1. Some reactions occur is several sequential steps.
A mechanism is a step by step description of how a reaction occurs.
A mechanism must agree with all kinetic information and this is one of the main purposes of the study of kinetics-to determine the mechanism for a reaction.
An example of a multi step mechanism is shown below.
Step 1: NO → N2O (fast)
Step 2: N2O H2 → N2O H2O (slow)
Step 3: N2O H2 → N H2O (fast)
The slowest step in a mechanism is the rate determining step. The kinetics of a reaction are always based on the rate determining step.
How can we find the overall reaction being described by this mechanism?

2. Adding the reactions together allows us to identify the overall reaction.
Step 1: NO → N2O (fast)
Step 2: N2O H2 → N2O H2O (slow)
Step 3: N2O H2 → N H2O (fast)
Overall Reaction: NO H2 → N H2O
Notice that some compounds that are present in the mechanism steps do not appear in the overall reaction equation.
Any substance that is created in one step and is consumed in a later step is called an intermediate.
If a substance is consumed in one step and is recreated in a later step, it would be called a catalyst and not an intermediate.
If all reactants, intermediates, catalysts and products are in the same phase, the reaction is said to be Homogeneous.

3. Identify any Intermediates in the mechanism below:
Identify any Catalysts in the mechanism below:
Step 1: A BC → AB C (slow)
Step 2: AB D → AD B (fast)
Step 3: B + C → BC (fast)
Overall Reaction: A D → AD
Remember: Catalysts and Intermediates do not appear as reactants or products in the overall reaction.
Remember: Intermediates are created in an early step and then are consumed in a later step while catalysts are consumed in an early step and are then recreated in a later step.
Intermediate:
Catalyst:
AB C B BC

4. Collision Theory:
In order for two molecules to react, three key things must happen.
The two molecules must collide with each other.
The orientation of the collision must cause the correct atoms to hit each other.
The collision must provide enough energy to break any bonds that must be broken.

5. Based on the concepts of collision theory, what factors can influence how fast a reaction occurs.
Since two molecules must collide with each other, anything that
affects how often molecules collide will influence the rate of reaction.
concentration, temperature, the surface area (for solids)
Since the collision must provide enough energy to break any bonds that must be broken, anything that affects the energy of a collision or the strength of a bond will influence the rate of reaction.
temperature, nature of the reactants (strength of the bonds or how reactive)
Since the orientation of the collision must cause the correct atoms to hit each other, anything that affects the orientation of molecules will influence the rate of reaction.
shape of the molecule, some catalysts (enzymes)

6. Which orientation or orientations can lead to a reaction?
Example of the importance of Collision Orientation:
2 ONBr (g) → 2 NO (g) Br2 (g)
The ONBr molecules can theoretically collide with many different orientations.
Since Br ends up bonded to Br, the ONBr molecules must hit each other in such a way to ensure that the Br atoms collide in order for the reaction to have a chance to occur.
Three possible orientations are shown below.
Which orientation or orientations can lead to a reaction?

7. A Reaction Coordinate Diagram (Reaction Pathway) shows the energy changes associated with a reaction and identifies reactants, intermediates (if any) and products for the reaction.
The distance from the energy of the reactants to the top of the hill is called the energy of activation (Ea). Ea is the minimum energy that must be provided by a collision in order for the reaction to actually happen.
The distance from the energy of the reactants to the energy of the products is called the heat of reaction (DHRxn). If the products are lower than the reactants, the reaction is exothermic.
The substance shown at the top of the hill is called the transition state or activated complex.

8. The lower Ea is, the faster the reaction is because more collisions will provide enough energy to cause the reaction to happen.

9. Temperature: the temperature of a reactions impacts how fast the reaction goes because the temperature impacts how fast the molecules are moving. If molecules are moving faster, they will hit harder and therefore a collision is more likely to have an energy greater than Ea.
The area under the red curve and past the Ea is greater than the area under the blue curve that is past Ea.
Higher temperature means that there are more collisions with sufficient energy to cause a reaction.
Increasing the temperature by 10oC approximately doubles the rate of a reaction.

10. Catalysts lower the activation energy, so the reaction goes faster.
Catalysts: Catalysts make reactions go faster by lowering the activation energy for a reaction. If the Ea is lower, more collisions will have sufficient energy to cause the reaction to occur.
Catalysts can affect:
Strength of bonds in the reactants;
The orientation of molecules;
The relative concentration of molecules.
Any or all of these can be used to explain why a reaction goes faster with a catalyst than without. All of these mean that the activation energy is being lowered because of the presence of the catalyst.
Catalysts lower the activation energy, so the reaction goes faster.

