1. Chemical Kinetics Chapter 14
Copyright © The McGraw-Hill Companies, Inc.  Permission required for reproduction or display.

2. Chemical Kinetics Thermodynamics – does a reaction take place?
Kinetics – how fast does a reaction proceed?
Reaction rate is the change in the concentration of a reactant or a product with time (M/s).
A B
rate = -
D[A] Dt
D[A] = change in concentration of A over
time period Dt
rate =
D[B] Dt
D[B] = change in concentration of B over
time period Dt
Because [A] decreases with time, D[A] is negative.

3. I- + ClO- -----) Cl- + IO-
Rates
Example of average rate calculation for:
I- + ClO ) Cl- + IO-
Given the [I-] and time in seconds, then what is the average rate?
Time (s) [I-] Molarity
Rate = - ΔM / ΔT
Rate = - ( ) M / ( ) s = 6.8x10-4 M / 6.00s
Rate = 1.1 x 10-4 M/s

4. A B
rate = -
D[A] Dt
rate =
D[B] Dt

5. Br2 (aq) + HCOOH (aq) 2Br- (aq) + 2H+ (aq) + CO2 (g)
red-brown
Br2 (aq) + HCOOH (aq) Br- (aq) + 2H+ (aq) + CO2 (g)
time
393 nm
light
Detector
t1< t2 < t3
D[Br2] a D Absorption

6. Reaction Rates and Stoichiometry
2A B
Two moles of A disappear for each mole of B that is formed.
rate = -
D[A] Dt 1 2
rate =
D[B] Dt
aA + bB cC + dD
rate = -
D[A] Dt 1 a
= -
D[B] Dt 1 b =
D[C] Dt 1 c =
D[D] Dt 1 d

7. Expressing reaction rates
For a chemical reaction, there are many ways to express the reaction rate. The relationships among expressions depend on the equation.
Note the expression and reasons for their relations for the reaction
2 NO + O2 (g) = 2 NO2 (g)
[O2] [NO] [NO2] Reaction rate = – ——— = – — ———— = — ———  t  t  t
Make sure you can write expressions for any reaction and figure out the relationships. For example, give the reaction rate expressions for
2 N2O5 = 4 NO2 + O2
How can the rate expression be unique and universal?
15 Chemical Kinetics

8. Calculating reaction rate
The concentrations of N2O5 are 1.24e-2 and 0.93e-2 M at 600 and 1200 sec after the reactants are mixed at the appropriate temperature. Evaluate the reaction rates for
2 N2O5 = 4 NO2 + O2
Solution: (0.93 – 1.24)e – 0.31e-2 M Decomposition rate of N2O5 = – ———————— = – —————— – s
= 5.2e-6 M s-1.
Note however, rate of formation of NO2 = 1.02e-5 M s-1. rate of formation of O2 = 2.6e-6 M s-1.
The reaction rates are expressed in 3 forms
Be able to do this type problems

9. Reaction rates
Ex: Balance the following reaction, then determine how the rates of each compound are related:
N2O5 (g) → NO2 (g) + O2 (g) 2 4
[] = molarity unless otherwise noted
If D[O2]/Dt = 5.0 M/s, what is D[N2O5]/Dt?

10. The Rate Law
The rate law expresses the relationship of the rate of a reaction to the rate constant and the concentrations of the reactants raised to some powers.
aA + bB cC + dD
Rate = k [A]x[B]y
Reaction is xth order in A
Reaction is yth order in B
Reaction is (x +y)th order overall

11. F2 (g) + 2ClO2 (g) FClO2 (g)
rate = k [F2]x[ClO2]y
Double [F2] with [ClO2] constant (compare exp1 and exp3)
Rate doubles
x = 1
Quadruple [ClO2] with [F2] constant (compare exp1 and exp2)
Rate quadruples
y = 1
rate = k [F2][ClO2]

12. Rate Laws Rate laws are always determined experimentally.
Reaction order is always defined in terms of reactant (not product) concentrations.
The order of a reactant is not related to the stoichiometric coefficient of the reactant in the balanced chemical equation.
F2 (g) + 2ClO2 (g) FClO2 (g) 1
rate = k [F2][ClO2]

13. Ex: Determining the rate law using the following data:
Exp # [NH4+] [NO21-] Initial rate (M/s)
x 10-3
x 10-3
x 10-3
x 1
x 2
x 2
x 3
x 1
x 9
Rate = k [NH4+]m [NO2-]n
3m = rate = 9
, m = 2
2n = rate = 2
, n = 1
Rate = k [NH4+]2 [NO2-]1
3.0 x 10-3 = k [0.50]2 [0.20]1
k = 0.060
Rate = [NH4+]2 [NO2-]1

