1. Two Types of Rate Laws
Differential- Data table contains RATE AND CONCENTRATION DATA. Uses “table logic” or algebra to find the order of reaction and rate law
Integrated- Data table contains TIME AND CONCENTRATION DATA. Uses graphical methods to determine the order of the given reactant. K=slope of best fit line found through linear regressions

2. Integrated Rate Law
Can be used when we want to know how long a reaction has to proceed to reach a predetermined concentration of of some reagent

3. Graphing Integrated Rate Law
Time is always on x axis
Plot concentration on y axis of 1st graph
Plot ln [A] on the y axis of the second graph
Plot 1/[A] on the y axis of third graph
Your are in search of a linear graph

4. Results of linear graph
Zero order: time vs concentration= line
y= mx+ b
[A]= -kt + [A0 ]
A- reactant A,
A0 - initial concentration of A at t=0
l slope l= k, since k cannot be negative, and k will have a negative slope
Rate law will be rate=k[A]0

5. Results of linear graph
First order: time vs ln [ ]= line
y= mx+ b
ln [A]= -kt + ln [A0 ]
A- reactant A,
A0 - initial concentration of A at t=0
l slope l= k, since k cannot be negative, and k will have a negative slope
Rate law will be rate=k[A]1

6. Results of linear graph
second order: time vs 1/ [ ]= line
y= mx+ b
1/[A]= kt + 1/ [A0 ]
A- reactant A,
A0 - initial concentration of A at t=0
k=slope
Rate law will be rate=k[A]1

7. First-Order Processes
Consider the process in which methyl isonitrile is converted to acetonitrile.
CH3NC
CH3CN
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8. First-Order Processes
CH3NC
CH3CN
This data were collected for this reaction at C.
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9. First-Order Processes
When ln P is plotted as a function of time, a straight line results.
Therefore,
The process is first-order.
k is the negative of the slope: 5.1  105 s1.
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10. Second-Order Processes
The decomposition of NO2 at 300 °C is described by the equation
NO2(g)
NO(g) O2(g) 1 2
and yields data comparable to this table:
Time (s)
[NO2], M
0.0
50.0
100.0
200.0
300.0
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11. Using the graphing calculator
L1=time
L2=concentration-- if straight line, zero order
L3=ln concentration-- if straight line- 1st order
L4= 1/concentration– if straight line—2nd order
Perform 3 linear regressions

12. Second-Order Processes
The plot is a straight line, so the process is second-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
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13. Second-Order Processes1
[NO2]
Graphing vs. t, however, gives this plot Fig. 14.9(b).
Because this is a straight line, the process is second-order in [A].
Time (s)
[NO2], M
1/[NO2]
0.0
100
50.0
127
100.0
154
200.0
208
300.0
263
Add problem after this slide.
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14. Determine the rate law and calculate k
What is the concentration of N2 O5 at 600s?
At what time is the concentration equal to M?

15. The decomposition of N2 O5 was studied at constant temp 2 N2 O5 (g)  4 NO2(g) + O2(g)
Time (s)
0.1000
0.0707 50
0.0500
100
0.0250
200
0.0125
300
400

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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.
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17. Half-Life For a first-order process, this becomes 0.5 [A]0 [A]0 ln
= kt1/2
ln 0.5 = kt1/2
0.693 = kt1/2
= t1/2
0.693 k
Note: For a first-order process, then, the half-life DOES NOT DEPEND ON CONCENTRATION!!!!!!!!

