1. Ch #16: Kinetics
Kinetics: Rates and Mechanisms
of Chemical Reactions

2. Chemical kinetics:
Study of reaction rates, changes in concentrations of reactants or products as a function of time.

3. Factors that influence reaction rate:
1.Concentration - molecules must collide to react;
Rate α collision frequency α conc
2.Physical state - molecules must mix to collide;
Smaller the particle size, greater the surface area and more the collisions.
3.Temperature - molecules must collide with enough energy to react;
Rate α collision frequency α conc.

4. The effect of surface area on reaction rate.
Figure 16.3
The effect of surface area on reaction rate.

5. Collision energy and reaction rate.
Figure 16.4
Collision energy and reaction rate.

6. Factors that influence reaction rate:
4.The use of a catalyst.

7. Expressing the Reaction Rate
reaction rate - changes in the concentrations of reactants or products per unit time
reactant concentrations decrease while product concentrations increase
For a reaction A → B
Rate of reaction = -∆[A]/∆t
[A]= conc of A in mol/L
∆=change

8. Expressing the Reaction Rate
Units= moles per liter per second.
Mol L-1s-1 or mol/ L.s
Change I product conc is positive so the rate is Rate of reaction = ∆[B]/∆t
Rate decreases during course of reaction.
Instantaneous rate: Rate at a particular instant: considering closer values.

9. Expressing the Reaction Rate
Use initial rate as soon as the reactants are present( no products at this time)
For a reaction aA + bB →cC+ dD,
Rate =-1/a∆[A]/∆t=-1/b∆[B]/∆t==1/c∆[C]/∆t==1/d∆[A]/∆t

10. Table 16.1 Concentration of O3 at Various Time in its
Reaction with C2H4 at 303K
Time (s)
Concentration of O3 (mol/L)
0.0
3.20x10-5
 (conc A) - t
10.0
2.42x10-5
20.0
1.95x10-5
30.0
1.63x10-5
40.0
1.40x10-5
50.0
1.23x10-5
60.0
1.10x10-5

11. The concentrations of O3 vs. time during its reaction with C2H4
Figure 16.5
The concentrations of O3 vs. time during its reaction with C2H4
rate = -
 [C2H4] t = -
 [O3] t

12. Plots of [C2H4] and [O2] vs. time.
Figure 16.6
Plots of [C2H4] and [O2] vs. time.
Tools of the Laboratory

13. Problem 1
P.1.2H2 (g) +O2 (g)→ 2H2O (g) Write the rate equation in terms of reactants and products. If [O2] is decreasing at 0.23 mol/L .s at what rate is [H2O] increasing?

14. The Rate Law and its components:
Rate = k[A]m [B]n
K= rate constant(does not change as reaction proceeds)
M and n are reaction orders.
Coefficients are not related to reaction orders.
Rate constant and orders can only be found by experimental data.

15. Determining the Initial Rate:
By performing expt and collecting data.

16. Reaction order: Rate=k[A]- 1st order Rate = k[A]2- 2nd order
Rate = k[A]0- Zero order OR Rate=k(1)=k (not dependent on conc of A)

17. Problem 2
P.2.For each of the following reactions, determine the reaction order with respect to each reactant and the overall order from the given rate law.
2NO(g) + O2(g) → 2NO2(g); rate = k[NO]2[O2]
CH3CHO(g) → CH4(g) + CO(g); rate = k[CH3CHO]3/2

18. Problem 2
H2O2(aq) + 3I-(aq) + 2H+(aq) → I3-(aq) + 2H2O(l); rate = k[H2O2][I-]

19. Determining Reaction Orders:
P.3.NO2(g) + CO (g) → NO (g) + CO2(g) rate=k[NO2]m [CO]n

20. Sample Problem 16.3
Determining Reaction Order from Initial Rate Data
PROBLEM:
Many gaseous reactions occur in a car engine and exhaust system. One of these is
rate = k[NO2]m[CO]n
Use the following data to determine the individual and overall reaction orders.
Experiment
Initial Rate(mol/L*s)
Initial [NO2] (mol/L)
Initial [CO] (mol/L) 1 2 3
0.0050
0.080
0.10
0.40
0.20
PLAN:
Solve for each reactant using the general rate law using the method described previously.
SOLUTION:
rate = k [NO2]m[CO]n
First, choose two experiments in which [CO] remains constant and the [NO2] varies.

