1. Chemical Kinetics
The area of chemistry that concerns reaction rates and reaction mechanisms.

2. Reaction Rate
The change in concentration of a reactant or product per unit of time

3. Reaction Rates: 1. Can measure disappearance of reactants
2NO2(g)  2NO(g) + O2(g)
Reaction Rates:
1. Can measure
disappearance of
reactants
2. Can measure
appearance of
products
3. Are proportional
stoichiometrically

4. Reaction Rates: 4. Are equal to the slope tangent to that point
2NO2(g)  2NO(g) + O2(g)
Reaction Rates:
4. Are equal to the
slope tangent to
that point
5. Change as the
reaction proceeds,
if the rate is
dependent upon
concentration
[NO2] t

5. Rate Laws
Differential rate laws express (reveal) the relationship between the concentration of reactants and the rate of the reaction.
The differential rate law is usually just called “the rate law.”
Integrated rate laws express (reveal) the relationship between concentration of reactants and time

6. Writing a (differential) Rate Law
Problem - Write the rate law, determine the value of the rate constant, k, and the overall order for the following reaction:
2 NO(g) + Cl2(g)  2 NOCl(g)
Experiment
[NO]
(mol/L)
[Cl2]
Rate
Mol/L·s 1
0.250
1.43 x 10-6 2
0.500
5.72 x 10-6 3
2.86 x 10-6 4
11.4 x 10-6

7. Writing a Rate Law
Part 1 – Determine the values for the exponents in the rate law:
R = k[NO]x[Cl2]y
Experiment
[NO]
(mol/L)
[Cl2]
Rate
Mol/L·s 1
0.250
1.43 x 10-6 2
0.500
5.72 x 10-6 3
2.86 x 10-6 4
1.14 x 10-5
In experiment 1 and 2, [Cl2] is constant while [NO] doubles.
The rate quadruples, so the reaction is second order with respect to [NO]
 R = k[NO]2[Cl2]y

8. Writing a Rate Law
Part 1 – Determine the values for the exponents in the rate law:
R = k[NO]2[Cl2]y
Experiment
[NO]
(mol/L)
[Cl2]
Rate
Mol/L·s 1
0.250
1.43 x 10-6 2
0.500
5.72 x 10-6 3
2.86 x 10-6 4
1.14 x 10-5
In experiment 2 and 4, [NO] is constant while [Cl2] doubles.
The rate doubles, so the reaction is first order with respect to [Cl2]
 R = k[NO]2[Cl2]

9. Writing a Rate Law
Part 2 – Determine the value for k, the rate constant, by using any set of experimental data:
R = k[NO]2[Cl2]
Experiment
[NO]
(mol/L)
[Cl2]
Rate
Mol/L·s 1
0.250
1.43 x 10-6

10. Writing a Rate Law
Part 3 – Determine the overall order for the reaction.
R = k[NO]2[Cl2] 2 + 1
= 3
 The reaction is 3rd order
Overall order is the sum of the exponents, or orders, of the reactants

11. Determining Order with Concentration vs. Time data
(the Integrated Rate Law)
Zero Order:
First Order:
Second Order:

12. Solving an Integrated Rate Law
Time (s)
[H2O2] (mol/L)
1.00
120
0.91
300
0.78
600
0.59
1200
0.37
1800
0.22
2400
0.13
3000
0.082
3600
0.050
Problem: Find the integrated rate law and the value for the rate constant, k
A graphing calculator with linear regression analysis greatly simplifies this process!!

13. Time vs. [H2O2] Regression results: y = ax + b a = -2.64 x 10-4
Time (s)
[H2O2]
1.00
120
0.91
300
0.78
600
0.59
1200
0.37
1800
0.22
2400
0.13
3000
0.082
3600
0.050
Regression results:
y = ax + b
a = x 10-4
b = 0.841
r2 =
r =

14. Time vs. ln[H2O2] Regression results: y = ax + b a = -8.35 x 10-4
Time (s)
ln[H2O2]
120
300
600
1200
1800
-1.514
2400
-2.04
3000
-2.501
3600
-2.996
Regression results:
y = ax + b
a = x 10-4
b = -.005
r2 =
r =

15. Time vs. 1/[H2O2] Regression results: y = ax + b a = 0.00460
Time (s)
1/[H2O2]
1.00
120
1.0989
300
1.2821
600
1.6949
1200
2.7027
1800
4.5455
2400
7.6923
3000
12.195
3600
20.000
Regression results:
y = ax + b
a =
b =
r2 =
r =

16. And the winner is… Time vs. ln[H2O2]
1. As a result, the reaction is 1st order
2. The (differential) rate law is:
3. The integrated rate law is:
4. But…what is the rate constant, k ?

