1. Chemical Kinetics Kinetics – how fast does a reaction proceed?
Reaction rate
Rate Law
Change of concentration with time
Rate and temperature
Reaction mechanism
Catalysis

2. I. Reaction Rate
A. Rate is the change in the concentration of a reactant or a product with time (M/s).
- Average rate
Instantaneous rate
Initial rate
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. A B
time
rate = –
D[A] Dt
rate =
D[B] Dt

4. Ex: Br2 (aq) + HCOOH (aq) -> 2Br- (aq) + 2H+ (aq) + CO2 (g)
slope of
tangent
slope of
tangent
slope of
tangent
instantaneous rate = rate for specific instance in time
average rate = -
D[Br2] Dt
= -
[Br2]final – [Br2]initial
tfinal - tinitial
average rate = -
( M M)
(400.0 s s)
=2.3 x 10-5 M/s
From 0.0 s s

5. B. Rate 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

6. Q: Write the rate expression for the following reaction:
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (g)
rate = –
D[CH4] Dt
= –
D[O2] Dt 1 2 =
D[CO2] Dt =
D[H2O] Dt 1 2
Q: If the CH4(g) is burning at a rate of 0.44 mol/s, what is the rate of consumption of O2(g)?
0.44 mol/s = –
D[CH4] Dt
= –
D[O2] Dt 1 2 –
D[O2] Dt
= 2 x 0.44 mol/s =0.88 mol/s

7. C. Rate involving gas measure DP over time
2H2O2 (aq) H2O (l) + O2 (g)
PV = nRT
P = RT = [O2]RT n V
[O2] = P RT 1
rate =
D[O2] Dt RT 1 DP Dt =

8. D. Factors affecting reaction rate
Physical states of reactants
Homogeneous reaction
Heterogeneous reaction
Concentration
Temperature
Catalyst

9. II. Rate law
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
k : rate constant

10. 2 Example: F2 (g) + 2ClO2 (g) 2FClO2 (g) rate = k [F2]x[ClO2]y
Rate data for the reaction between F2 and ClO2
rate = k [F2]x[ClO2]y
Double [F2] with [ClO2] constant
Rate doubles
x = 1
Quadruple [ClO2] with [F2] constant
Rate quadruples
y = 1
k =
rate
[F2][ClO2]
= 1.2 M-1 s-1
rate = k [F2][ClO2] 2
Overall reaction order :

11. II. Rate Laws Rate laws are always determined experimentally.
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]

12. S2O82- (aq) + 3I- (aq) 2SO42- (aq) + I3- (aq) Q:
Determine the rate law and calculate the rate constant for the following reaction from the following data:
S2O82- (aq) + 3I- (aq) SO42- (aq) + I3- (aq) Q:
Experiment
[S2O82-]
[I-]
Initial Rate (M/s) 1
0.08
0.034
2.2 x 10-4 2
0.017
1.1 x 10-4 3
0.16
rate = k [S2O82-]x[I-]y
y = 1
x = 1
rate = k [S2O82-][I-]
From experiment 1 & 2: double [I-], rate doubles
Rate1
Rate2
2.2x10-4M/s
1.1x10-4M/s
= = 2 =
k(0.08M)x(0.034M)y
k(0.08M)x(0.017M)y
= 2y
Double [S2O82-], rate doubles (experiment 2 & 3)
k =
rate
[S2O82-][I-] =
2.2 x 10-4 M/s
(0.08 M)(0.034 M)
= 0.08/M•s

13. III. Change in concentration with time (Integrated rate law)
A Product
Order
Rate Law
Integrated rate
law
Plot
rate = k
[A] = [A]0 - kt
[A] vs. t 1
rate = k [A]
ln[A] = ln[A]0 - kt
ln[A] vs. t 1
[A] =
[A]0
+ kt 1
[A] 2
rate = k [A]2

14. A. First-order reaction
[A] = [A]0exp(-kt)
A product
rate = -
D[A] Dt
rate = k [A]
ln[A] = ln[A]0 - kt
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

