1. Chapter 14 Chemical Kinetics

2. Introduction to Kinetics
Chemistry is concerned with change. It is concerned with how chemicals can react to form new substances.
Equally important, is how rapidly these chemical reactions occur. They can take fractions of a second (an explosion) or millions of years (diamond formation).

3. Kinetics
Kinetics is the area of chemistry concerned with the speed (or rate) of reactions (chemical processes).
Why are rates important? Some examples:
Rate of food spoiling
Rate of steel rusting
Rate of paint drying
Rate of aging

4. Kinetic Outline  You will learn:
1. Collision theory
What factors control rates.
2. Reaction Rates
How we measure rates.
3.Rate Laws
How the rate depends on amounts of reactants.
4. Integrated Rate Laws
How to calculate amount left or time to reach a given amount.
5. Half-life
How long it takes to react 50% of reactants.
6. Arrhenius Equation
How rate constant changes with temperature.
7. Reaction Mechanisms
How the reaction occurs at the molecular level.

5. Collision Theory In order for a chemical reaction to occur:
There must be collisions between the reactant molecules.
(More frequent collisions = faster rate)
2. The collisions must be affective.
3. To be affective, collisions must occur with:
a) sufficient energy.
b) suitable orientation.

6. Factors That Affect Reaction Rates
1. Physical State of the Reactants
The likelihood of collisions is higher if the reactants are mixtures of
homogeneous gases or liquids or solutions.
Solids react faster if their exposed surface area is increased.
2. Concentration of Reactants
As the concentration of reactants increases, so does the likelihood that reactant molecules will collide.
3. Temperature
At higher temperatures, reactant molecules have more kinetic energy, move faster, and collide more often and with greater energy.
4. Catalysts
Speed reactions without being used up by changing the reaction mechanism. (Examples: enzymes)

7. [ ] = concentration (Molarity) t = time (usually seconds)
Reaction Rates
Rates of reactions can be determined by monitoring the change in concentration of either reactants or products as a function of time.
Average Rate = [A] /t
= change
[ ] = concentration (Molarity)
A = the chemical
t = time (usually seconds)

8. Reaction Rates In this reaction, the concentration of butyl chloride, C4H9Cl, was measured at various times, t.
C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq)
Notice: The concentration of the reactant, C4H9Cl, decreases as time goes on.
What happens to the concentration of the products?

9. Average Rate
The average rate of the reaction over each interval is the change in concentration divided by the change in time:
average rate = M/s
You do the next time interval.

10. Average Rate = – M
100.0 – 50.0 s
Average Rate = M/s

11. Completed Chart
Notice that the average rate decreases as the reaction proceeds. Why? As the reaction goes forward, there are fewer collisions as the amount of reactant decreases.

12. Reaction Rates
A plot of concentration vs. time for this reaction yields a curve like this.
The slope of a line tangent to the curve at any point is the instantaneous rate at that time.

13. Reaction Rates
The reaction slows down with time because the concentration of the reactants decreases.
Instantaneous Rate = slope
Instantaneous Rate = Rise/Run
Instantaneous Rate = [A] /t

14. Reaction Rates and Stoichiometry
C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq)
In this reaction, the ratio of C4H9Cl to C4H9OH is 1:1.
Thus, the rate of disappearance of C4H9Cl is the same as the rate of appearance of C4H9OH.
Rate =
-[C4H9Cl] t =
[C4H9OH]

15. Reaction Rates and Stoichiometry
What if the ratio is not 1:1?
H2(g) + I2(g)  2 HI(g)
HI will appear twice as fast as H2 and I2 disappear.
To correct for this, you must equate the rates.
Negative sign for reactants! Why?

16. General Equation for Equating Rates
For the reaction
aA + bB
cC + dD
Reactants (decrease)
Products (increase)

17. You try: 4 A (g) + 3 B (g)  7 C (g) + 2 D (g)
Which chemical appears or disappears the fastest?
Why?

18. Given: 4 A (g) + 3 B (g)  7 C (g) + 2 D (g)
Sample Problem
Given: 4 A (g) B (g)  7 C (g) D (g)
If the rate of disappearance of A = 0.48 mol/s, what is the rate of appearance of C?
Easiest Way: set up proportion
4 moles A = 7 moles C
0.48 mol/s x
Cross-multiply 4 x = 7 (0.48)
4 x = 3.36
x = 0.84
answer = 0.84 mol/s