21. Sample Problem 16.3
Determining Reaction Order from Initial Rate Data
continued
[NO2] 2
[NO2] 1 m =
rate 2
rate 1
k [NO2]m2[CO]n2
k [NO2]m1 [CO]n1 =
The reaction is 2nd order in NO2.
0.40
0.10 = m
0.080
0.0050 ;
16 = 4m and m = 2
[CO] 3
[CO] 1 n =
rate 3
rate 1 =
k [NO2]m3[CO]n3
k [NO2]m1 [CO]n1
The reaction is zero order in CO.
0.20
0.10 n
0.0050 = ;
1 = 2n and n = 0
rate = k [NO2]2[CO]0 = k [NO2]2

22. Determining the Rate constant:
Calculated from collected data.
Units of Rate constant depend on order.

23. Table 16.4 An Overview of Zero-Order, First-Order, and
Simple Second-Order Reactions
Zero Order
First Order
Second Order
Rate law
rate = k
rate = k [A]
rate = k [A]2
Units for k
mol/L*s
1/s
L/mol*s
[A]t =
k t + [A]0
Integrated rate law in straight-line form
ln[A]t =
-k t + ln[A]0
1/[A]t =
k t + 1/[A]0
Plot for straight line
[A]t vs. t
ln[A]t vs. t
1/[A]t = t
Slope, y-intercept
k, [A]0
k, 1/[A]0
-k, ln[A]0
Half-life
[A]0/2k
ln 2/k
1/k [A]0

24. Integrated rate laws and reaction order
Figure 16.7
Integrated rate laws and reaction order
1/[A]t = kt + 1/[A]0
[A]t = -kt + [A]0
ln[A]t = -kt + ln[A]0

25. Problem 5
P.5At 1000 o C, cyclobutane (C4H8) decomposes in a first-order reaction, with the very high rate constant of 87s-1, to two molecules of ethylene (C2H4).
a) If the initial C4H8 concentration is 2.00M, what is the concentration after s?
(b) What fraction of C4H8 has decomposed in this time?

26. Rules for orders through graphs
If a straight line is produced with the foll:
Ln[reactant] vs time-1st
1/[reactant] vs time-2nd
[reactant]vs time -0 order.

27. Half Life
Half Life: Time required for the reactant concentration to reach half of its initial value.

28. A plot of [N2O5] vs. time for three half-lives.
Figure 16.9
A plot of [N2O5] vs. time for three half-lives.
t1/2 =
for a first-order process
ln 2 k
0.693 =

29. Problem 6
P.6 Cyclopropane is the smallest cyclic hydrocarbon. Because its 600 bond angles allow poor orbital overlap, its bonds are weak. As a result, it is thermally unstable and rearranges to propene at 10000C via a first-order reaction:
The rate constant is 9.2s-1; (a) What is the half-life of the reaction? (b) How long does it take for the concentration of cyclopropane to reach one-quarter of the initial value?

30. Effect of Temperature:
Temperature affects rate by affecting rate constant.
Arrhenius equ: k=Ae-Ea/RT
K=rate const, e=base of natural logs, T=absolute temp, R= Universal gas const, Ea=activation energy (minimum energy that molecules must have to react)

31. Effect of Temperature:
As T increases, k increases, rate increases.
Ln k =ln A- Ea/R(1/T)
A plot of ln k vs 1/T gives a straight line
Slope=-Ea/R
Y-int= ln A
Ln K2/k1=-Ea/R(1/T2-1/T1)

32. The Effect of Temperature on Reaction Rate
The Arrhenius Equation
where k is the kinetic rate constant at T
Ea is the activation energy
R is the energy gas constant
T is the Kelvin temperature
ln k = ln A - Ea/RT
A is the collision frequency factor ln k2 k1 = Ea RT - 1 T2 T1

33. Figure 16.11
Graphical determination of the activation energy
ln k = -Ea/R (1/T) + ln A

34. Problem 7
P.7.The decomposition of hydrogen iodide 2HI(g) → H2(g) + I2(g)
has rate constants of 9.51x10-9L/mol*s at 500. K and 1.10x10-5 L/mol*s at 600. K. Find Ea.

35. Figure 16.12
Information sequence to determine the kinetic parameters of a reaction.
Series of plots of concentra-tion vs. time
Initial rates
Determine slope of tangent at t0 for each plot
Compare initial rates when [A] changes and [B] is held constant and vice versa
Reaction orders
Rate constant (k) and actual rate law
Substitute initial rates, orders, and concentrations into general rate law: rate = k [A]m[B]n
Find k at varied T
Activation energy, Ea
Rearrange to linear form and graph
Rate constant and reaction order
Find k at varied T
Use direct, ln or inverse plot to find order
Integrated rate law (half-life, t1/2)
Plots of concentration vs. time

36. Effects of concentration and temperature:
Collision Theory: Reactant particles must collide with each other.
Concs are multiplied in rate law.
Ea: Energy required to activate the molecules into a state from which reactant bonds can change into product bonds.
Only those collisions with enough energy to exceed Ea can lead to reaction.