14. Rate = k [NH4+]2 [NO2-]1

15. Initial Rate of Formation of
Practice Problem: A + 2 B C + D
The following data about the reaction above were obtained from three experiments:
(a) What is the rate equation for the reaction?
(b) What is the numerical value of the rate constant k? What are its dimensions?
Experiment
[A]
[B]
Initial Rate of Formation of
C in M 1
0.60
0.15
6.3´10-3 2
0.20
2.8´10-3 3
7.0´10-4

16. rate = k [A]2[B]1
(b)
= 0.12 L2mol-2min-1

17. First-Order Reactions
rate = -
D[A] Dt
A product
rate = k [A]
D[A] Dt
= k [A] -
rate
[A]
M/s M =
k =
= 1/s or s-1
[A] is the concentration of A at any time t
[A]0 is the concentration of A at time t=0
[A] = [A]0e−kt
ln[A] = ln[A]0 - kt

18. Concentration and time of 1st order reaction
Describe the features of plot of [A] vs. t and ln[A] vs. t for 1st order reactions. Apply the technique to evaluate k or [A] at various times.
[A]
ln[A]
ln [A] = ln [A]o – k t
[A] = [A]o e – k t t½ t t

19. 13.4
The conversion of cyclopropane to propene in the gas phase is a first-order reaction with a rate constant of 6.7 × 10−4 s−1 at 500°C.
If the initial concentration of cyclopropane was 0.25 M, what is the concentration after 8.8 min?
How long (in minutes) will it take for the concentration of cyclopropane to decrease from 0.25 M to 0.15 M?
How long (in minutes) will it take to convert 74 percent of the starting material?

20. 13.4
Solution (a) In applying Equation (13.4), we note that because k is given in units of s−1, we must first convert 8.8 min to seconds:
We write
Hence,
Note that in the ln [A]0 term, [A]0 is expressed as a dimensionless quantity (0.25) because we cannot take the logarithm of units.
ln[A] = ln[A]0 - kt

21. 13.4
(b) Using Equation (13.3),
(c) From Equation (13.3),

22. 13.5
The rate of decomposition of azomethane (C2H6N2) is studied by monitoring the partial pressure of the reactant as a function of time:
The data obtained at 300°C are shown in the following table:
Are these values consistent with first-order kinetics? If so, determine the rate constant.

23. 13.5
Solution First we construct the following table of t versus ln Pt.
Figure 13.11, which is based on the data given in the table, shows that a plot of ln Pt versus t yields a straight line, so the reaction is indeed first order. The slope of the line is given by

24. 13.5
According to Equation (13.4), the slope is equal to −k, so k = 2.55 × 10−3 s−1 .
ln[A] = ln[A]0 - kt

25. First-Order Reactions
The half-life, t½, is the time required for the concentration of a reactant to decrease to half of its initial concentration.
t½ = t when [A] = [A]0/2 ln
[A]0
[A]0/2 k = t½
ln 2 k =
0.693 k =

26. 13.6
The decomposition of ethane (C2H6) to methyl radicals is a first-order reaction with a rate constant of 5.36 × 10−4 s−1 at 700°C:
Calculate the half-life of the reaction in minutes.

27. 13.6
For a first-order reaction, we only need the rate constant to calculate the half-life of the reaction. From Equation (13.6)

28. Second-Order Reactions
rate = -
D[A] Dt
A product
rate = k [A]2
rate
[A]2
M/s M2 =
D[A] Dt
= k [A]2 -
k =
= 1/M•s 1
[A] =
[A]0
+ kt
[A] is the concentration of A at any time t
[A]0 is the concentration of A at time t=0
t½ = t when [A] = [A]0/2
t½ = 1
k[A]0

29. 13.7 Iodine atoms combine to form molecular iodine in the gas phase
This reaction follows second-order kinetics and has the high rate constant 7.0 × 109/M · s at 23°C.
If the initial concentration of I was M, calculate the concentration after 2.0 min.
Calculate the half-life of the reaction if the initial concentration of I is 0.60 M and if it is 0.42 M.

