1. Answer the following questions:
Define the term “spontaneous reaction”.
Which of the following reactions do you think are spontaneous?
1) A nail rusting.
2) The decomposition of carbon in diamond to graphite.
3) The combination of hydrogen and oxygen to form water.
4) The decomposition of hydrogen peroxide.
5) The combustion of butane.
What does the spontaneity of a reaction indicate about the speed at which the reaction occurs?

2. Chemical Kinetics

3. Kinetics The study of reaction rates.
Spontaneous reactions are reactions that will happen - but we can’t tell how fast. (Spontaneity implies nothing about speed)
For example: Diamond will spontaneously turn to graphite – so slow it is not detectable.
Reaction mechanism- the steps by which a reaction takes place.

4. Factors that Affect Reaction Rates
Nature of the reactants
Concentration of the reactants
Temperature
Presence of a catalyst

5. Collision Theory of Reaction Rates
For a reaction to occur, particles must collide.
Increasing the concentration or reactants results in a greater number of collisions and therefore, a faster rate.
Not all collisions are effective collisions.

6. Effective Collisions For a collision to be effective:
1) particles must possess a necessary minimum energy to break bonds and form new ones (activation energy)
2) particles must have proper orientation at the time of the collision. (see page 554)

7. Reaction Rate
Defined as the change in the concentration of a reactant or product per unit time.
Rate=Conc.of A at t2-Conc. of A at t1
t2 – t1
Rate = D[A] Dt

8. For this reaction: N2 + 3 H2  2 NH3
As the reaction progresses the concentration H2 goes down
C O N C E N T R A T I O N
[H2]
TIME

9. As the reaction progresses the concentration N2 goes down 1/3 as fast
TIME

10. As the reaction progresses the concentration NH3 goes up.

11. Calculating Rates Average rates are taken over long intervals
Instantaneous rates are determined by finding the slope of a line tangent to the curve at any given point because the rate can change over time

12. Defining Rate
We can define rate in terms of the disappearance of the reactant or in terms of the rate of appearance of the product.
In our example: N2 + 3H2  2NH D[N2] = -3D[H2] = 2D[NH3] Dt Dt Dt

13. Rate Laws Reactions are reversible.
As products accumulate they can begin to turn back into reactants.
Initially, the rate will depend on only the amount of reactants present.
Therefore, reactions must be studied at a point soon after the reactants are mixed.
This is called the Initial rate method.

14. Rate Laws
The concentration of the products do not appear in the rate law because this is an initial rate.
The “order” must be determined experimentally, and can’t be obtained from the equation
The “order” indicates to what degree the concentration affects the rate.

15. Types of Rate Laws
Differential Rate law - describes how rate depends on concentration (often simply called rate law)
Integrated Rate Law - describes how concentration depends on time.
For each type of differential rate law there is an integrated rate law and vice versa.
Rate laws can help us better understand reaction mechanisms.

16. Reaction Mechanism
The step by step pathway by which a reaction occurs is called its mechanism.
The individual steps are called elementary steps.
The rate law for an elementary step can be written from its molecularity.
Molecularity is defined as the number of species that must collide to produce the reaction indicated by that step.

17. Examples of Elementary Steps
Molecularity
Rate Law
A  products
Unimolecular
Rate = k[A]
A + A  products
Bimolecular
Rate =k[A]2
A + B products
Rate =k[A][B]
A + A + B products
Termolecular
Rate=k[A]2[B]
A + B + Cproducts
Rate=k[A][B][C]

18. 2 NO2  2 NO + O2
The rate will only depend on the concentration of the reactants.
Rate = k[NO2]n
This is called a rate law expression or differential rate law.
k is called the rate constant.
n is the “order” of the reactant -usually a positive integer and must be determined experimentally
Overall reaction order is the sum of the orders for each individual reactant.

19. Integrated Rate Laws
Integrated Rate Laws express the reactant concentration as a function of time (instead of rate as a function of the reactant concentration).
We will only be considering reactions involving a single reactant when determining integrated rate laws.

20. First-Order Rate Laws Consider the following reaction:
2N2O5  4NO O2
The rate law (differential) is determined to be
Rate = k[N2O5]
This means that if the concentration of N2O5 is doubled, the rate of production of the products is also doubled.

