1. Chemical Kinetics Chapter 15
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2. Chapter 15; Kinetics Factors Controlling Rate
Two Mathematical Rate Descriptions
Change Concentration/ Change in Time
Rate Law
Experimental Determination
Reaction Order

3. Kinetics – how fast does a reaction proceed?
Chemical Kinetics
Kinetics – how fast does a reaction proceed?
Chemical Reaction; Reactant Bonds Broken, Product Bonds Made
NO2 (g) + CO (g) -> NO (g) + CO2 (g) N O C +
13.1

4. How to increase the rate of reaction?

5. Molecular Collision Theory
Incorrect Orientation
Reactants must collide with
correct orientation
enough energy, Ea O C N O O
Correct Orientation O C N

6. 13.4 A + B C + D Exothermic Reaction Endothermic Reaction
The activation energy (Ea) is the minimum amount of energy required to initiate a chemical reaction.
13.4

7. Activation Energy..
..is energy needed so there is enough energy to break reactant bonds.
..is the energy needed to get molecules in the correct orientation.
..is the energy needed to reach the transition state or activated complex

8. Factors for Reaction Rate
Temperature
Concentration

9. Concentration
Reaction rate is the change in the concentration of a reactant or a product with time (M/s).
A B
rate = -
D[A] Dt
D[A] = change in concentration of A over
time period Dt
rate =
D[B] Dt
D[B] = change in concentration of B over
time period Dt
13.1

10. Reaction Rates and Stoichiometry
2A B
Two moles of A disappear for each mole of B that is formed.
Rate = -
D[A] Dt 1 2
rate =
D[B] Dt
aA + bB cC + dD
rate = -
D[A] Dt 1 a
= -
D[B] Dt 1 b =
D[C] Dt 1 c =
D[D] Dt 1 d
13.1

11. Write the rate expression for the following reaction:
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (g)
rate = -
D[CH4] Dt
= -
D[O2] Dt 1 2 =
D[CO2] Dt =
D[H2O] Dt 1 2
13.1

12. A B
time
rate = -
D[A] Dt
rate =
D[B] Dt
13.1

13. Algebra Review of Slopes of Straight Lines y = mx +b; m = slope Dy/Dx
(+) positive slope
(-) negative slope y y x x

14. Algebra Review of Slopes of Straight Lines y = mx +b; m = slope Dy/Dx
More vertical; higher slope
More horizontal; lower slope y y x x
Note; both are positive slopes

15. Curved Rather than Straight Line Tell Us That.....
Rate is not Constant Throughout Reaction
Reaction Rate is Higher with Higher Concentration
(-) slope means concentration
is decreasing.

16. Two Mathematical Expressions to Describe Reaction Rate
A B
Rate = -A/ t =; + B/ t
Determined from stoichiometry
Uses both reactants & products
Rate Law; rate =k[A]x
Determined by experimental data- Stoichiometry of equation is irrelevant
Only reactants in rate law

17. The Rate Law
The rate law expresses the relationship of the rate of a reaction to the rate constant and the concentrations of the reactants raised to some powers.
aA + bB cC + dD
Rate = k [A]x[B]y
k is the Rate Law Constant
x and y are determined experimentally,
and do not depend on stoichiometric
coefficients from balanced equation
13.2

18. The Rate Law
The rate law expresses the relationship of the rate of a reaction to the rate constant and the concentrations of the reactants raised to some powers.
aA + bB cC + dD
Rate = k [A]x[B]y
Reaction Order
reaction is xth order in A
reaction is yth order in B
reaction is (x +y)th order overall
Reaction order tells how quickly rate
will increase when concentration increases
13.2

19. Reaction Order Tells How Changing Reactant Concentration Affects Rate
Oth Order; rate =k[A]0; double [A], rate same
1st Order; rate = k[A]1; double [A], double rate
2nd Order; rate =k[A]2; double [A]; quadruple rate
3rd Order; rate =k[A]3; double [A]; rate 8X

20. Rate Laws Rate laws are always determined experimentally.
Reaction order is always defined in terms of reactant (not product) concentrations.
The order of a reactant is not related to the stoichiometric coefficient of the reactant in the balanced chemical equation.
F2 (g) + 2ClO2 (g) FClO2 (g) 1
rate = k [F2][ClO2]
13.2

