1. Kinetics: Rates and Mechanisms of Reactions
Chemical Kinetics tells us:
…how fast a reaction will occur
…how molecules react (MECHANISM)
a mechanism is a sequence of steps that lead to the product

2. Factors Affecting the rate:
1. concentration:
2. temperature:
generally a 10oC increase will double the rate
3. nature of the reactant: i.e. surface area
4. catalyst: (two types: homogenous and heterogeneous)
5. mechanism: (orientation, shape, & order)
COLLISION THEORY = CAPPING A MARKER

3. Rate of RXN =
The increase in concentration of a product per unit time. or
The decrease in concentration of a reactant per unit time.

4. Conc. is usually measured in M (Molarity= mol/L)
for solutions.
Rate = M
time
= mol
L • s
or mol•L-1•s-1 or M•s -1
Square brackets [ ] are often used to express molarity (i.e. [HCl] means Molarity of HCl)
Since many reactions involve gases, P is often used for concentration.
moles/L

5. 3ClO- (aq)  2Cl-(aq) + ClO3-(aq)
Consider the reaction (net ionic eq.):
3ClO- (aq)  2Cl-(aq) + ClO3-(aq)
Rate could be defined in at least 3 ways: (3 coefficients and ions)
1. appearance of ClO3-
2. appearance of Cl-
Disappearance
3. disappearance of ClO-

6. 3ClO- (aq)  2Cl-(aq) + ClO3-(aq)
Consider the reaction (net ionic):
3ClO- (aq)  2Cl-(aq) + ClO3-(aq)
1. appearance of ClO3-
2. appearance of Cl-
Disappearance
3. disappearance of ClO-
Question: Are these three rates equal?

7. Note the use of coefficients and the - sign
Consider the reaction (net ionic):
3ClO- (aq)  2Cl-(aq) + ClO3-(aq)
Let’s make these three rates equal.
Note the use of coefficients and the - sign

8. aA + bB  cC + dD General Form: = - [A] = - [B] = [D] rate =  [C]
atime
= [B]
btime
rate =  [C]
ctime
= [D]
dtime
“REACTANTS”
“PRODUCTS”

9. Average rate = slope (over time period)
negative sign

10. instantaneous rate = tangent slope
(changing)
WHY?
Collision Theory!

12. IF we can now somehow get a linear plot in the form of:
y = mx + b.
The slope would be a constant independent of concentration!

13. We could call the slope the rate constant and assign it the letter k!
rate constant = k  instantaneous rate or rate
“call in the mathematicians”

14. rate constant: k
is conc. independent
Is still temperature dependent
and mechanism dependent!
General form of rate law :
for RXN: A  products
m = RXN order according to A.
Determined by experiment only!
conc. of A
rate constant

15. General form of rate law :
RXN orders (w, x, y, and z)must be determined by exp. only!
Total (overall) order =  individual orders

16. General Equation Forms:
0 order: rate = k
1st order: rate = k[A]
2nd order: rate = k[A]2 or rate = k[A][B]
3rd order: rate = k[A]3 or
Simple experiments are done by doubling 1 conc. at a time
and looking at the effect.

17. General Equation Forms:
0 order: rate = k
1st order: rate = k[A]
2nd order: rate = k[A]2 or rate = k[A][B]
3rd order: rate = k[A]3 or
Simple experiments are done by doubling 1 conc. at a time
and looking at the effect.
order doubling effect on rate
0 [2]0 = 1 none
1/2 [2]1/2 = increase by
1 [2]1 = 2 doubles
2 [2]2 = 4 quadruples
Question: suppose rate = k[A]2[B] what is the effect of
doubling both A and B?

18. NH4+(aq) + NO2-(aq)  N2(g) + 2H2O(l)
Let’s look a some rate data for the RXN:
NH4+(aq) + NO2-(aq)  N2(g) + 2H2O(l)

19. NH4+(aq) + NO2-(aq)  N2(g) + 2H2O(l)
doubles
double
doubles
double

20. NH4+(aq) + NO2-(aq)  N2(g) + 2H2O(l)
for NO2- : [2]y = 2
y = 1 (1st Order)
for NH4+ : [2]x = 2
x = 1 (1st Order)

21. first order with respect to each reactant, 2nd order overall.
orders usually have integer values, but can be fractional.
Can also be (-) (inhibitors).
Since rate has units, k must also have units.
so units of k must work with [ ] to match units.

