1. Summary of the Kinetics of Zero-Order, First-Order
and Second-Order Reactions
Order
Rate Law
Concentration-Time Equation
Half-Life
t½ =
[A]0 2k
rate = k
[A] = [A]0 - kt t½
ln2 k = 1
rate = k [A]
ln[A] = ln[A]0 - kt 1
[A] =
[A]0
+ kt
t½ = 1
k[A]0 2
rate = k [A]2
13.3

2. What is the order of the reaction? Units on rate constant
The decomposition of a certain antiobiotic in water has a rate constant at 20°C of 1.29 yr-1.
A. Determine the molarity of an antiobiotic solution that sits for 1.00 month, if its original concentration was 5.00 x 10-3 M?
What is the order of the reaction? Units on rate constant
tell us it is first order
ln[A] = ln[A]0 - kt
[A]0 = 5.00 x 10-3 M
t = (1.00 month)(1 yr/12 months) = yr
k = 1.29 yr-1
ln[A] = ln (5.00 x 10-3 M) - (1.29 yr-1)( yr)
ln[A] =
take antilog
[A] = 4.53 x 10-3 M
13.3

3. The decomposition of a certain antiobiotic in water has a rate constant at 20°C of 1.29 yr-1.
B. If the antiobiotic solution is considered to be “no longer effective” at concentrations lower than 5.00 x 10-4 M, how long will the solution be effective?
first order reaction
ln [A]0 = kt
[A] t
[A]0 = 5.00 x 10-3 M
[A]t = 5.00 x 10-4 M
k = 1.29 yr-1
ln (5.00 x 10-3 M/ 5.00 x 10-4 M) = (1.29 yr-1) t
t = ln(10)/ (1.29 yr-1 ) = yr
13.3

4. The decomposition of a certain antiobiotic in water has a rate constant at 20°C of 1.29 yr-1.
C. A 70. % depletion of the initial concentration of the antiobiotic solution results in a significantly lower effectivness. How long before this lower effectiveness will be noticed?
first order reaction
ln [A]0 = kt
[A] t
[A]t = 0.30 [A]0
k = 1.29 yr-1
ln [A] = (1.29 yr-1) t
0.30[A]0
t = ln(3.3)/ (1.29 yr-1 ) = yr
13.3

5. D. Determine the half-life of the antiobiotic.
The decomposition of a certain antiobiotic in water has a rate constant at 20°C of 1.29 yr-1.
D. Determine the half-life of the antiobiotic.
first order reaction
k = 1.29 yr-1
t 1/2 = ln 2 / k
t 1/2 = ln(2)/ (1.29 yr-1 ) = yr
13.3

6. k = 1 / (6.23 x 10-2 M)(2.54 x 103 seconds) = 6.32 x 10-3 /Ms
A second order decomposition reaction takes 2.54 x 103 seconds for the initial concenteration to fall to one half of its origianal value. What is the value of the rate constant if the initial concentration was 6.23 x 10-2 M?.
Second order reaction
t 1/2 = 2.54 x 103 seconds
[A]0 = 6.23 x 10-2 M
t 1/2 = 1 / k [A]0
k = 1 / [A]0 t 1/2
k = 1 / (6.23 x 10-2 M)(2.54 x 103 seconds) = x 10-3 /Ms
13.3

7. A + B C + D Exothermic Reaction Endothermic Reaction
The activation energy (Ea) is the minimum amount of energy required to initiate a chemical reaction.
13.4

8. Temperature Dependence of the Rate Constant
k = A • exp( -Ea/RT )
(Arrhenius equation)
Ea is the activation energy (J/mol)
R is the gas constant (8.314 J/K•mol)
T is the absolute temperature
A is the frequency factor
lnk = - Ea R 1 T
+ lnA
13.4