11. The same concept from the perspective of the number of effective collisions.
Since the Ea is lower in the catalyzed reaction, more collisions have sufficient energy to cause the reaction-the reaction is faster.

12. Practice Drawing Reaction Coordinate diagrams (Reaction Pathway) for each of the following situations: One step reaction that is endothermic; one step reaction that is exothermic; two step reaction with both steps being exothermic; etc… Label all important locations in each diagram.

13. Completely label the diagram below: you should label the following: Reactants, Intermediates, Products, Activation Energies, Transition States, and DH’s

14. Rate: how much of something happens per unit time.
In other words, how fast something happens.
Example:
a car is traveling at 55 mi/hr
So the car covers 55 mi of distance in 1 hour of time.
Reaction Rate: how fast a reaction occurs.
This is measured by determining how much reactant disappears in a unit of time or by determining how much product appears in a unit of time. This is expressed as a change in concentration per unit of time.
Concentration is usually in molarity (M or mol/L) and time is usually in seconds,
so the units for rate are usually M/s or mol/L·s.
Reaction rates frequently change as the reaction proceeds, so rate is usually expressed as an average rate.
Average Rate =
D (Concentration)
D (time)

15. Consider the following reaction:
2 NO2 (g) → NO (g) O2 (g)
At the instant the reaction begins, there is no NO or O2 in the container and the rate of reaction of NO2 is at its highest.
As the reaction continues, what happens to the concentration of NO2?
The concentration of NO2 decreases because it is a reactant.
As the reaction continues, what happens to the concentrations of NO and O2?
The concentration of NO and O2 increase because they are products.
Based on the stoichiometry of the reaction, which product’s concentration will increase the fastest, or will they increase at the same rate?
The concentration of NO increases twice as fast as the concentration of O2.

16. 2 NO2 (g) → NO (g) O2 (g)
Notice that the concentration of NO2 is decreasing over time, the concentrations of NO and O2 are both increasing over time, but NO is increasing twice as fast as O2.
The triangles represent the change in concentration over the time interval indicated (110s for NO2, and 70s for NO and O2 respectively). This is used to calculate the average rate for each of these compounds.

17. D(Concentration) = Concentration at time 2 – Concentration at time 1
D(time) = time 2 – time where time 2 > time 1
For a reactant, Concentration at time 2 < Concentration at time 1,
so a negative sign is used to make the rate a positive value.
D (Concentration)
For a reactant:
Average Rate = -
D (time)
For a product, Concentration at time 2 > Concentration at time 1,
so no negative sign is needed to make the rate a positive value.
D (Concentration)
For a product:
Average Rate =
D (time)

18. Example of Average Rate
C4H9Cl (l) H2O (l) → C4H9OH (l) HCl (aq)
At the start of the reaction, the concentration of C4H9Cl was M After 4.00 seconds, the concentration of C4H9Cl was M. Calculate the average rate of reaction of C4H9Cl.
D (Concentration)
Average Rate = -
D (time)
- (0.100 M – M)
Average Rate =
( 4.00 s – 0.00 s)
Average Rate = M/s or mol/L·s

19. Why would the average rate decrease over time?
How do the ideas of collision theory support the fact that the reaction slows down as it proceeds?

20. Rate Law: an equation that expresses the rate of reaction in terms of the concentration
k is a specific rate constant
[A] is the concentration of substance “A”
Rate = k[A]
If “A” is a reactant, this equation implies that the rate goes down as the reaction proceeds because the concentration of a reactant decreases as a reaction occurs. Below are some examples of reactions and the rate laws that have been determined by kinetic experiments.
3 NO → N2O O Rate = k[NO]2
NO2 + CO → NO CO Rate = k[NO2]2
2 NO2 → 2 NO O Rate = k[NO2]2
2 H2O2 → 2 H2O O Rate = k[H2O2]
Notice that the numbers (superscripts) in the rate laws do not always agree with the coefficients in the reaction.

