1. Chemical Kinetics lecture no.8
CHE 101:Chemistry
Chemical Kinetics lecture no.8
Dr Harpreet Kaur
Lovely School of Engineering & School of Technology

2. Kinetics Studies the rate at which a chemical process occurs.
Besides information about the speed at which reactions occur, kinetics also sheds light on the reaction mechanism (exactly how the reaction occurs).

3. Outline: Kinetics Reaction Rates Rate Laws Integrated Rate Laws
How we measure rates.
Rate Laws
How the rate depends on amounts of reactants.
Integrated Rate Laws
How to calc amount left or time to reach a given amount.
Half-life
How long it takes to react 50% of reactants.
Arrhenius Equation
How rate constant changes with T.
Mechanisms
Link between rate and molecular scale processes.

4. Factors That Affect Reaction Rates
Concentration of Reactants
As the concentration of reactants increases, so does the likelihood that reactant molecules will collide.
Temperature
At higher temperatures, reactant molecules have more kinetic energy, move faster, and collide more often and with greater energy.
Catalysts
Speed rxn by changing mechanism.

5. Reaction Rates
Rxn Movie
Rates of reactions can be determined by monitoring the change in concentration of either reactants or products as a function of time. [A] vs t

6. The factors mentioned will affect the composition of the reaction mixture at any given time.
Therefore
The change in composition of the reaction mixture with time
is the rate of reaction, denoted by R, r or .
R is the same whether monitoring reactants or products
Generally,

7. Rate laws, rate constants, reaction order
R = ƒ ([A][B])
And, R  [A]m [B]n
With the use of a proportionality constant k, which is the rate constant (independent of conc. but dependent on temp.), R = -d[A]/dt = k [A]m [B]n
Such an equation is called the rate law

8. The order of the reaction = m + n
The exponents m and n are the order of the reaction with respect to reactant A and the order of the reaction with respect to reactant B respectively.
The order of the reaction = m + n
If m = n =1, then the reaction is first-order in A and first-order in B, but second-order overall, therefore: R = k [A][B]
Hence,
Units for rate constant
for 2nd order reaction
If first-order overall????
Units for rate constant
for 1st order reaction

9. Lecture no.9-10 9

10. S2O82- (aq) + 3I- (aq) 2SO42- (aq) + I3- (aq)
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)
rate = k [S2O82-]x[I-]y
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
y = 1
x = 1
rate = k [S2O82-][I-]
Double [I-], rate doubles (experiment 1 & 2)
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
13.2

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

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

13. Second-Order Reactions
rate = -
D[A] Dt
A product
rate = k [A]2
D[A] Dt
= k [A]2 -
rate
[A]2
M/s M2 =
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

14. Reaction Orders For the reaction: A →B, the rate law is: rate = k [A]m
Order (m)
Δ[A] by a factor of:
Effect on rate
Zero (0)
2, 4, 15, ½, etc.
None
1st (1) 2 2X 3 3X
2nd (2) 4X 9X ¼X

15. First-Order Processes
If a reaction is first-order, a plot of ln [A]t vs. t will yield a straight line with a slope of -k.
So, use graphs to determine rxn order.

16. First-Order Processes
Consider the process in which methyl isonitrile is converted to acetonitrile.
CH3NC
CH3CN
How do we know this is a first order rxn?

17. First-Order Processes
CH3NC
CH3CN
This data was collected for this reaction at 198.9°C.
Does
rate=k[CH3NC]
for all time intervals?

18. First-Order Processes
When ln P is plotted as a function of time, a straight line results.
The process is first-order.
k is the negative slope: 5.1  10-5 s-1.

19. Second-Order Processes
Similarly, integrating the rate law for a process that is second-order in reactant A:
Rearrange, integrate:
y = mx + b
also in the form

20. Second-Order Processes
So if a process is second-order in A, a plot of 1/[A] vs. t will yield a straight line with a slope of k.
First order:
If a reaction is first-order, a plot of ln [A]t vs. t will yield a straight line with a slope of -k.

21. Half-Life
Half-life is defined as the time required for one-half of a reactant to react.
Because [A] at t1/2 is one-half of the original [A],
[A]t = 0.5 [A]0.