22. Determine the units of k in each of the following:
Since rate has units, k must also have units.
(so units of k must work with [ ] to match units.)
1. rate = k[A] M
kunits must = time-1 ?
2. rate = k[A]2 M2
kunits must = time-1 M-1 ?
so k must have units of M-1t-1

23. 4. rate = k[A][B]2[C]
kunits=M-3time-1
5. rate = k[A]0

24. 14.38 In a kinetic study of the reaction:
Ebbing 4th ed. P 490
In a kinetic study of the reaction:
2NO(g) + O2(g)  2NO2(g)
the following data were obtained for the initial rates disap-
pearance of NO:
Obtain the rate law. What is the value of the rate constant? .

25. Write overall rate equation: rate = k[NO]x[O2]y
Ebbing 4th ed. P 490
In a kinetic study of the reaction:
2NO(g) + O2(g)  2NO2(g)
the following data were obtained for the initial rates disap-
pearance of NO:
Obtain the rate law. What is the value of the rate constant?
Write overall rate equation:
rate = k[NO]x[O2]y
For NO: select a pair of experiments in which the conc. of [NO] is changed, but other concentrations are unchanged.
Let’s use Exp. 1 and 2

26. “Let’s divide Exp.1 by Exp.2 to allow us to cancel terms.
Ebbing 4th ed. P 490
In a kinetic study of the reaction:
2NO(g) + O2(g)  2NO(g)
Obtain the rate law. What is the value of the rate constant?
overall rate equation:
rate = k[NO]x[O2]y
“Let’s divide Exp.1 by Exp.2 to allow us to cancel terms.

27. overall rate equation: rate = k[NO]x[O2]y
Ebbing 4th ed. P 490
In a kinetic study of the reaction:
2NO(g) + O2(g)  2NO(g)
Obtain the rate law. What is the value of the rate constant?
overall rate equation:
rate = k[NO]x[O2]y

28. How do we solve for x? Use logarithms
Ebbing 4th ed. P 490
In a kinetic study of the reaction:
2NO(g) + O2(g)  2NO(g)
overall rate equation:
rate = k[NO]x[O2]y
How do we solve for x?
Use logarithms

29. Therefore the rate equation becomes:
Ebbing 4th ed. P 490
In a kinetic study of the reaction:
2NO(g) + O2(g)  2NO(g)
overall rate equation:
rate = k[NO]x[O2]y
How do we solve for x?
Use logarithms
Therefore the rate equation becomes:
rate = k[NO]2[O2]y

30. Now let’s determine the reaction order with respect to [O2]

31. Write overall rate equation: rate = k[NO]2[O2]y
Ebbing 4th ed. P 490
In a kinetic study of the reaction:
2NO(g) + O2(g)  2NO2(g)
the following data were obtained for the initial rates disap-
pearance of NO:
Write overall rate equation:
rate = k[NO]2[O2]y
For O2: select a pair of experiments in which the conc. of [O2] is changed, but all other concentrations are unchanged.
Let’s use Exp. 1 and 3

32. Write overall rate equation: rate = k[NO]2[O2]y
Ebbing 4th ed. P 490
In a kinetic study of the reaction:
2NO(g) + O2(g)  2NO2(g)
the following data were obtained for the initial rates disap-
pearance of NO:
Write overall rate equation:
rate = k[NO]2[O2]y
Let’s use Exp. 1 and 3

33. Write overall rate equation: rate = k[NO]2[O2]y
Ebbing 4th ed. P 490
In a kinetic study of the reaction:
2NO(g) + O2(g)  2NO2(g)
Write overall rate equation:
rate = k[NO]2[O2]y

34. rate = k[NO]2[O2]1 Write overall rate equation: rate = k[NO]2[O2]y
Ebbing 4th ed. P 490
In a kinetic study of the reaction:
2NO(g) + O2(g)  2NO2(g)
Write overall rate equation:
rate = k[NO]2[O2]y
y = 1
rate = k[NO]2[O2]1
Therefore:
Now solve for k.