21. Determining the Form of the Rate Law

22. How Data is created

23. Method of Initial Rates
Used to find the form of the rate law
Choose one reactant to start with
Find two experiments where the concentration of that reactant changes but all other reactants stay the same
Write the rate laws for both experiments
Divide the two rate laws
Solve for the order
Follow the same technique for other reactants

24. Example Choose one reactant to start with NH4+
Find two experiments where the concentration of that reactant changes but all other reactants stay the same
Exp 2 & 3

25. Example Write the rate laws for both experiments
Exp 2: Rate = 2.70x10-7 = k(0.100)x(0.010)y
Exp 3: Rate = 5.40x10-7 = k(0.200)x(0.010)y
Divide the two rate laws
0.50 = 0.50x
Use log rules to solve for the order
x = 1 so the order for NH4+ is one

26. Example Follow the same technique for other reactants NO2-: Exp 1 & 2
Exp 1: Rate = 1.35x10-7 = k(0.100)1(0.0050)y
Exp 2: Rate = 2.70x10-7 = k(0.100)1(0.010)y
0.5 = 0.5y
y = 1
So Rate = k[NH4+]1[NO2-]1
Overall Reaction Order – sum of orders of reactants

27. Finding k We can find k using values from any of the experiments given
Units will be different for k depending on order of reactants

28. Example
BrO3- : Exp 1 & 2
Exp 1: Rate = 8.0x10-4 = k(0.10)x(0.10)y(0.10)z
Exp 2: Rate = 1.6x10-3 = k(0.20)x(0.10)y(0.10)z
0.50 = 0.50x
x = 1

29. Example
Br- : Exp 2 & 3
Exp 2: Rate = 1.6x10-3 = k(0.20)1(0.10)y(0.10)z
Exp 3: Rate = 3.2x10-3 = k(0.20)1(0.20)y(0.10)z
0.50 = 0.50y
y = 1

30. Example H+ : Exp 1 & 4 Exp 1: Rate = 8.0x10-4 = k(0.10)1(0.10)1(0.10)z
0.25 = 0.50z OR ¼ = (½)z
z = 2

31. Example So Rate = k[BrO3-]1[Br-]1[H+]2 Solve for rate constant, k
Overall order of reaction = 4
Solve for rate constant, k

32. Double [F2] with [ClO2] constant
F2 (g) + 2ClO2 (g) FClO2 (g)
rate = k [F2]x[ClO2]y
Double [F2] with [ClO2] constant
Rate doubles
x = 1
rate = k [F2][ClO2]
Quadruple [ClO2] with [F2] constant
Rate quadruples
y = 1
13.2

33. 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)
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
rate = k [S2O82-]x[I-]y
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

34. Types of Rate Laws
Differential Rate law - describes how rate depends on concentration.
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.

35. Integrated Rate Laws Consider a simple 1st order rxn: A  B
Differential form:
How much A is left after time t? Integrate:

36. Integrated Rate Laws The integrated form of first order rate law:
Can be rearranged to give:
[A]0 is the initial concentration of A (t=0).
[A]t is the concentration of A at some time, t, during the course of the reaction.

37. Integrated Rate Laws Manipulating this equation produces… y = mx + b
…which is in the form
y = mx + b

38. 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.

39. 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?

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

41. 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.

42. 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

43. 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.

45. Determining rxn order
The decomposition of NO2 at 300°C is described by the equation
NO2 (g)
NO (g) + 1/2 O2 (g)
and yields these data:
Time (s)
[NO2], M
0.0
50.0
100.0
200.0
300.0

46. Determining rxn order Graphing ln [NO2] vs. t yields:
The plot is not a straight line, so the process is not first-order in [A].
Time (s)
[NO2], M
ln [NO2]
0.0
-4.610
50.0
-4.845
100.0
-5.038
200.0
-5.337
300.0
-5.573
Does not fit:

47. Second-Order Processes
A graph of 1/[NO2] vs. t gives this plot.
This is a straight line. Therefore, the process is second-order in [NO2].
Time (s)
[NO2], M
1/[NO2]
0.0
100
50.0
127
100.0
154
200.0
208
300.0
263

48. 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.

49. 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.

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

51. Outline: Kinetics First order Second order complicated Rate Laws
Integrated Rate Laws
complicated
Half-life