15. Example: Decomposition of N2O5

16. Q: The reaction 2A –> B is first order in A with a rate constant of 2.2 x 10-2 s-1 at 80oC. (a) How long will it take for A to decrease from 0.72 M to 0.14 M. (b) What is the concentration of A after 20 s if the initial concentration of A is 0.42M?
1st order reaction : ln[A] = ln[A]0 - kt
(a) [A]0 = 0.72 M
[A] = 0.14 M
t = ?
kt = ln[A]0 – ln[A] ln
[A]0
[A] k = ln
0.72 M
0.14 M
2.2 x 10-2 s-1 =
ln[A]0 – ln[A] k
t =
= 74 s
(b) [A]0 = 0.42 M
t = 20 s
[A] = ?
ln[A] = ln[A]0 - kt
= ln(0.42) x10-2 s-1 x 20 s
=-1.3
[A] = exp(-1.3) = e-1.3
=0.27 M

17. Half-life, t1/2, for first-order reaction
The half-life, t1/2, is the time required for the concentration of a reactant to decrease to half of its initial concentration.
t1/2= t when [A] = [A]0/2 ln
[A]0
[A]0/2 k =
t1/2
ln2 k =
0.693 k =
(*indep of concentration)
Q: What is the half-life of N2O5 if it decomposes with a rate constant of 5.7 x 10-4 s-1?
t1/2
ln2 k =
0.693
5.7 x 10-4 s-1 =
= 1200 s = 20 minutes
How do you know decomposition is first order?
units of k (s-1)

18. First-order reaction
A product
# of
half-lives
[A] = n[A]0 1
1/2 2
1/4 3
1/8 4
1/16

19. B. Second-order reaction1
[A] t
A product
rate = -
D[A] Dt
rate = k [A]2 1
[A] =
[A]0
+ kt
rate
[A]2
M/s M2 =
k =
= 1/M•s
t1/2= t when [A] = [A]0/2
t1/2 = 1
k[A]0
[A] is the concentration of A at any time t
[A]0 is the concentration of A at time t=0

20. C. Zeroth-order reaction
A product
rate = -
D[A] Dt
rate = k [A]0= k
[A] = [A]0 - kt
rate
[A]0
k =
= M/s
t1/2= t when [A] = [A]0/2
t1/2 =
[A]0 2k
[A] is the concentration of A at any time t
[A]0 is the concentration of A at time t=0

21. Integrated rate law – summary
Order
Rate Law
Concentration-Time Equation
Half-Life
t/2 =
[A]0 2k
rate = k
[A] = [A]0 - kt
t1/2
ln2 k = 1
rate = k [A]
ln[A] = ln[A]0 - kt 1
[A] =
[A]0
+ kt
t1/2 = 1
k[A]0 2
rate = k [A]2

22. Determine graphically the rate law for the reaction of
4 PH3(g) -> P4 (g) + 6 H2(g) Q:
Time(s)
[PH3]
(M/s)
ln[PH3]
1/[PH3] 10 30 50 90
150
200
1.000
0.819
0.549
0.368
0.165
0.050
0.018
0.0
-0.2
-0.6
-1.0
-1.8
-3.0
-4.0
1.00
1.22
1.82
2.72
6.06
20.0
55.6
Straight line is obtained when
ln[PH3] is plotted against time
1st order reaction

23. IV. Reaction Rate and Temperature
A. Collision theory
Based on the kinetic-molecular theory
Collisions between reactants leads to reaction
Requires proper orientation for reaction to occur
Requires minimum energy (activation energy) for reaction to occur

25. The energy change during the chemical reaction
B. Activation energy, Ea
Ea: The minimum amount of energy required to initiate a chemical reaction
Transition state, activated complex
The energy change during the chemical reaction
Transition state
Transition state
Exothermic Reaction
Endothermic Reaction

26. C. Temperature dependence of rate constant
Arrhenius equation
k = A • exp( -Ea/RT )
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

