1. Unit 4 How do we model chemical change?
The central goal of this unit is to help you understand and apply ways of thinking that can be used to model chemical change in a system.
M1. Understanding Proportions
Determining the amount of substance formed or consumed.
M2. Tracking Energy
Predicting the amount of energy absorbed or released.
M3. Analyzing Rate and Extent
Identifying the factors that affect chemical transformations.

2. How do we model chemical change? Module 3: Analyzing Rate and Extent
Unit 4
How do we model chemical change?
Module 3: Analyzing Rate and Extent
Central goal:
To explore the factors that determine the rate and extent of chemical reactions.

3. Modeling How do I explain it?
The Challenge
Modeling How do I explain it?
We would like to generate models that allow us to answer questions such as this:
What amounts of reactants and products are involved?
How much energy will be needed or produced?
How fast will the process go and how can I control it?
To what extent will the reactants be changed into products?

4. Time Issues
When analyzing chemical reactions we need to pay attention to how substances change and how much energy is transferred, but also to how fast they go. T
Why do we need a spark? Why is this reaction fast?
Let′s think!
Let’s go back to our original system :
Spark
Fast

5. A Chemical Model
All these phenomena can be explained using the following model:
For a reaction to occur, the reactant particles must collide.
The more collisions in a given time, the faster the process.
A + B  AB
We can control collision frequency changing T, P, V, and N.

6. Effective Interactions
Colliding particles must be positioned so that the reacting groups interact effectively.
Effective collision +
NO2 O2
Consider this reaction:
NO + O3  NO2 + O2
Ineffective collision NO O3
The higher the % of configurations that lead to effective collisions (configuration effectiveness), the faster the process.

7. Effective Interactions
Relative Speeds?
Let’s Think

8. For example: NO2(g) + O2(g)  NO(g) + O3(g)
Reaction Steps
It is important to keep in mind that most chemical reactions don’t occur in one step, but several.
For example: NO2(g) + O2(g)  NO(g) + O3(g)
STEP 1 NO2  NO + O
STEP O + O2  O3
Despite this fact, the more complex a reacting mixture (#, type, size, geometry, distribution of molecules), the fewer the expected % of effective configurations that can lead to new products.

9. The smaller the Activation Energy, the faster the process.
3. Colliding particles must have enough energy to reach a transition state that leads to the formation of the new products.
The energy required to attain the transition state is called the Activation Energy (Ea).
Products
Reactants Ep
Reaction Coordinate
Energy Profile
Transition State Ea
The smaller the Activation Energy, the faster the process.

10. The lower Ea, the more particles have enough energy to react
Activation Energy
Typical kinetic energy distribution among particles at a fixed temperature
The lower Ea, the more particles have enough energy to react
Ea Rate

11. Build a complete energy profile for the process.
Let’s Think
Spark
Fast T
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g)
Based on our model, propose hypotheses to explain why the combustion of CH4 needs a spark and why it goes fast.
Build a complete energy profile for the process.
Methane CH4

12. A Combination of Factors
According to our model, the rate of a reaction is determined by these three factors:
Collision Frequency Configuration Effectiveness Activation Energy
Based on this model, predict the effect of increasing temperature (T), volume (V), and amount of reactants (NA, NB) on the rate of this reaction.
Let′s think!

13. Available at: http://www.chem.arizona.edu/chemt/C21/sim
Let’s Think
Available at:

14. Temperature
The kinetic energy distribution among particles changes with changing temperature.

15. Let’s Think The following two reactions have the same Ea: A + B  C
D + E  F Ep Ep
A + B F C
D + E
End first day
Reaction Coordinate
Reaction Coordinate
If [A]o= [B]o= [D]o= [E]o and T is held constant, what reaction will be faster?

16. The Challenge
Up to now, our models have allowed to generate qualitative answers to these questions:
How much product will we get if the process goes to completion?
How much energy will be needed or produced?
To what extent will the reactants be changed into products?
Let’s now consider the “extent or completion” issue.
How fast will the process go and how can I control it?

17. Reaction Extent
The extent of a reaction refers to the relative amounts of reactants and products that we find in the system once the process ends.
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g)
N2(g) + O2(g)  2 NO(g)
Consider these two reactions in a combustion engine:
People think that all reactions end when at least one of the reactants is consumed, but that is not always the case.
~100% completion
Around % stays as N2 and O2.

18. The concentration of N2O4 increases at low temperatures.
Reaction Extent
In some cases, the reaction extent depends on temperature or the relative concentrations of the different species in the system.
2 NO2(g)  N2O4(g)
Brown White
The concentration of N2O4 increases at low temperatures.
How do we explain this behavior?