9. lnk = - Ea R 1 T
+ lnA
13.4

10. 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
13.5

11. Unimolecular reaction – elementary step with 1 molecule
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
13.5

12. 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.
13.5

13. What is the equation for the overall reaction?
The experimental rate law for the reaction between NO2 and CO to produce NO and CO2 is rate = k[NO2]2. The reaction is believed to occur via two steps:
Step 1:
NO2 + NO NO + NO3
Step 2:
NO3 + CO NO2 + CO2
What is the equation for the overall reaction?
NO2+ CO NO + CO2
What is the intermediate?
NO3
What can you say about the relative rates of steps 1 and 2?
rate = k[NO2]2 is the rate law for step 1 so
step 1 must be slower than step 2
13.5

14. But what happens if the first step is NOT the rate determining step?
1: NO + O2  NO fast
2: NO3  NO + O fast
3: NO3 + NO  2 NO slow
rate = k3 [NO3][NO], but NO3 is an intermediate
Since the first two steps are fast,
the rate for creating NO3 = rate for destroying NO3
k1 [NO] [O2] = k2 [NO3] + k3 [NO3][NO]
negligibly small rate
=> very small number
solve for [NO3]: [NO3] = k1 [NO] [O2] k2
substitute into rate expression: rate = k1k3 [NO]2 [O2]
=> observed rate = kobs [NO]2 [O2]

15. What does the potential energy diagram look like for a reaction with multiple steps ?
Reaction Progress
A + B
C + D Ea Ea
Potential Energy
A + B
C + D
Reaction Progress
Reaction with first step as slow step
Reaction with last step as slow step
Each hill represents a reaction
Any dip or valley represents an intermediate

16. Consider the following mechanism for
a gas phase reaction:
1: Cl Cl fast
2: Cl . + CHCl3  HCl + .CCl slow
3: Cl CCl3  CCl fast
A. What is the overall reaction?
Cl2 + CHCl3  HCl + CCl4
B. What are the intermediates in the reaction?
Cl & CCl3
C. What is the molecularity of each step in the mechanism?
Step 1 = unimolecular; Steps 2 & 3 = bimolecular
D. What is the rate determining step?
Step 2 = slow step = rate determing step

17. E. What is the rate law predicted by this mechanism?
Step 2 = slow step = rate determing step
rate = k2 [Cl .][CHCl3 ]
But Cl . is an intermediate formed in Step 1 and , thus, cannot appear in the rate law.
Cl . is in equilibrium with Cl2
By definition, this means that the forward and reverse rates of this reaction are equal.
k1 [Cl2 ] = k-1[Cl .]2
Solving for [Cl .] in terms of [Cl2],
[Cl .]2 = k1/ k-1 [Cl2 ] => [Cl .] = {k1/ k-1 [Cl2 ]} 1/2
Substituting into the overall rate law:
rate = k2 {k1/ k-1 [Cl2 ]} 1/2 [CHCl3 ]
rate = kobs [Cl2 ]1/2 [CHCl3 ]
overall reaction order is 3/2

18. Chain Reactions: a series of reactions in which one reaction initiates the next
Initiation: I2  2 I . slow - produces reactive intermediate
Propagation: I . + Br2  IBr + Br . fast
Br . + I2  IBr + I fast
Termination: Br Br .  Br rare reactions since
I I .  I2 concentrations of species
Br I .  IBr are small

19. 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 ‘
13.6

20. Haber synthesis of ammonia
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 catalyses
Base catalyses
A catalyst can change the mechanism of a reaction, but it will not cause a reaction to proceed that is not favorable.
13.6

21. Haber Process
Fe/Al2O3/K2O
catalyst
N2 (g) + 3H2 (g) NH3 (g)
13.6

22. Ostwald Process 4NH3 (g) + 5O2 (g) 4NO (g) + 6H2O (g)
Pt catalyst
4NH3 (g) + 5O2 (g) NO (g) + 6H2O (g)
2NO (g) + O2 (g) NO2 (g)
Hot Pt wire
over NH3 solution
2NO2 (g) + H2O (l) HNO2 (aq) + HNO3 (aq)
Pt-Rh catalysts used
in Ostwald process
13.6

23. Catalytic Converters CO + Unburned Hydrocarbons + O2 CO2 + H2O
NO + NO2
N2 + O2
13.6

24. Enzyme Catalysis
13.6

25. uncatalyzed
enzyme
catalyzed
13.6