1. Chapter 12 - Kinetics
DE Chemistry
Dr. Walker

2. Chemical Kinetics The area of chemistry that concerns reaction rates
How fast the reaction goes!

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

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

5. 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
[NO2]
5. Change as the
reaction proceeds,
if the rate is
dependent upon
concentration t

6. Rate Laws For the reaction, 2NO2(g) 2NO (g) + O2(g) Rate = k[NO2]n
k is a proportionality constant called the rate constant
n is the order of the reactant (an integer, including zero, or a fraction)
k and n must be determined experimentally
Concentrations of products do not appear in the rate law
Product can only form as fast as reactants can react

7. Rate Laws
Differential rate laws aka rate laws express (reveal) the relationship between the concentration of reactants and the rate of the reaction. Time is NOT a variable.
Rate = k[NO2]n
Integrated rate laws express (reveal) the relationship between concentration of reactants and time

8. 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

9. Writing a Rate Law 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
1.14 x 10-5
Compare the rate change to the concentration change
Zero order – concentration increases, rate stays the same
First order – concentration and rate increase by the same amount
Second order – rate increase is DOUBLE the concentration increases
Negative order – concentration goes up, reaction gets slower!

10. 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

11. 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]

12. 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

13. Writing a Rate Law R = k[NO]2[Cl2] 2 + 1 = 3 The reaction is 3rd order
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
The exponents frequently work out to be the coefficients, but it must be experimentally proven!!

14. Another Example
Data from a reaction

15. Another Example Data from a reaction
Experiment 1 and 2 – [NH4+] is constant
Experiment 2 and 3 – [NO2-] is constant
Basic rate law: Rate = k [NH4+]n [NO2-]m

16. Another Example Data from a reaction
Experiment 1 and 2 – [NH4+] is constant
When [NO2-] doubles, rate doubles
[NO2-] is first order, m = 1

17. Another Example Data from a reaction
Experiment 2 and 3 – [NO2-] is constant
When [NH4+] doubles, rate doubles
[NH4+] is first order, n=1

18. Another Example Overall reaction order = sum of m + n = 1 + 1 =2
Rate equation = k [NH4+][NO2-]

19. Another Example Rate equation = k [NH4+][NO2-]
Calculate k using rate and concentrations
1.35 x 10-7 = k[0.100][0.0050]
Solve for k, k =

20. Integrated Rate Law
Not all reactions have multiple reactants, so this frequently represents single reactant rates
Deals with concentration vs. time graphs
Can be first or second order typically
[A] = concentration of A
[A]0 = concentration of A at time = 0
Nothing has reacted yet!

21. Integrated Rate Laws [A] = -kt + [A]0 Zero order
A B
Zero order
Rate is constant, does not change with changing concentration
[A] = -kt + [A]0
[A] = concentration of A
[A]0 = concentration of A at time = 0
k = rate constant
t = time

22. Integrated Rate Laws First order
Reaction slows down with loss of reactant
[A] = concentration of A
[A]0 = concentration of A at time = 0
k = rate constant
t = time

23. Graphing The Data

24. Integrated Rate Laws Second order
Graph of 1/[A] versus t is a straight line with slope k
[A] = concentration of A
[A]0 = concentration of A at time = 0
k = rate constant
t = time

25. Graphing The Data
The graph of 1/[NO2] vs. time is linear for a second order reaction

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

27. Determining Concentrations at a Point in Time
A first order reaction starts with an initial concentration of reactant of 0.50 M and has a rate constant of 3.0 x 10-3 M/s. What is the concentration of the reactant after 30 seconds?

28. Determining Concentrations at a Point in Time
A first order reaction starts with an initial concentration of reactant of 0.50 M and has a rate constant of 3.0 x 10-3 M/s. What is the concentration of the reactant after 30 seconds?
Plug and chug for a 1st order reaction
ln[A] = - (3.0 x 10-3 M/s)(30 s) + ln[0.50]
ln[A] = (-0.693)
ln[A] = [A] = 0.457

29. Determining Concentrations at a Point in Time
A zero order reaction starts with an initial concentration of reactant of 0.01 M and has a rate constant of 2.0 x 10-4 M/s. What is the time required to achieve a reactant concentration of M?

30. Determining Concentrations at a Point in Time
A zero order reaction starts with an initial concentration of reactant of 0.01 M and has a rate constant of 2.0 x 10-4 M/s. What is the time required to achieve a reactant concentration of M?
[A] = -kt + [A]0
[0.005 M] = -(2.0 x 10-4)t M
M = -(2.0 x 10-4)t
t = 25 s

31. Determining Concentrations at a Point in Time
A second order reaction starts with an initial concentration of reactant of 0.01 M and has a rate constant of 2.0 x 10-4 M/s. What is the time required to achieve a reactant concentration of M?

32. Determining Concentrations at a Point in Time
A second order reaction starts with an initial concentration of reactant of 0.01 M and has a rate constant of 2.0 x 10-4 M/s. What is the time required to achieve a reactant concentration of M?
1/[0.005] = (2.0 x 10-4 M/s)t + 1/[0.01]
200 = (2.0 x 10-4 M/s)t + 100
100 = (2.0 x 10-4 M/s)t
s = t (or 5.78 days – that’s a slow reaction!)

33. What Order Is It?

34. Plot Data 1st order – time vs. ln[C4H6] is linear 2nd order – time vs.

35. Results
What order is it?

36. Results
Notice the time vs 1/[C4H6] is linear, making this a 2nd order reaction

37. Half-Lives Zero Order First Order Second Order t1/2 = 2k [A]0 t1/2 = k
0.693
= -kt1/2 1 2 ln or =
t1/2
k[A]0 1

38. Reaction Mechanism
The reaction mechanism is the series of elementary steps by which a chemical reaction occurs.
The sum of the different steps must give the overall balanced equation
Think back to Hess’s Law…
The mechanism must agree with the experimentally determined rate law.

39. A classic nucleophilic substitution mechanism

40. Molecularity
The number of species that must collide to produce the reaction indicated by that step
Unimolecular step - a reaction involving one molecule
Bimolecular step - reaction involving the collisions of two species
Termolecular step - reaction involving the collisions of three species

41. Typical Rate Laws and Molecularity
Notice the relationship between reaction coefficients and reaction order
You would still need to prove the relationship experimentally

42. Rate – Determining Step
In a multi-step reaction, the slowest step is the rate-determining step. It therefore determines the rate of the reaction
Think about a relay team – If the team has 3 top sprinters and 1 slower runner, the team will LOSE – the slower runner will hold them back!!

43. 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 # H2(g) + 2NO(g)  N2O(g) + H2O(g)
Step # N2O(g) + H2(g)  N2(g) + H2O(g)
Step #1 agrees with the experimental rate law

44. Intermediates
A species that is neither a reactant nor a product, but that is formed and consumed during a chemical reaction
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)

45. 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

46. Find The Intermediates
CH4  C + 2H kJ
C + O2  CO kJ
2H2 + O2  2 H2O kJ

47. Find The Intermediates
CH4  C + 2H kJ
C + O2  CO kJ
2H2 + O2  2 H2O kJ
CH4 + 2O2  CO2 + 2H2O kJ
Carbon and hydrogen are the intermediates because they are formed and consumed during the process.

48. 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 (or pressure in a gas phase reaction) USUALLY increases the rate of a reaction
Presence of Catalysts, which lower the activation energy by providing alternate pathways