21. Integrated First-Order Rate Law
The previous differential rate law can be integrated and as a result, written in the following form:
ln[N2O5] = -kt + ln[N2O5]0
Where:
ln indicates the natural log
t is the time
[N2O5] is the concentration at time “t”
[N2O5]o is the initial concentration

22. Integrated First-Order Rate Laws
All integrated first order rate laws take the following general form:
ln[A] = -kt + ln[A]o
Items to note:
1) The equation shows how the concentration of A depends on time

23. Integrated First-Order Rate Laws ln[A] = -kt + ln[A]o
2) The above equation is of the form y=mx + b, where a plot of y vs. x is a straight line with slope m and intercept b.
In the above rate law:
y=ln[A] x=t m=-k b=ln[A]o
As a result, plotting ln[A] vs. time always gives a straight line. (This fact is often used to test whether a reaction is 1st order or not)
3) The integrated rate law for a 1st order reaction can also be written as:
ln([A]o/[A]) = kt

24. Half-Life
The time required for a reactant to reach half of its original concentration is called the half-life of a reactant and is designated with the symbol t1/2.
The general equation for the half-life of a first order reaction is
t1/2 =0.693/k
(Half-life does not depend on concentration)

25. 2nd Order Rate Laws Consider the following reaction:
2C2H6 (g)  C8H12(g)
The rate law (differential) is determined to be
Rate = k[C2H6]2
This means that if the concentration of C2H6 is doubled, the rate of production of the products is quadrupled.

26. Integrated Second-Order Rate Laws
The previous differential rate law can be integrated and as a result, written in the following form:
1/[C4H6] = kt + 1/[C4H6]o
All integrated second order rate laws have the form:
1/[A] = kt + 1/[A]o
Note:
1) A plot of 1/[A] vs. time produces a straight line with a slope=k.
2) The half-life equation for a 2nd order reaction is: t1/2 = 1/k[A]o

27. Zero-Order Rate Laws
The rate law for a zero order reaction is: rate = k
This means that changing the concentration has no effect on the rate of the reaction.
Example: N2O5  2NO /2O2
The integrated rate law for a zero order reaction is
[A] = -kt + [A]o
Note:
1) A plot of [A] vs. time results is a straight line
2) the half life equation is:
t1/2 = [A]0/2k

28. Interpreting Differential Rate Laws

29. A + B + C  products The rate law is determined to be rate = k[A][B]2
What happens to the reaction rate
when we make the following changes
to the concentrations?

30. The concentration of A is doubled, while B and C remain unchanged.
The rate remains the same.
The rate doubles.
The rate is cut in half.
The rate is quadrupled.
The rate is increased by a factor of 8. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32
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31. The concentration of B is doubled while A and C remain unchanged.
The rate remains the same.
The rate doubles.
The rate is cut in half.
The rate quadruples.
The rate increases by a factor of 8. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32
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32. The concentration of C is doubled while A and B remain unchanged.
The rate remains the same.
The rate is doubled.
The rate is cut in half.
The rate is quadrupled.
The rate is increased by a factor of 8. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32
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33. The concentration of all three reactants are doubled simultaneously.10
The rate remains the same.
The rate is doubled.
The rate is cut in half.
The rate is quadrupled.
The rate increases by a factor of 8. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32