35. Overall rate equation: rate = k[NO]2[O2]1
Ebbing 4th ed. P 490
In a kinetic study of the reaction:
2NO(g) + O2(g)  2NO2(g)
Overall rate equation:
rate = k[NO]2[O2]1
Choose any Exp. and substitute in experimental to obtain k.
i.e. Exp1. : rate = k[NO]2[O2]1
so : = k[0.0125]2[0.0253]
k = 7100 M-2s-1

36. 14.40 Iodine ion is oxidized to hypoiodite ion, IO-, by hypochlorite
Ebbing 4th ed. P 490
Iodine ion is oxidized to hypoiodite ion, IO-, by hypochlorite
ion, ClO-, in basic solution.
the following initial rate experiments were run:
Obtain the rate law. What is the value of the rate constant? .

37. k = 6.1 s-1

38. This data isn’t linear! What can we do?
Integrated Rate Laws

39. IF we can now somehow get a linear plot in the form of:
y = mx + b.
The slope would be a constant, independent of concentration!

40. We could call the slope the rate constant and assign it the letter k!
rate constant = k  rate
“call in the mathematicians”

41. Key Equations:
*Since the units of rate are concentration/time, the units of k (the rate constant) must dimensionally agree. So for each order, k will have different units and those units can tell one which equation to use. [ ] means the concentration of the enclosed species in Molarity (M).

42. The data below was collected for the reaction:
NOCl(g)  NO(g) + 1/2Cl2(g)
Time (s) [NOCl] (M)
Prepare THREE graphs to determine if the RXN is ZERO, 1st, or 2nd order. Then determine the value and units of the rate constant k.

43. Zero Order Plot
[A]t vs. time
rate = k
[A]t = -kt + [A]0
y = mx + b

44. First Order Plot
ln[A]t vs. time
rate =k[A]
ln[A]t = -kt + ln[A]0
y = mx + b

45. 2nd Order Plot
1/[A]t vs. time
rate = k[A]2
1/[A]t = kt + 1/[A]0
y = mx + b
Plot is linear so 2nd Order
k = slope = M-1s-1

46. Zero Order: Integrated rate laws: rate = k[A]0 = k
rate = k integrated gives:
[A]t = -kt + [A]0
y = mx + b
slope = -k
If a RXN is zero order, a plot of [A] vs. time should be linear and the slope = -k.

47. Integrated rate laws:
1st order rate laws:
rate = k[A] integrated gives:
rearranged to : y = mx + b gives:
ln[A]t = -kt + ln[A]0
slope = -k
If reaction data is 1st order, a plot of ln[A] vs. time should be linear.

48. 2nd Order Integrated Rate Equations:
rate = k[A]2
integrated gives:
y = mx + b
slope = k
If a RXN is 2nd order, a plot of 1/[A] vs. time should be linear and the slope = k.

49. RXN is first order with respect to CH3NC
Zero Order Plot
1st Order Plot
slope = -k
ln[CH3NC]t = -kt + ln[CH3NC]0

50. First order plot 2nd order Plot RXN is 2nd order with respect to [NO2]
Slope = k

51. General form 1st order: ln[A]t = -kt + ln[A]0
Note: This is a formula that can be used to solve (1st order) problems. (If all but one of the variables are given)
1. Given the RXN: C3H6  CH2=CHCH3
Where k = 6.0 x 10-4
Looking at the units of k, determine the order.
Problem: if [C3H6]0 = M, find 955 s.
ln[A]t = -kt + ln[A]0
ln[A]t =
[A]t = M

52. The other equations can be used in a similar fashion.

53. Half-life: the time it takes to decrease the concentration to 1/2
its initial value.
“fold paper to view subsequent half-lives”

54. Half-lives: Formulas: 1st order: t1/2 = 0.693/k Derivation: @t1/2 :
and:
so:

55. Half-life formulas:
Zero Order First Order nd Order
These 2 are conc. dependent (and not very useful).

56. Energy Diagrams:
All chemical and physical changes are accompanied by energy changes.
Ea = Activation energy
reactants E
Exothermic
energy 
products
time 
Question: What keeps the reactants from rolling down the hill?

62. rate constant (k) varies with temperature.
but not with concentration

63. The Arrhenius Equation:
Temp in K
8.31 J/mol•K
activation energy
base e (natural ln)
frequency factor (1/time), fraction of
collisions with correct geometry.
rate constant
-Ea/RT is always <1 and refers to the fraction of molecules having minimum energy for a RXN.