18. Half-Life First order decay is what is seen in radioactive decay
0.693 k
This is the equation used to calculate the half-life of a radioactive isotope

19. Problem
Exercise
A certain first-order reaction has a half-life of 20.0 minutes.
a. Calculate the rate constant for this reaction.
b. How much time is required for this reaction to be 75% complete?
3.47 × 10−2 min−1; minutes

20. Half-Life For a second-order process, 1 0.5 [A]0 = kt1/2 + [A]0 2 [A]0
Add problem after this slide.
2  1
[A]0
= kt1/2 1 =
= t1/2 1
k[A]0
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21. Half life zero order

22. Half life first order

23. Half life second order

24. Summary table order zero First second Rate Law Rate=k rate-=k[ A]
Integrated rate law in form of y=mx+b
[A]t = -kt + [Ao ]
ln[A]t = -kt + ln[Ao ]
Rate law in data packet on AP exam
Does not appear
ln[A]t -ln[Ao ] = -kt
1/[A]t -1/[Ao ] = kt
slope
Slope = -k
Slope= -k
slope=k
Half-life
[A0 ]/2k
0.693/k
1/kA0

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Reaction Mechanisms
The molecularity of a process tells how many molecules are involved in the process.
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Reaction Mechanisms
Chemical reactions proceed via a sequence of distinct stages.
The sequence is known as the mechanism and each part of the mechanism is known as a “step”.
The rate of the reaction is only dependant upon the slowest step, also known as the rate determining step or RDS.
only reactants that appear in the rate determining step appear in the rate equation and vice-versa
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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.
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28. Example #1 The reaction below W + Y  Z Has the following mechanism
W  R slow
R + Y Q fast
Q  Z fast
Here the rate only depends on the concentration of _______________
and therefore the rate equation only contains this reactant
The rate equation is, ______________ in the rate determining step is 1.
Note: R and Q are not reactants or products, but are rather they are called ________________________, produced in one step, but then are used up in a subsequent step.
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Note:
1. In all valid mechanisms the sum of the individual fast and slow steps must be the same as the overall chemical equation.
2.The stoichiometric coeffiecient of a substance that appears in the slow step is the power that the
concentration of that substance is raised to in the rate equation.
3. If a substance is present at the beginning of a reaction AND present in the same form at the
end of the reaction, it can be identified as a catalyst
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30. 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.
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The reaction below
A + B C + D
Has the following mechanism
A  Q fast equilibrium
Q + B  C + D slow
Here the slow step contains Q and B, and _____is an intermediate
Can intermediates be featured the rate equation? _____________
Since the formation of Q is dependent on A, Q can be replaced by A in the rate equation. Therefore the rate equation is given as_________________ The orders w.r.t A and B are________________________________
since the stoichiometric coefficient of B in the rate determining step is 1, and the stoichiometric coefficient of A (which Q depends upon) is also 1.
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Temperature and Rate
Generally, as temperature increases, so does the reaction rate.
This is because k is temperature-dependent.
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33. Reaction Coordinate Diagrams
The diagram 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.
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34. Maxwell–Boltzmann Distributions
Temperature is defined as a measure of the average kinetic energy of the molecules in a sample.
At any temperature there is a wide distribution of kinetic energies.
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35. Maxwell–Boltzmann Distributions
As the temperature increases, the curve flattens and broadens.
Thus, at higher temperatures, a larger population of molecules has higher energy.
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36. Maxwell–Boltzmann Distributions
If the dotted line represents the activation energy, then as the temperature increases, so does the fraction of molecules that can overcome the activation-energy barrier.
As a result, the reaction rate increases.
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37. Maxwell–Boltzmann Distributions
This fraction of molecules can be found through the expression
where R is the gas constant and T is the Kelvin temperature.
−Ea/RT
f = e
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Arrhenius Equation
Svante Arrhenius developed a mathematical relationship between k and Ea:
k = Ae
where A is the frequency factor, a number that represents the likelihood that collisions would occur with the proper orientation for reaction.
−Ea/RT
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Arrhenius Equation
Taking the natural logarithm of both sides, the equation becomes
ln k =  ( ) + ln A Ea R 1 T
Add problem(s) after this slide.
y = mx + b
Therefore, if k is determined experimentally at several temperatures, Ea can be calculated from the slope of a plot of ln k vs. . 1 T
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Catalysts
Catalysts increase the rate of a reaction by decreasing the activation energy of the reaction.
Catalysts change the mechanism by which the process occurs.
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