22. Half-Life
For a first-order process, set [A]t=0.5 [A]0 in integrated rate equation:
NOTE: For a first-order process, the half-life does not depend on [A]0.

23. Half-Life- 2nd order For a second-order process, set
[A]t=0.5 [A]0 in 2nd order equation.

24. Lecture no. 11

25. Activation Energy
In other words, there is a minimum amount of energy required for reaction: the activation energy, Ea.
Just as a ball cannot get over a hill if it does not roll up the hill with enough energy, a reaction cannot occur unless the molecules possess sufficient energy to get over the activation energy barrier.

26. Temperature and Rate
Generally, as temperature increases, so does the reaction rate.
This is because k is temperature dependent.

27. Temperature Effects
Maxwell-Boltzman Distribution
At higher temperatures, more molecules will have adequate energy to react.
This increases the reaction rate.

28. The Collision Model
In a chemical reaction, bonds are broken and new bonds are formed.
Molecules can only react if they collide with each other.

29. The Collision Model
Furthermore, molecules must collide with the correct orientation and with enough energy to cause bond breakage and formation.

30. Reaction Coordinate Diagrams
It is helpful to visualize energy changes throughout a process on a reaction coordinate diagram like this one for the rearrangement of methyl isonitrile.

31. Reaction Coordinate Diagrams
It shows the energy of the reactants and products (and, therefore, E).
The high point on the diagram is the transition state.
The species present at the transition state is called the activated complex.
The energy gap between the reactants and the activated complex is the activation energy barrier.

32. Arrhenius Equation
Svante Arrhenius developed a mathematical relationship between k and Ea:
where A is the frequency factor, a number that represents the likelihood that collisions would occur with the proper orientation for reaction.

33. Arrhenius Equation
Taking the natural logarithm of both sides, the equation becomes 1 RT
y = mx + b
When k is determined experimentally at several temperatures, Ea can be calculated from the slope of a plot of ln k vs. 1/T.

34. Outline: Kinetics First order Second order complicated Rate Laws
Integrated Rate Laws
complicated
Half-life
k(T)

35. Lecture no.12

36. Reaction Mechanisms
The sequence of events that describes the actual process by which reactants become products is called the reaction mechanism.

37. Reaction Mechanisms
Reactions may occur all at once or through several discrete steps.
Each of these processes is known as an elementary reaction or elementary process.

38. Reaction Mechanisms
The molecularity of a process tells how many molecules are involved in the elementary process.
The rate law for an elementary step is written directly from that step.

39. Multistep Mechanisms
In a multistep process, one of the steps will be slower than all others.
The overall reaction cannot occur faster than this slowest, rate-determining step.

40. Slow Initial Step NO2 (g) + CO (g)  NO (g) + CO2 (g)
The rate law for this reaction is found experimentally to be
Rate = k [NO2]2
CO is necessary for this reaction to occur, but the rate of the reaction does not depend on its concentration.
This suggests the reaction occurs in two steps.

41. Slow Initial Step A proposed mechanism for this reaction is
Step 1: NO2 + NO2  NO3 + NO (slow)
Step 2: NO3 + CO  NO2 + CO2 (fast)
The NO3 intermediate is consumed in the second step.
As CO is not involved in the slow, rate-determining step, it does not appear in the rate law.

42. Fast Initial Step
The rate law for this reaction is found (experimentally) to be
Because termolecular (= trimolecular) processes are rare, this rate law suggests a two-step mechanism.

43. Fast Initial Step A proposed mechanism is Step 1 is an equilibrium-
it includes the forward and reverse reactions.

44. Fast Initial Step
The rate of the overall reaction depends upon the rate of the slow step.
The rate law for that step would be
But how can we find [NOBr2]?

45. Fast Initial Step NOBr2 can react two ways:
With NO to form NOBr
By decomposition to reform NO and Br2
The reactants and products of the first step are in equilibrium with each other.
Therefore,
Ratef = Rater

46. Fast Initial Step [NO] [Br2] = [NOBr2] Because Ratef = Rater ,
k1 [NO] [Br2] = k−1 [NOBr2]
Solving for [NOBr2] gives us k1
k−1
[NO] [Br2] = [NOBr2]

47. Fast Initial Step
Substituting this expression for [NOBr2] in the rate law for the rate-determining step gives