64. How can we make this a linear equation in the form of
y = mx + b?
Take the ln of each side.
ln k = ln A - Ea/RT
or:
y = b - mx
A plot of ln k vs. 1/T gives a straight line with the
slope = -Ea/R (Ea = x slope)

65. At two diff. temps. we get:

66. Reaction Mechanisms: Reaction is broken into steps with intermediates being formed.
“some RXNS occur in one step, but most occur in in multiple steps.”
Each Step is called an elementary step, and the number of molecules involved in each step defines the molecularity of the step.
uni-molecular: = 1 i.e. O3*  O2 + O
bi-molecular: = 2 (these are the most common)
i.e. HI + HI  activated complex  H2 + I2
ter-molecular: = 3 (rare, due to probability of orientation and energy both being correct.)
i.e. O(g) + O2(g) + N2(g)  O3(g) + “energetic” N2(g)

67. The Raschig process for the preparation of hydrazine (N2H4)
Overall RXN: 2NH3(g) + NaOCl(aq)  N2H4(aq) + NaCl(aq) + H2O(l)
Proposed Mechanism:
(Only from experiment)
Step 1:
NH3(aq) + OCl-(aq)  NH2Cl(aq) ‡ + OH-‡ (aq)
Step 2:
NH2Cl(aq) ‡ + NH3(aq)  N2H5+‡ + Cl-(aq)
Step 3:
N2H5+(aq) ‡ + OH-‡ (aq)  N2H4(aq) + H2O(l)
“Cancel intermediates and “add steps” to give overall RXN.”
2NH3(g) + OCl-(aq)  N2H4(aq) + Cl-(aq) + H2O(l)
The overall rate law, mechanism, and the total order can’t be predicted from the stoichiometry, only by experiment.

68. The following is only true for individual steps:
The rate law of an elementary step is given by the product of a rate constant and the conc. of the reactants in the step.
Step Molecularity rate law
uni
A  Product(s)
rate = k[A] bi
rate = k[A][B]
A + B  Product(s) bi
rate = k[A]2
A + A  Product(s)
rate = k[A]2[B]
2A + B  Product(s)
ter
The overall mechanism must match the observed rate law.
Usually one STEP is assumed to be the rate determining step.

69. Example:
Overall RXN:
2NO2(g) + F2(g)  2NO2F(g)
Observed Experimental rate law:
rate = k[NO2][F2]
Question:
Why does this rule out a single step RXN?
Answer:
rate law for single step process would be:
rate = k[NO2]2[F2]
“Let’s try to work out a Mechanism that matches the
observed rate law.”

70. Example:
Overall RXN:
2NO2(g) + F2(g)  2NO2F(g)
Observed Experimental rate law:
rate = k[NO2][F2] k1
slow:
NO2(g) + F2(g)  NO2F(g) + F(g) ‡
rate = k[NO2][F2] k2
fast:
NO2(g) + F(g) ‡  NO2F(g)
2NO2(g) + F2(g)  2NO2F(g)
The rate law is dependent upon the slow step.

71. The rate law is dependent upon the slow step.
Let’s try making the 2nd RXN slow and the first fast
Overall RXN:
2NO2(g) + F2(g)  2NO2F(g)
Observed Experimental rate law:
rate = k[NO2][F2] k1
fast:
NO2(g) + F2(g)  NO2F(g) + F(g) ‡ k2
rate = k[NO2][F ‡]
slow:
NO2(g) + F(g) ‡  NO2F(g)
2NO2(g) + F2(g)  2NO2F(g)

73. 1. At low temperatures, the rate law for the reaction:
CO(g) + NO2(g)  CO2(g) + NO(g) is:rate = k[NO2]2
Which mechanism is consistent with the rate law?
a. CO + NO2  CO2 + NO
rate = k[CO][NO2]
b. 2NO2  N2O4‡ (fast)
N2O4 ‡ + 2CO  2CO NO (slow)
c. 2NO2  NO3 ‡ + NO (slow)
NO3 ‡ + CO  NO2 + CO2 (fast)
d. 2NO2  2NO + O2 ‡ (slow)
2CO + O2 ‡  2CO2 (fast)

74. For the following mechanism:
2NO  N2O2 (fast)
N2O2 + H2  H2O + N20 (slow)
N2O + H2  N2 + H2O (fast)
a. Determine the overall reaction
b. Does the mechanism agree with the rate law:
rate = k[NO]2[H2]