21. Generic Rate Law: a generic rate law can be written for any reaction.
a A + b B + c C → d ABC
Rate = k[A]m[B]n[C]p
Again, k is a rate constant that is specific to the reaction and the temperature. The superscripts (m, n, p) are called “order” of reaction and are usually integer numbers 0, 1, or 2. The order for a given reactant can not be determined by looking at the overall reaction (i.e. m does not have to equal a and n does not have to equal b and p does not have to equal c). If there were only two reactants in a reaction, then the [C]p term would not be needed. If there were four reactants in a reaction, then an additional term ([D] q) would be needed.
The overall order for a reaction is the sum of the superscripts
Overall Order = m + n + p

22. Write a generic rate law for the reaction below:
2 NO (g) O2 (g) → NO2 (g)
Generic Rate Law: Rate = k[NO]m[O2]n
Notice that “m” and “n” are unknown and can not be predicted from the coefficients of the balanced reaction.
Once you have the generic rate law, you must have experimental data to determine the values for the order.
We can use this data to determine what happens to the rate when the concentration of a reactant is changed. Doing this allows us to determine order.

23. ( ) ( ) ( ) Rate1 = k[NO]1m[O2]1n Rate2 = k[NO]2m[O2]2n
An alternative to the “look and see” method used by the book is to determine the order from the experimental data using mathematics.
Rate1 = k[NO]1m[O2]1n
Generic rate law for experiment 1
Rate2 = k[NO]2m[O2]2n
Generic rate law for experiment 2
Divide one equation by the other equation:
This simplifies to: (
Rate [NO]2 m [O2] n
Rate [NO] [O2]1 = )
Rate k[NO]2m[O2]2n
Rate k[NO]1m[O2]1n =
Now, if [O2]1 = [O2]2, this simplifies to:
which simplifies to:
Rate [NO] n
Rate [NO]1 = ( ) (
Rate [NO] m n
Rate [NO]1 = ) 1
Choosing two experiments where the initial concentration of a specific reactant is held constant will let you determine the order of reaction with respect to the other reactant.

24. (
Rate [NO] m n
Rate [NO]1 = ) 1
Rate [NO] m
Rate [NO]1 = ( ) m = ( )
2 = 2m so the only value of m that works is 1 and m = 1 (
Rate [O2] n
Rate [O2]2 = ) n = ( )
4 = 2n so the only value of n that works is 2 and n = 2

25. 2 NO (g) O2 (g) → NO2 (g)
From the previous slide, we know that: Rate = k[NO][O2]2
Once the orders are known, the data from the table to find the rate constant for the reaction. We can pick any trial and each one should give the same value for k. Below we will use the data from trial 3.
Rate = k[NO][O2]2
0.016 = k (0.200)(0.200)2
So k = 2.0 L2/mol2 s
Once the orders and k are known for a reaction, we can predict the initial rate of a reaction assuming that we know the initial concentrations.
What would be the rate of reaction if the initial concentration of NO was M and the initial concentration of O2 was M?

26. An example: Find the order of reaction with respect to NH4+ and NO2-1
A generic rate law can be written for the reaction: Rate = k[NH4+]m[NO2-]n
Dividing reaction 2 by reaction 1 would allow you to determine the value of “n” because the concentration of NH4+ is the same in these two reactions.
Dividing reaction 3 by reaction 2 would allow you to determine the value of “m” because the concentration of NO2- is the same in these two reactions.

27. For the reaction: 2A + 2B → 2C + D
The rate law was determine to be: Rate = k[A] 2[B]
What is the order of reaction with respect to A?
What is the order of reaction with respect to B?
What is the overall order of reaction?
If the initial rate of disappearance of A was mol/Ls, what would the initial rate of appearance of D be?
If in a new experiment the concentration of A was doubled and the concentration of B was doubled, what would happen to the rate of reaction?
If in a new experiment the concentration of A was tripled while the concentration of B was unchanged, what would happen to the rate of reaction?

28. A mechanism is a step by step description of how a reaction occurs.
A mechanism must agree with all kinetic information and this is one of the main purposes of the study of kinetics-to determine the mechanism for a reaction.
An example of a multi step mechanism is shown below.
Step 1: A B ↔ AB (slow)
Step 2: AB C → AC D (fast)
The slowest step in a mechanism is the rate determining step. The kinetics of a reaction are always based on the rate determining step.
We can write rate laws for mechanism steps (we know what molecules are actually colliding). The experimental rate law must agree with the mechanism we have proposed.
Is the rate law: Rate = k[A]2 consistent with the mechanism?

29. Summary of mechanism steps and their associated rate laws.
The “Molecularity” is the same concept as order. A unimolecular step is first order.
A bimolecular step is second order, and a termolecular step is third order.
The molecularity is the number of molecules that must collide in a given reaction step.

30. Write the rate law for the reaction below and find the value of the rate constant.