27. lnk = - Ea R 1 T
+ lnA
ln = ( ) Ea R 1 T2 k1 k2 T1

28. Q: The rate constant of a certain reaction triples when the temperature increases from 25oC to 35oC. What is the activation energy of this reaction
ln = ( ) Ea R 1 T2 k1 k2 T1
T1 = 25oC = 298 K
k2 = 3 k1
Ea=?
T2 = 35oC = 308 K ln k1
3k1 = 1
308 K
8.314 J/K.mol Ea
( )
298 K
8.314 J/K.mol Ea
(-1.09 x /K )
-1.10 =
-1.10 x J/K.mol
(-1.09 x /K)
Ea =
= 83.9 kJ/mol

29. (a) Slowest reaction to fastest: (1) < (2) < (3) (fastest)
Q: Consider the following potential energy profile. (a) Rank the rate of the reaction. (Assume frequency factor is about the same). (b) Which reaction(s) are endothermic?
(1)
(2)
(3)
34 kJ/mol
23 kJ/mol
17 kJ/mol
30 kJ/mol
56 kJ/mol
Potential energy
40 kJ/mol R R P P R P
reaction progress
reaction progress
reaction progress
Ea= 40 kJ/mol
Ea= 34 kJ/mol
Ea= 23 kJ/mol
DE= 10 kJ/mol
DE=-22 kJ/mol
DE= 6 kJ/mol
(a) Slowest reaction to fastest:
(1) < (2) < (3) (fastest)
(b) Endothermic reactions:
(1) and (3)

30. N2O2 is detected during the reaction!
V. Reaction Mechanism
Detailed molecular process by which reaction occurs
Consists of 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

31. V. Reaction Mechanism
Elementary step: each step in a reaction mechanism
Elementary step:
NO + NO N2O2
N2O2 + O NO2
Overall reaction:
2NO + O NO2 +
Intermediates
Species that appear in a reaction mechanism but not in the overall balanced equation
Always formed in an early elementary step and consumed in a later elementary step
Ex. _____ is the intermediate in the above reaction.
N2O2
Molecularity
Number of molecules participate in an elementary step
Unimolecular, bimolecular, termolecular reactions

32. Rate laws for elementary steps
exponent in the rate law = coefficient in the elementary step
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
Ex. Write the rate law for the the elementary step
NO (g) + O3(g) NO2 (g) + O2 (g)
rate = k [NO][O3]

33. Writing plausible reaction mechanisms
aA + bB cC + dD ?
When the exponent in rate law ≠ coeff in chemical reaction => multistep reaction
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.

34. Q:
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

35. rate = k[NO2][F2] is the rate law if step 1 is rdsQ:
The experimental rate law for the reaction between NO2 and F2 to produce NO2F is rate = k[NO2][F2]. What is the reaction mechanism?
What is the equation for the overall reaction?
2 NO2+ F NO2F
One possible mechanism?
rate = k[NO2][F2] is the rate law if step 1 is rds
rds
Step 1:
NO2 + F NO2F + F
Step 2:
NO2 + F NO2F
Overall reaction:
2 NO2+ F NO2F
What is the intermediate? F

36. Change the reaction rate without being used in the reaction
VI. Catalysis
Change the reaction rate without being used in the reaction
Homogeneous catalysis
Heterogeneous catalysis
Enzyme
Biological catalyst

37. 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 ’

38. 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 or gas.
- Acid catalysis
- Base catalysis

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

40. Enzyme Catalysis

41. Enzyme Catalysis E +S ES ES P + E D[P] rate = Dt rate = k [ES] enzyme
uncatalyzed
enzyme
catalyzed
rate =
D[P] Dt
E +S ES
ES P + E
rate = k [ES]

42. Rate law of enzyme catalysis
uncatalyzed
enzyme
catalyzed
E +S ES (fast equilibrium) k1
k-1
k1[E][S]= k-1[ES]
[ES] =
k1[E][S]
k-1 k2
ES P + E
rate = = k2 [ES] =
D[P] Dt
k1k2[E][S]
k-1