17. Finding the Rate Constant, k
Method #1: Calculate the slope from the
Time vs. ln[H2O2] table.
Time (s)
ln[H2O2]
120
300
600
1200
1800
-1.514
2400
-2.04
3000
-2.501
3600
-2.996
Now remember:
 k = -slope
k = 8.32 x 10-4s-1

18. Finding the Rate Constant, k
Method #2: Obtain k from the linear regresssion analysis.
Regression results:
y = ax + b
a = x 10-4
b = -.005
r2 =
r =
Now remember:
 k = -slope
k = 8.35 x 10-4s-1

19. Rate Laws Summary Rate = k Rate = k[A] Rate = k[A]2 [A] = -kt + [A]0
Zero Order
First Order
Second Order
Rate Law
Rate = k
Rate = k[A]
Rate = k[A]2
Integrated Rate Law
[A] = -kt + [A]0
ln[A] = -kt + ln[A]0
Plot that produces a straight line
[A] versus t
ln[A] versus t
Relationship of rate constant to slope of straight line
Slope = -k
Slope = k
Half-Life

20. Reaction Mechanism
The reaction mechanism is the series of elementary steps by which a chemical reaction occurs.
The sum of the elementary steps must give the overall balanced equation for the reaction
The mechanism must agree with the experimentally determined rate law

21. Rate-Determining Step
In a multi-step reaction, the slowest step is the rate-determining step. It therefore determines the rate of the reaction.
The experimental rate law must agree with the rate-determining step

22. Identifying the Rate-Determining Step
For the reaction:
2H2(g) + 2NO(g)  N2(g) + 2H2O(g)
The experimental rate law is:
R = k[NO]2[H2]
Which step in the reaction mechanism is the rate-determining (slowest) step?
Step #1 H2(g) + 2NO(g)  N2O(g) + H2O(g)
Step #2 N2O(g) + H2(g)  N2(g) + H2O(g)
Step #1 agrees with the experimental rate law

23. Identifying Intermediates
For the reaction:
2H2(g) + 2NO(g)  N2(g) + 2H2O(g)
Which species in the reaction mechanism are intermediates (do not show up in the final, balanced equation?)
Step #1 H2(g) + 2NO(g)  N2O(g) + H2O(g)
Step #2 N2O(g) + H2(g)  N2(g) + H2O(g)
2H2(g) + 2NO(g)  N2(g) + 2H2O(g)
 N2O(g) is an intermediate

24. Collision Model Key Idea: Molecules must collide to react.
However, only a small fraction of collisions produces a reaction. Why?

25. Collision Model
Collisions must have sufficient energy to produce the reaction (must equal or exceed the activation energy). 1.
Colliding particles must be correctly oriented to one another in order to produce a reaction. 2.

26. Factors Affecting Rate
Increasing temperature always increases the rate of a reaction.
Particles collide more frequently
Particles collide more energetically
Increasing surface area increases the rate of a reaction
Increasing Concentration USUALLY increases the rate of a reaction
Presence of Catalysts, which lower the activation energy by providing alternate pathways

27. Endothermic Reactions

28. Exothermic Reactions

29. The Arrhenius Equation
k = rate constant at temperature T
A = frequency factor
Ea = activation energy
R = Gas constant, J/K·mol

30. The Arrhenius Equation, Rearranged
Simplifies solving for Ea
-Ea / R is the slope when (1/T) is plotted against ln(k)
ln(A) is the y-intercept
Linear regression analysis of a table of (1/T) vs. ln(k) can quickly yield a slope
Ea = -R(slope)

31. Catalysis
Catalyst: A substance that speeds up a reaction without being consumed
Enzyme: A large molecule (usually a protein) that catalyzes biological reactions.
Homogeneous catalyst: Present in the same phase as the reacting molecules.
Heterogeneous catalyst: Present in a different phase than the reacting molecules.

32. Lowering of Activation Energy by a Catalyst

33. Catalysts Increase the Number of Effective Collisions

34. Heterogeneous Catalysis
Carbon monoxide and nitrogen monoxide adsorbed on a platinum surface
Step #1: Adsorption and activation of the reactants.

35. Heterogeneous Catalysis
Carbon monoxide and nitrogen monoxide arranged prior to reacting
Step #2:
Migration of the adsorbed reactants on the surface.

36. Heterogeneous Catalysis
Carbon dioxide and nitrogen form from previous molecules
Step #3:
Reaction of the adsorbed substances.

37. Heterogeneous Catalysis
Carbon dioxide and nitrogen gases escape (desorb) from the platinum surface
Step #4:
Escape, or desorption, of the products.