19. Let’s Think Ep
1.2 kJ/mol
2 NO2(g)
57.2 kJ/mol
Observe the effect of different variables (# of molecules, V, T) on the concentrations of NO2 and N2O4 at equilibrium.
Discuss how we can use our model to explain what is going on in this system?
N2O4(g)

20. Reactants can become products, but products can also become reactants.
Back and Forth
Chemical reactions are dynamic processes.
Reactants can become products, but products can also become reactants.
A + B  AB
AB  A + B
A + B AB

21. End Point Ep
1.2 kJ/mol
2 NO2(g)
57.2 kJ/mol
2 NO2(g) N2O4(g)
As the reaction begins with pure NO2, the rate forward is higher than the rate backwards.
As more N2O4 is produced and NO2 is consumed, there is a point at which both rates become equal.
N2O4(g)

22. Let’s Think
t (s)
A (mol/L)
B (mol/L) 1
A  B
B A
B  A 2 3 4
Etc.
0.6 
 0
The extent of the reaction depends on the likelihood of the forward process compared to the backwards process.
0.4
0.6
0.4 x 0.6 = 0.24 
 0.6 x 0.3 = 0.18
0.34
0.66 ?
?  ?
?  ?
Imagine 60% of A goes to B, but only 30% B goes to A. When does it stop?
A B

23. Kinetic Factors Eaforward It is determined by: Activation Energy;Ep
It is determined by:
Activation Energy;
Collision Effectiveness;
in each direction.
Eaback
2 NO2(g)
N2O4(g)
The lower Ea, the more likely the transformation
The higher the % of effective configurations, the more likely the transformation.
2 NO2(g)  N2O4(g) favored by low Ea
N2O4  2 NO2(g) favored by effective configurations

24. The Outcome 2 NO2(g)  N2O4(g) favored by low Ea
N2O4  2 NO2(g) favored by effective configurations
Who wins?
Eaforward Ep
At low T, the process with lower Ea is more likely ( N2O4)
At high T, when there is plenty of energy available, the process with more effective configurations may be more likely ( NO2).
Eaback
2 NO2(g)
N2O4(g)

25. Reactant- or Product-Favored
Based on our analysis, exothermic reactions are likely to be product-favored.
A + 2 B C + D
Eaforward Ep
Eaback
A+ 2 B
However, the relative configuration effectiveness of the forward and backward processes may change the outcome.
C + D
End second day
Although Ea and configuration effectiveness cannot be changed without adding other substances, we can influence extent by changing reaction conditions (affecting the frequency and energy of collisions).

26. Let’s Think 2 NO2(g) N2O4(g)
Imagine this systems has reached equilibrium:
Predict what would happen if:
The volume is reduced;
More NO2 is added to the system;
T is increased.
Eaforward
Eaback
2 NO2(g)
N2O4(g)

27. Let’s Think

28. Shifting Equilibrium
Changing the temperature, pressure, or number of particles of the different species may affect the collision frequency of reactants and products in different ways. Ep
H2(g) + I2(g) HI(g)
2 HI
H2 + I2
In general, chemical equilibrium shifts in the direction of the process that becomes faster due to the change, until the forward and backward rates become the same again.

29. Shifting Equilibrium Ea
Changing temperature has a larger effect on the process with higher Ea (the endothermic process) Ep
2 HI
Thus, the endothermic process becomes more likely at high T and less likely at low T.
H2 + I2

30. Let’s Think
Consider two of the most important reactions happening inside a combustion engine:
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g)
N2(g) + O2(g)  2 NO(g) Ep Ep
CH4 + 2 O2
2 NO
CO2 + 2 H2O
N2 + O2
Would you expect them to be product-favored or reactant-favored? Would you expect the equilibrium to shift by changing T or V?

31. Let′s apply!
Assess what you know

32. Ozone Layer
“Ozone: Good Up High, Bad Nearby”
Ozone is a naturally occurring substance. Most of it (90%) is found in the lower parts of the stratosphere.
Ozone Layer
12 ppm
As an illustration of the low relative abundance of ozone in our atmosphere, one can imagine bringing all the
ozone molecules in the troposphere and stratosphere down to Earth’s surface and uniformly distributing these molecules into a gas layer over the globe. The resulting layer of pure ozone would have a thickness of less than half a centimeter (about one-quarter inch)
ppb
Stratospheric ozone screens out much of the Sun’s harmful UV radiation.
How does it work?

33. Predict
Let′s apply!
Ozone is naturally formed in the stratosphere through this chemical reaction:
3 O2(g) O3(g) Ep
Reaction Coordinate
2 O3
3 O2
DHrxn = 284 kJ/mol
Ea = kJ/mol
Based on this approximate energy profile, discuss whether you would expect this reaction to be product or reactant-favored (based on Ea and configuration effectiveness).

34. The formation of O3 occurs in two main steps:
Represent
Let′s apply!
The formation of O3 occurs in two main steps:
DHrxn = 498 kJ/mol Ea = kJ/mol
DHrxn = -107 kJ/mol Ea ~ 0 kJ/mol
per O3 formed.
Build the Energy Profile for this process. Discuss how increasing T, P, and O2 concentration (which happens as we move down the atmosphere) may affect the equilibrium.