37. The dependence of possible collisions on the product
Figure 16.13
The dependence of possible collisions on the product
of reactant concentrations. A B
4 collisions A B A B
Add another molecule of A
6 collisions A B A A B
Add another molecule of B A B A B

38. Effects of concentration and temperature:
Rise in temp enlarges the fraction of collisions with enough energy to exceed Ea.
F= e-Ea/RT e-base of natural log, T=temp, R=gas const.
In exothermic reaction: Eaf > Ear
In endothermic reaction: Eaf < Ear
: k=Ae-Ea/RT A is the frequency factor, A=pZ, Z=collision frequency and p=orientation probability factor.

39. The effect of temperature on the distribution of collision energies
Figure 16.14
The effect of temperature on the distribution of collision energies

40. Energy-level diagram for a reaction
Figure 16.15
Energy-level diagram for a reaction
Collision Energy
Collision Energy
ACTIVATED STATE
Ea (forward)
Ea (reverse)
REACTANTS
PRODUCTS
The forward reaction is exothermic because the reactants have more energy than the products.

41. Figure 16.17
The importance of molecular orientation to an effective collision.
NO + NO NO2
A is the frequency factor
Z is the collision frequency
p is the orientation probability factor
A = pZ where

42. Transition state theory:
Transition state/activated complex: neither reactant nor product present, a transitional species with partial bonds is present.
Ea is the energy required to stretch and deform bonds in order to reach transition state.
Reaction energy diagram : TB. Pg.699

43. Nature of the transition state in the reaction between CH3Br and OH-.
Figure 16.18
Nature of the transition state in the reaction between CH3Br and OH-.
CH3Br + OH CH3OH + Br -
transition state or activated complex

44. Reaction energy diagram for the reaction of CH3Br and OH-.
Figure 16.19
Reaction energy diagram for the reaction of CH3Br and OH-.

45. Reaction energy diagrams and possible transition states.
Figure 16.20
Reaction energy diagrams and possible transition states.

46. Molecularity of a reaction:
Elementary steps make up the reaction mechanism.
An elementary step is not made up of simpler steps.
Molecularity means the number of reactant particles involved in the step.

47. Molecularity of a reaction:
In an elementary step equation coeff= order.
Reaction order= molecularity
A→ product Unimolecular Rate = k[A]
2A→ product Bimolecular Rate = k[A]2
A + B→ product Bimolecular Rate = k[A][B]
2A + B→ product Termolecular Rate =k [A]2[B]

48. Table 16.6 Rate Laws for General Elementary Steps
REACTION MECHANISMS
Table Rate Laws for General Elementary Steps
Elementary Step
Molecularity
Rate Law
A product
Unimolecular
Rate = k [A]
2A product
Bimolecular
Rate = k [A]2
A + B product
Bimolecular
Rate = k [A][B]
2A + B product
Termolecular
Rate = k [A]2[B]

49. Problem 8
P.8. The following two reactions are proposed as elementary steps in the mechanism of an overall reaction: NO2Cl(g) → NO2(g) + Cl(g) NO2Cl(g) + Cl(g)→ NO2(g) + Cl(g) (a) Write the overall balanced equation.
NO2Cl(g) → NO2(g) + Cl(g)
NO2Cl(g) + Cl(g)→ NO2(g) + Cl(g)

50. Problem 8 (a) Write the overall balanced equation.
Determine the molecularity of each step.
Write the rate law for each step

51. Rate-Determining step:
Also called as rate limiting step.
It represents rate law for overall reaction.
Slow reaction is a rate determining reaction.
The elementary steps must add up to the overall equation.
The elementary steps must be physically reasonable.
The mechanism must correlated with the rate law.

52. Catalysis: Each catalyst has its own specific way of functioning.
In general a catalyst lowers the energy of activation.
Lowering the Ea increases the rate constant, k, and thereby increases the rate of the reaction
A catalyst increases the rate of the forward AND the reverse reactions.

53. Catalysis:
A catalyzed reaction yields the products more quickly, but does not yield more product than the uncatalyzed reaction.
A catalyst lowers Ea by providing a different mechanism, for the reaction through a new, lower energy pathway