42-(aq) + I31-(aq)
Determine the rate law:
S2O82-(aq) I1-(aq)  2 SO42-(aq) + I31-(aq)
Experiment
[NO]
[H2]
Initial Rate (M/s) 1
0.080
0.034
2.2 x 10-4 2
0.017
1.1 x 10-4
0.160

26. Change in Conc. with time
So far, we have considered rate based on the change in concentration and rate constants.
Using calculus, we can convert these same equations to more useful forms.
This is the graphing method to determining the order of a reaction.
Specific for only one reactant: [A]

27. Rate Laws Differential Rate Law: Integrated Rate Law
Expressed how rate depends on concentration.
Integrated Rate Law
Integrated form of the differential. Has specific variables.

28. Zero Order Reaction
Rate only depends on the rate constant…not on the concentration of A
Differential:
Integrated:

29. First Order Reaction
Rate only depends on the rate constant and on the concentration of A
Differential:
Integrated:

30. First Order
Plot of [A] vs. t
Plot of ln[A] vs. t

31. Second Order Reaction
Rate only depends on the rate constant and on the square concentration of A
Differential:
Integrated:

32. Second Order
Plot of ln[A] vs. t
Plot of 1/[A] vs. t

33. Usefulness of the Integrated Rate Laws
[A]t = -kt + [A]0
We can know the concentration at any time point for a given reaction.
We can determine the order of a reaction.
We can determine the half life of a reaction.

34. Determining the Order [A]t = -kt + [A]0 y = mx + b
This tells us that a plot of concentration of A vs time will yield a straight line.
Because this is the zero order rate equation, a plot of [A] vs. t will yield a straight line.
y = mx + b

35. Determining the Order First Order: ln[A]t = -kt + ln[A]0 Second Order:
A plot of ln[A] vs. t will give a straight line.
Second Order:
A plot of 1/[A] vs. t will give a striaght line.

36. Half life, t1/2
The time required for the concentration of a reactant to reach one-half its value:
[A]t1/2 = ½[A]0
This is a convenient way to describe the rate of a reaction.
A fast reaction will have a short half life.

37. Derivation: t1/2 of First Order
ln[A]t = -kt + ln[A]0

38. Summary
** Note: The half life of a first order says that it does NOT depend the concentration of the reactant A. So, the concentration decreases by ½ each regular time interval, t1/2.

39. A first order reaction has k = 6.7 x 10-4 s-1.
How long will it take for the conc to go from 0.25M to 0.15M?
If the initial conc is 0.25M, what is the conc after 8.8 min?

40. A first order reaction has [A] = 2. 00M initially
A first order reaction has [A] = 2.00M initially. After 126 min, [A] = M.
What is the rate constant k?
What is the half life?

41. A second order reaction has k = 7.0 x 10-9 M-1s-1.
If the initial conc is 0.086M, what is the conc after 2.0 min?

42. A first order reaction has t1/2 = 35.0 sec.
What is the rate constant k?
How long would it take for 95% decomposition of the reactant?

43. A first order reaction has a half life of 19. 8 min
A first order reaction has a half life of 19.8 min. What is the reaction rate when [A] = 0.750M?

44. Temperature

45. Collision Model Based on Kinetic Molecular Theory
Molecules must collide to react
Greater the collisions, greater the rate
As concentration increases, rate increases
As temperature increases, rate increases

46. Orientation Most collisions do not lead to reactions
Molecular orientation of collision is important

47. Still not enough
Usually, a collision in the correct orientation is still not enough to cause a reaction.
Kinetic energy of a collision must cause bonds to break.
For a reaction to occur, there must be enough kinetic energy to be greater than some energy.
Activation Energy, Ea, is the minimum energy required to initiate a reaction.

48. Activation Energy

49. Transition State Transition state is also called the activated complex
High energy intermediate state
The activation energy represents the higher energy state of the transition state
A A B B
A A B B
A B A B ‡

51. Arrhenius Equation k = A e−Ea/RT
This equation combines all factors contributing to the reaction rate.
k = A e−Ea/RT
k = rate constant
Ea = activation energy
R = Gas constant
T = Temperature in Kelvin
A = frequency factor “constant”
probability of correctly oriented collisions

52. Other versions y = m * x + b Original Equation: k = A e−Ea/RT
Linear Equation:
If we know the Ea and k2 at T2, we can calculate k1 at T1:
y = m * x + b

53. Catalysis Catalyst – speed up a reaction without being consumed
Beneficial or harmful, ex. Body
Homogenous catalyst – same phase as reactant
Phase transfer catalyst
Heterogeneous catalyst – different phase than reactant
Saturation of alkenes and alkynes

54. Catalyst
Lowers the activation energy
Potential energy diagram is 3-D

55. Enzymes Biological Catalyst Reactants called “substrate”
S + E  SE  P + E
Active site – binding location
SE = enzyme-substrate complex
Lock and Key Model vs. Induced Fit Model

56. Reaction Mechanism
Balanced rxn tells us the species before a rxn starts and after the rxn ends.
Does NOT show how the reaction occurs.
Reaction Mechanism – the steps a reaction progresses.
The steps can be of varying speeds

57. Elementary Reactions Reactions occur because of collisions
Must be correctly oriented
Must have enough energy to reach TS
A + B  C + D
This is a single collision reaction
Elementary reactions – rxn occurring in a single step

58. Molecularity
Number of molecules that participate in a single reaction.
Unimolecular, bimolecular, termolecular
Probability decreases with molecularity
A  P
2A  P
A + B  P
A + 2B  P

59. Multiple Steps Some reactions occur in multiple steps
Multistep Mechanism – a reaction consisting of a series of elementary reactions
Must add to give the overall reaction.
Items that are produced within a mechanism and consumed within a mechanism are intermediates.
Intermediates are not R or P.

61. NO2(g) + CO(g)  NO(g) + CO2(g)

62. Rate Laws
Earlier we stated that rate laws can not be determined from a chemical equation.
Why?
There is a possibility for multistep mechanism
We must consider the speed of each step
However, we can derive the rate law from a the mechanism of a reaction.

63. It’s Elementary, my dear chemists!
If a rxn is an elementary rxn, the rate law is based off the molecularity (coefficients)
A  P Rate =
2 A  P Rate =
A + B  P Rate =
3 A  P Rate =
A + 2 B  P Rate =
A + B + C  P Rate =

64. Rate-Determining Step (RDS)
Let’s go shopping.
Mechanisms can have a slow step.
RDS = slowest step – governs rate law
If first step is slow, intermediates short lived
If later step is slow, there is a buildup if intermediates.
Each step has its own rate constant

65. Slow Initial Step
NO2(g) + CO(g)  NO(g) + CO2(g) Experimentally Determined Rate Law: Rate = k[NO2]2[CO]0

66. Fast Initial Step
The slow step rate law still governs the rate law as before.
However, because the slow step is the second step, there are intermediates in the rate law.
Intermediates are short lived and can not be measured easily – not good in rate law

67. Fast Initial Step
Make assumptions: there is a dynamic equilibrium established in the fast step
The intermediate is more likely to decompose (reverse of fast step is also fast = k-1) than be consumed in step 2 (k2)
The forward rxn rate in step one (k1) equals the rate of the reverse reaction: Rate forward = Rate Reverse

68. 2 NO(g) + Br2(g0  2 NOBr(g)
Experimental Rate Law: Rate = k[NO]2[Br2]