30. 13.7
(a) To calculate the concentration of a species at a later time of a second−order reaction, we need the initial concentration and the rate constant. Applying Equation (13.7) 1
[A] =
[A]0
+ kt

31. 13.7
where [A]t is the concentration at t = 2.0 min. Solving the equation, we get
(b)
For [I]0 = 0.60 M
t½ = 1
k[A]0

32. 13.7
For [I]0 = 0.42 M

33. Zero-Order Reactions rate = - D[A] Dt A product rate = k [A]0 = k rate
= M/s
[A] is the concentration of A at any time t
[A] = [A]0 - kt
[A]0 is the concentration of A at time t = 0
t½ = t when [A] = [A]0/2
t½ =
[A]0 2k

34. Summary of the Rate Laws

35. Activation energy ( Ea ): minimum energy required to initiate a chemical reaction
Activated complex Ea
Reactants
Energy
DErxn
Products
Rxn pathway (or rxn coordinate)
Note that: DErxn, forward = - DErxn, backward
Ea, forward ≠ Ea, backward

36. Figure 2: Change in Potential Energy

37. A = frequency factor (related to # of collisions)
Arrhenius equation
Relationship between rate and T
A = frequency factor (related to # of collisions)
R = J/(mol•K)
Swedish, father was a surveryor, taught himself to read at age 3; In 1884, based on this work, he submitted a 150-page dissertation on electrolytic conductivity to Uppsala for the doctorate. It did not impress the professors, and he received the lowest possible passing grade. Later this very work would earn him the Nobel Prize in Chemistry. Worked with Boltzmann and van’t Hoff.
Svante Arrhenius ( )

38. How to determine Ea: perform rate experiments using various T (and keep concentrations constant.)
ln A k
Temp (K)
ln k
1/T (K-1)

39. Ex: Determine the activation energy using the following data:
T (K)
k (s-1)
190.
2.50 x 10-2
200.
4.50 x 10-2
210.
7.66 x 10-2

40. Application of Arrhenius Equation
From k = A e – Ea / R T, calculate A, Ea, k at a specific temperature and T.
The reaction:
2 NO2(g) -----> 2NO(g) + O2(g)
The rate constant k = 1.0e-10 s-1 at 300 K and the activation energy Ea = 111 kJ mol-1. What are A, k at 273 K and T when k = 1e-11?
Method: derive various versions of the same formula
k = A e – Ea / R T
A = k e Ea / R T
A / k = e Ea / R T
ln (A / k) = Ea / R T
Make sure you know how to transform the formula into these forms.

41. Application of Arrhenius Equation (continue)
The reaction:
2 NO2(g) -----> 2NO(g) + O2(g)
The rate constant k = 1.0e-10 s-1 at 300 K and the activation energy Ea = 111 kJ mol-1. What are A, k at 273 K and T when k = 1e-11?
Use the formula derived earlier:
A = k eEa / R T = 1e-10 s-1 exp ( J mol-1 / (8.314 J mol-1 K –1*300 K)) = 2.13e9 s-1
k = 2.13e9 s-1 exp (– J mol-1) / (8.314 J mol-1 K –1*273 K) = 1.23e-12 s-1
T = Ea / [R* ln (A/k)] = J mol-1 / (8.314*46.8) J mol-1 K = 285 K

42. Practice Problem:
The rate constant for the reaction H2(g)  + I2(g) ---> 2HI(g) is 5.4 x 10-4 M-1s-1 at 326 oC.     At 410 oC the rate constant was found to be 2.8 x 10-2 M-1s-1. Calculate the a) activation energy
and
b) high temperature limiting rate constant
for this reaction.

43. We know the rate constant for the reaction at two different temperatures and thus we can calculate the activation energy from the above relation. First, and always, convert all temperatures to Kelvin, an absolute temperature scale. Then simply solve for Ea in units of R.
ln(5.4 x 10-4 M-1s -1/ 2.8 x 10-2 M-1s-1) = (-Ea /R ){1/599 K - 1/683 K}
= - Ea/R {2.053 x 10-4 K-1}
Ea= (1.923 x 104 K) (8.314 J/K mol)
Ea= 1.60 x 105 J/mol

44. Now that we know Ea, the pre-exponential factor, A, (which is the largest rate constant that the reaction can possibly have) can be evaluated from any measure of the absolute rate constant of the reaction. so
5.4 x 10-4 M -1s-1 = A exp{-(1.60 x 105 J/mol)/((8.314 J/K mol)(599K))}
(5.4 x 10-4 M-1s-1) / (1.141x10-14) = 4.73 x 1010 M-1s-1
The infinite temperature rate constant is 4.73 x 1010 M-1s-1

45. Alternate Form of the Arrhenius Equation
At two temperatures, T1 and T2 or

46. 13.9
The rate constant of a first-order reaction is 3.46 × 10−2 s−1 at 298 K.
What is the rate constant at 350 K if the activation energy for the reaction is 50.2 kJ/mol?

