1. Chapter 12
Chemical Kinetics

2. Reactions In order to use chemical reactions commercially…
Stoichiometry
Energetics
Rate*
Spontaneity*

3. Spontaneity Defined as the inherent tendency for a process to occur
Does NOT mean fast - there are slow, spontaneous reactions

4. Rate Rate helps determine usefulness of reactions
Chemical Kinetics is the area of chemistry that deals with reaction rates
Rate Laws
Reaction Mechanisms
Simple Models for chemical reactions

5. Reaction Rates
Change in a given quantity over a specific period of time (for reactions, it deals with concentrations…mol/L…of reactants/products)
Rate = ∆[A] ∆t
∆[A] -> change in concentration of reactant or product
∆t -> change in time

6. A B
time
rate = -
D[A] Dt
rate =
D[B] Dt

7. Change in concentrations over change in time shows rate = slope of the line tangent to the curve.

8. For REACTANTS
Reactant concentrations will decrease with time, so rate expressions with reactants will end up with a negative sign…
…rates are positive

9. For PRODUCTS
Coefficients of balanced equation must be taken into account
Calculate the difference in rates of forward and reverse reactions
Generally, we study reactions soon after they reactants are mixed to avoid this problem
2NO2(g)  2NO(g) + O2(g)
Rate of consumption of NO2  Rate of production of NO + 2(rate of production of O2)

10. Neglecting the Reverse Reaction
Rate = k[x]n
k -> rate constant (determined experimentally)
[x] -> concentration of x
n -> reaction order (determined experimentally), generally positive whole numbers, can be fractions

11. 2H2(g) + O2(g) --> 2H2O(l)
Time(s)
[H2]
[O2]
[H2O]
0.0020
0.050
0.080
0.0040
0.025
0.0675
0.0250
0.0060
0.018
0.064
0.032
0.0080
0.0125
0.0612
0.0375
Example
2H2(g) + O2(g) --> 2H2O(l)
1. Based on the coefficients of the chemical equation alone, what is the rate of oxygen reaction between and seconds? (1.6 mol/L s)
2. What is the rate of water production during the same time period? (3.2 mol/L s)

12. Types of Rate Laws
Differential Rate Law (Rate Law): Expresses how the rate depends on concentration
Integrated Rate Law: Expresses how the concentration depends on time
The two are related - once we know one, we know the other

13. Which to Choose?
Depends on the experimental information available, ability to collect data, etc.

14. Determining the Form of A Rate Law (Differential Rate Law)
Determine what the relationship is between the concentration and rate using given data…
When the concentration of one reactant doubles, what happens to the rate?
If multiple reactants, each needs to be taken into account
Called the Method of Initial Rates
Rate = k[A]n[B]n…
K= rate constant, n= order of specific reactant

15. Data…

16. Results…
If both the [ ] and rate double or half, the value of n is 1 and it is a first order reaction for that particular reactant
If the [ ] doubles while the rate quadruples (rate increases with [ ]2, the value of n is 2 and it is a second order reaction for that particular reactant
If the [ ] doubles while the rate remains the same, the value of n is 0 and it is a zero order reaction for that particular reactant

17. More Results…
Overall reaction order is the sum of all reactants’ orders
Once you have the exponent values, plug any of the experiment’s data in for variables to solve for k.

18. Example

19. Initial Rate of Reaction (mol/L s)
Example #2
Determine n, p, and k for
rate = k[A]n[B]p
n=2, p=1, k=0.153L2/mol2s
Rxn
[A]0
[B]0
Initial Rate of Reaction (mol/L s) 1
0.100
1.53 X 10-4 2
0.300
4.59 X 10-4 3
0.200
6.12 X 10-4 4
3.06 X 10-4 5
0.600
8.26 X 10-3

20. Initial Rate of Reaction (mol/L s)
Rxn
[C]0
[D]0
[E]0
Initial Rate of Reaction (mol/L s) 1
0.400
0.300
0.560
7.14 X 10-4 2
0.100
0.500
0.200
4.55 X 10-5 3 4
0.750
1.28 X 10-3 5
3.57 X 10-4
Example #3
Determine the rate law, and solve for the order and value of the rate constant

21. Units for Rate Law Constant (k)
In a first order reaction, unit = 1/s
In a second order reaction, unit = L/mol s
In a third order reaction, unit = L2/mol2 s
In general, unit = L(n-1)/mol(n-1) s
Try with the three former example problems