34. The concentration of A is cut in half while B and C are doubled.
The rate remains the same.
The rate is doubled.
The rate is cut in half.
The rate is quadrupled.
The rate increases by a factor of 8. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32
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35. Which of the following is the rate law for a first order reaction?
Rate = k[A]
Rate = k[A]2
Rate = k[A][B]
Rate = k[A]2[B]
Rate = k[A]2[B]2 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32
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36. Which is a rate law that is second order with respect to B?
Rate = k[A]
Rate = k[A]2
Rate = k[A][B]
Rate = k[A]2[B]
Rate = k[A]2[B]2 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32
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37. Which is the rate law for a third order reaction?
Rate = k[A]
Rate = k[A]2
Rate = k[A][B]
Rate = k[A]2[B]
Rate = k[A]2[B]2 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32
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38. Rate = k[A] Rate = k[B]2 Rate = k[A][B] Rate = k[A]2[B]
Which rate law represents a reaction in which the initial rate will increase by a factor of eight when [A] and [B] are doubled?
Rate = k[A]
Rate = k[B]2
Rate = k[A][B]
Rate = k[A]2[B]
Rate = k[A]2[B]2 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32
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39. Which rate law represents a reaction in which the rate will increase by a factor of two when [A] and [B] are doubled?
Rate = k[A]
Rate = k[B]2
Rate = k[A][B]
Rate = k[A]2[B]
Rate = k[A]2[B]2 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32
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40. Which rate law represents a reaction in which the rate will not change when [A] is doubled and [B] is held constant?
Rate = k[A]
Rate = k[B]2
Rate = k[A][B]
Rate = k[A]2[B]
Rate = k[A]2[B]2 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32
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41. The rate of reaction and the rate constant will increase.
A reaction occurs according to the following rate law: rate = k[A]. If the temperature of the reaction chamber were increased, which of the following would be true?
The rate of reaction and the rate constant will increase.
The rate of reaction and the rate constant will not change.
The rate of reaction will increase and the rate constant will decrease.
The rate of reaction will increase and the rate constant will not change.
The rate of reaction will not change and the rate constant will increase. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32
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42. Determining the form of a rate law (Method of Initial Rates)
The initial rate is the instantaneous rate right after the reaction begins (t=0)
Eliminates the effect of the reverse reaction.
The initial concentrations of the reactants are varied.
The reaction rate is determined for each trial.
See the example problem on p. 536

43. A + 2B  C Trial Initial [A] Initial [B]
Initial Rate of formation of [C] 1
0.010 M
1.5 x 10-6 M/s 2
0.020 M
3.0 x 10-6 M/s 3
6.0 x 10-6 M/s

44. The rate law for the reaction is
Rate = k [A][B]
Rate = k [A]2[B]
Rate = k [A][B]2
Rate = k [A]2[B]2 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32
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45. Reaction Mechanism (continued)
In most mechanisms, one step is slower than the others.
A reaction can never occur faster than its slowest step.
The slow step is the rate-determining step.

46. Determining Possible Reaction Mechanisms
The balanced equation for the overall reaction is equal to the sum of all of the individual steps.
The rate law expression matches the coefficients of the rate determining step.

47. Example NO2 + CO  NO + CO2 rate = k[NO2]2 One proposed mechanism:
NO2 + NO2  N2O4 (slow)
N2O4 + CO  NO + CO2 + NO2 (fast)
N2O4 is an intermediate (forms in one step and is consumed in a later step)
This proposed mechanism is consistent with the experimentally determined rate law.

48. Another possible mechanism
NO2 + NO2  NO3 + NO (slow)
NO3 + CO  NO2 + CO2 (fast)
NO2 + CO  NO + CO2
In order to be a possible mechanism the following two criteria must be met:
1) The individual steps must add up to the overall reaction
2) The mechanism is consistent with the experimentally determined rate law expression

49. Equilibrium Step
An equilibrium step is a step in which a product can rapidly re-form the reactants to reach equilibrium.
The concentration of products is equal to the concentration of the reactants at equilibrium.
An equilibrium step is indicated by a double arrow. (< -- > )
Substitutions can be made in the rate law expression between the reactants and the products.

50. Example 2NO + Br2  2NOBr rate = k[NO]2[Br2] Proposed mechanism:
NO + Br2 < --> NOBr2 (fast)
NOBr2 + NO  2NOBr (slow)
2NO + Br2  2NOBr
Rate = k[NOBr2][NO] can be substituted with
Rate =k[NO][Br2][NO] or Rate=k[NO]2[Br2]

51. A multi step reaction takes place by the following mechanism: A + B  C + D A + C  D + E Which of the species shown is an intermediate in the reaction? A B C D E 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32
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52. Rate = k[NO][O2] Rate = k[NO3][NO] Rate = k[NO]2[O2] Rate = k[NO2]2
Which of the following is the rate law for the predicted mechanism for the production of NO2? NO + O2 <-> NO3 (fast) NO3 + NO  2NO2 (slow)
Rate = k[NO][O2]
Rate = k[NO3][NO]
Rate = k[NO]2[O2]
Rate = k[NO2]2 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32
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53. Homework problems to be completed by tomorrow
Chapter 12: problems 51-53, 55, 65, 75, and 82

54. Example Problem p
Complete homework problems 32,34,39-42, 44, 46, 48, and 89.