35. Ozone Cycle Ozone is also naturally destroyed in the stratosphere.
Let′s apply!
Ozone is also naturally destroyed in the stratosphere.
Destruction
This photolytic step requires l <320 nm
Chapman’s Cycle
O2  O + O
O + O2  O3
Formation
O3  O2 + O
O3 + O  2 O2
Destruction
Every day, 3 x108 tons of O3 form and an equal mass decomposes (steady state)
UV light  Heat

36. UV-B Tanning and sunburn UV-A Wrinkling and aging
Ozone Protection
Let′s apply!
Energy
95% 5%
UV-B Tanning and sunburn
UV-A Wrinkling and aging

37. An ozone hole? We know the amount of O3 is changing
Let′s apply!
Concern about thinning of the ozone layer over Antarctica arose in the 1970’s but it was not monitored until in the 1980s
1 Dobson Unit = 0.01 mm at 0o C, 1 atm
We know the amount of O3 is changing
Observe carefully what happens over time.

38. Some Facts
Let′s apply!
The size of the ozone hole over Antarctica increased dramatically between
Radical Cl
STRONG CORRELATION:

39. CFCs
UV - C
Let′s apply!
Chlorofluorocarbons (CFCs) are the main source of Cl radicals in the atmosphere.
Cl
Originally used as:
Foaming agents for plastics;
Solvents for cleaning;
Aerosol propellants;
Refrigerants;
Fire extinguishers.

40. One single Cl radical can destroy over 100,000 O3 molecules!
Analyze
O3 Destruction
Let′s apply!
Natural
One single Cl radical can destroy over 100,000 O3 molecules! Ep
Ea = 17 kJ/mol
DHrxn = kJ/mol
Ea = 2.1 kJ/mol
O2 + ClO
Which of these routes is most favored?
O + O3
Cl + O3
Cl induced
+ O
2 O2 + Cl
Ea = 0.4 kJ/mol
DHrxn = kJ/mol
DHrxn = kJ/mol
2 O2
Reaction Coordinate

41. The Future
Let′s apply!
CFCs take fifteen years on average to reach the stratosphere, and they can remain in the atmosphere for up to one hundred years.
Used banned in 1987 Montreal Protocol
Estimated Recovery Up to 100 years from 2000.

42. Identify one topic you would have liked to learn more about related to this module’s topic.

43. Analyzing Rate and Extent
Summary
The rate of a reaction is determined by these three main factors: Collision frequency, configuration effectiveness, and activation energy. The higher the first two factors, and the lower the activation energy, the faster the process.
Thus, reactions tend to speed up with both increasing temperature and concentration of the reactive species.

44. Analyzing Rate and Extent
Summary
The extent of a chemical reaction depends on the likelihood of the conversion of reactants to products compared to the backwards process. The system reaches a state of chemical equilibrium when these two processes occur at the same rate.
The extent of a reaction is affected by temperature, pressure, and concentration of reactants and products. Chemical equilibrium shifts in the direction of the process that becomes faster due to the change

45. Are You Ready?

46. The Quest for Ammonia
Ammonia, NH3, is one of the most important industrial chemical substances.
It is widely used in the production of fertilizers, pharmaceuticals, refrigerants, explosives, and cleaning agents.
It ranks as one of the 10 top chemicals substances produced annually in the world.
“Let’s think” designed for students to brainstorm ideas.

47. Ammonia is mainly produced via this simple chemical reaction:
The Synthesis
Ammonia is mainly produced via this simple chemical reaction:
N2(g) H2(g) NH3(g)
Let′s think!
Balance this chemical reaction, complete the following particulate representation of the process, and identify the limiting reactant.
Limiting

48. How many tons of N2 are extracted from air each year?
The Amounts
Nitrogen for the synthesis of ammonia is obtained by separating it from air by liquefaction.
N2(g) + H2(g) NH3(g)
If approximately 100 million metric tons of ammonia are produced annually worldwide,
How many tons of N2 are extracted from air each year?
Let′s think!
1 ton = 1x103 kg = 1 x 106 g

49. Dissoc. Energy (KJ/mol)
Energy Transfer
The optimization of the industrial production of ammonia requires a good understanding of energy transfer during the process:
N2(g) + H2(g) NH3(g)
Estimate the heat of reaction for this process and determine whether the reaction is exo or endothermic.
Let′s think!
Bond
Dissoc. Energy (KJ/mol)
N-H
389
H-H
436
N≡N
945

50. Let’s Think N2(g) + H2(g) NH3(g)
Sketch the energy profile for this process and discuss whether the forward or backward processes can be expected to be favored.

51. The forward process has a lower Ea.
Competing Factors
N2(g) + H2(g) NH3(g)
The forward process has a lower Ea.
The backward process is expected to have a higher % of effective configurations.
Analyze the effect of increasing T and P on both reaction rate and reaction extent.
Let′s think!

52. WHAT TO DO? HIGH or LOW T? (Extent effects) HIGH or LOW P? (Costs)
Simulation
WHAT TO DO? HIGH or LOW T? (Extent effects) HIGH or LOW P? (Costs)
Let′s think!