47. 13.9
The data are

48. 13.9

49. N2O2 is detected during the reaction!
Reaction Mechanisms
The overall progress of a chemical reaction can be represented at the molecular level by a series of simple elementary steps or elementary reactions.
The sequence of elementary steps that leads to product formation is the reaction mechanism.
2NO (g) + O2 (g) NO2 (g)
N2O2 is detected during the reaction!
Elementary step:
NO + NO N2O2
Overall reaction:
2NO + O NO2 +
Elementary step:
N2O2 + O NO2

50. 2NO (g) + O2 (g) NO2 (g)
Mechanism:

51. Intermediates are species that appear in a reaction mechanism but not in the overall balanced equation.
An intermediate is always formed in an early elementary step and consumed in a later elementary step.
Elementary step:
NO + NO N2O2
N2O2 + O NO2
Overall reaction:
2NO + O NO2 +
The molecularity of a reaction is the number of molecules reacting in an elementary step.
Unimolecular reaction – elementary step with 1 molecule
Bimolecular reaction – elementary step with 2 molecules
Termolecular reaction – elementary step with 3 molecules

52. Rate Laws and Elementary Steps
Unimolecular reaction
A products
rate = k [A]
Bimolecular reaction
A + B products
rate = k [A][B]
Bimolecular reaction
A + A products
rate = k [A]2
Writing plausible reaction mechanisms:
The sum of the elementary steps must give the overall balanced equation for the reaction.
The rate-determining step should predict the same rate law that is determined experimentally.
The rate-determining step is the slowest step in the sequence of steps leading to product formation.

53. A catalyst is a substance that increases the rate of a chemical reaction without itself being consumed.
k = A • exp( -Ea / RT ) Ea k
Uncatalyzed
Catalyzed
ratecatalyzed > rateuncatalyzed
Ea < Ea ‘

55. Figure: 14-20

56. In heterogeneous catalysis, the reactants and the catalysts are in different phases.
Haber synthesis of ammonia
Ostwald process for the production of nitric acid
Catalytic converters
In homogeneous catalysis, the reactants and the catalysts are dispersed in a single phase, usually liquid.
Acid catalysis
Base catalysis

57. Half-Life
Sugar is fermented in a 1st order process (using an enzyme as a catalyst).
sugar + enzyme --> products
Rate of disappear of sugar = k[sugar]
k = 3.3 x 10-4 sec-1
What is the half-life of this reaction?

58. Half-Life Section 15.4 and Screen 15.8
Rate = k[sugar] and k = 3.3 x 10-4 sec-1. What is the half-life of this reaction?
Solution
[A] / [A]0 = 1/2 when t = t1/2
Therefore, ln (1/2) = - k • t1/2
= - k • t1/2
t1/2 = / k
So, for sugar,
t1/2 = / k = sec = 35 min

59. Half-Life
Rate = k[sugar] and k = 3.3 x 10-4 sec-1. Half-life is 35 min. Start with 5.00 g sugar. How much is left after 2 hr and 20 min?
Solution
2 hr and 20 min = 4 half-lives
Half-life Time Elapsed Mass Left
1st 35 min g
2nd g
3rd g
4th g

60. Half-Life Radioactive decay is a first order process.
Tritium ---> electron helium
3H 0-1e 3He
If you have 1.50 mg of tritium, how much is left after 49.2 years? t1/2 = years
Solution

61. Half-Life Solution ln [A] / [A]0 = -kt
Start with 1.50 mg of tritium, how much is left after 49.2 years? t1/2 = years
Solution
ln [A] / [A]0 = -kt
[A] = ? [A]0 = mg t = years
Need k, so we calc k from: k = / t1/2
Obtain k = y-1
Now ln [A] / [A]0 = -kt = - ( y-1) • (49.2 y) =
Take antilog: [A] / [A]0 = e-2.77 =
is the fraction remaining!

62. Half-Life Solution [A] / [A]0 = 0.0627
Start with 1.50 mg of tritium, how much is left after 49.2 years? t1/2 = years
Solution
[A] / [A]0 =
is the fraction remaining!
Because [A]0 = 1.50 mg, [A] = mg
But notice that 49.2 y = 4.00 half-lives
1.50 mg ---> mg after 1
---> mg after 2
---> mg after 3
---> mg after 4

63. Half-Lives of Radioactive Elements
Rate of decay of radioactive isotopes given in terms of 1/2-life.
238U --> 234Th + He x 109 y
14C --> 14N + beta y
131I --> 131Xe + beta d
Element seaborgium 263Sg s