1. Chemical Kinetics and the Nucleus, a Chemist’s View
Chapter 12 and 19

2. Homework
Due with test
AP Practice Test at the end of chapter 12 between

3. Kinetics
Chemical kinetics is the study of the changes in concentrations of reactants or products as a function of time.
It is the study of reaction rates.
Each reaction has its own characteristic rate.
The reactions may be slow, fast, or reversible.

4. What affects reactions rates
Concentration: molecules must collide in order to react.
 Physical state: molecules must mix in order to collide.
Temperature: molecules must collide with enough energy to react.
The use of a catalyst/inhibitor

5. Reaction Rate
Change in concentration of a reactant or product per unit time.
[A] means concentration of A in mol/L; A is the reactant or product being considered.

6. The Decomposition of Nitrogen Dioxide

7. Decomposition of Nitrogen Dioxide
NO2 NO + O2

8. Rate terms
A reaction rate is the change in concentration of a reactant or a product per unit of time.
An average rate is the change in concentration of reactants (or products) over a finite time period.
An instantaneous rate is the reaction rate at a particular time, given by the slope of a tangent to a plot of reactant concentration vs. time. 

9. Rate terms
An initial rate is the instantaneous rate at the point at which the reactants are mixed, that is, at t = 0.

10. Rate Law
Rate law- An expression showing how the rate depends on the concentrations of reactants.
For the decomposition of nitrogen dioxide:
2NO2(g) → 2NO(g) + O2(g)
Rate = k[NO2]n:
k = rate constant
n = order of the reactant

11. Rate Law
Rate = k[NO2]n
The concentrations of the products do not appear in the rate law because this particular reaction rate is being studied under conditions where the reverse reaction does not contribute to the overall rate.

12. Rate Law
Rate = k[NO2]n
The value of the exponent n must be determined by experiment; it cannot be written from the balanced equation.

13. Units of k
For whatever reason, the AP test likes to ask for the correct units of k.
This changes depending on the order of the reaction!
Just remember, the units must cancel out in the rate law equation!
Rate is M/t (where t is some unit of time, normally seconds), concentration is molarity

14. Units of k, if t is seconds
1st order rate = k [A]
M/s = ? M
k is s-1 or 1/s
2nd order rate = k [A]2
M/s = ? M2
k is M-1s-1 or 1/(Ms)
0 order rate = k
M/s = ?
k is M/s

15. question Determine k with the correct units the following rate law
Rate = k [NO]2[Cl2]
k= 180 M-2min-1 or
k= 180 L2/mol2 min
rate
[NO]
[Cl2]
.18 M/min
.10 M

16. Reaction order The “n” variable in the expression
A reaction order is a positive or negative exponent, for a reactant, for which the concentration is raised to in a rate law.
For most problems we will deal with, it will be positive integer.

17. 2 types of rate laws
The differential rate law (often called simply the rate law) shows how the rate of reaction depends on concentration.
The integrated rate law shows how the concentrations of species in the reaction depend on time.

18. Because the differential and integrated rate laws for a given reaction are related in a well-defined way, the experimental determination of either of the rate laws is sufficient.
 Experimental convenience usually dictates which type of rate law is determined experimentally.

19. Knowing the rate law for a reaction is important mainly because we can usually infer the individual steps involved in the reaction from the specific form of the rate law.

20. Determining the form of the rate law
The first step in understanding how a given chemical reaction occurs is to determine the form of the rate law.
We must determine experimentally the power to which each reactant concentration must be raised in the rate law.
An exponent of “1” is referred to as first order.
An exponent of “2” is referred to as second order.

21. Determining the form of a rate law
For 2N2O5(aq) → 4NO2 (aq) + O2(g)
Rate = k[N2O5]n
So x10-4 = k(.90)n
And 2.7x10-4 = k(.45)n
[N2O5]
Rate (M/s)
.90 M
5.4 x10-4
.45 M
2.7x10-4

22. Solving Using substitution 5.4x10-4 / (.90)n = k 2.7x10-4 = k(.45)n
2.7x10-4 = (.45)n 5.4x10-4 / (.90)n
2.7x10-4 /5.4x10-4= (.45/.90)n
.5 = .5n

23. solving
.5 = .5n
n=1
You can use logarithms to solve for n, but since it is 1 it is easy to see.
log (.5) = n log (.5)
n = log (.5)/ log (.5)= 1
All problems should have easy numbers to work with (i.e. 1, 2, 3)

24. To solve for k Just plug your value for n into either equation
5.4x10-4 M/s = k(.90M)1
2.7x10-4 M/s = k(.45M)1
Either way
k = 6.0x10-4 s-1

25. Determine the form of a rate law
For 2NOCl(g) → 2NO2 (g) + Cl2(g)
[NOCl]
Rate (M/s)
5.0 x10-4 M
1.35 x10-4
1.67 x10-4 M
5.0x10-6

26. Method of Initial Rates
Initial rate is the instantaneous rate at t = 0.
The overall reaction order is the sum of the orders for the various reactants.

27. Overall Reaction Order
The sum of the exponents in the reaction rate equation.
Rate = k[A]n[B]m
Overall reaction order = n + m
k = rate constant
[A] = concentration of reactant A
[B] = concentration of reactant B

28. Example Problem
From the experimental data below, determine the differential rate law.
NH4+(aq) + NO2- (aq) N2(g) + H2O(g)
Experiment
Initial rate
M/s
[NH4+]
[NO2-] 1
1.35x10-7
.100 M
.0050 M 2
2.70x10-7
.010 M 3
5.40x10-7
.200 M

29. Solving Using Rate = k[A]n[B]m We know 1.35x10-7 =k(.100)n (.0050)m
Now with substitution

30. Solving for m 1.35x10-7 / (.0050)m =k(.100)n
2.70x10-7 =k(.100)n (.010)m
2.70x10-7 = (.010)m 1.35x10-7 / (.0050)m
2.70x10-7 / 1.35x10-7= (.010 / .0050)m
2 = 2m
m = 1

31. Solving for n 2.70x10-7 /(.100)n =k (.010)m
5.40x10-7 = (.200)n 2.70x10-7 /(.100)n
2 = 2n
n = 1

32. Order of the reaction Order of reaction = n + m 1 + 1 = 2
To solve for k plug these values in
1.35x10-7 M/s =k(.100M)1(.0050M)1
2.70x10-7 M/s =k(.100M)1 (.010M)1
k = 2.7x10-4 M-1s-1

33. Example For the reaction below, determine the experimental rate law.
NO2(g) + CO(g) NO(g) + CO2(g)

34. Another
Determine the rate law for
 H2 + I2 2HI