1. Unit 7 (Chp 14): Chemical Kinetics (rates)
Chemistry, The Central Science, 10th edition
Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten
Unit 7 (Chp 14): Chemical Kinetics (rates)
John D. Bookstaver
St. Charles Community College
St. Peters, MO
 2006, Prentice Hall, Inc.

2. Chemical Kinetics Chemical Kinetics is the study of:
Rates of substances consumed or produced per time in a chemical reaction.
Factors that affect the rate of a reaction according to Collision Theory (temp, concentration, surface area, catalyst)
Mechanisms, or sequences of steps, for how a reaction actually occurs.
Rate Laws (equations) used to calculate: rates, rate constants, concentrations, time

3. Reaction Rates
Reaction rates are described by the change in concentration (DM) of reactants (consumed) or of products (produced) per change in time.

4. [C4H9Cl] measured at various times
Reaction Rates
C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq)
[C4H9Cl] measured at various times
recall:
[ ] brackets represent concentration in molarity (M)

5. Reaction Rates C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq) [C4H9Cl] t
“change in”

6. fewer reactant collisions
Reaction Rates
C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq)
average rate slows
WHY?
fewer reactant collisions

7. Reaction Rates C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq) [C4H9Cl] t
(rise)
(run)
The slope of a line tangent to the curve at any point is the instantaneous rate at that time.

8. Reaction Rates C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq)
All reactions slow down over time.
The initial rate of reaction is commonly chosen for analysis and comparison.

9. Reaction Rates and Stoichiometry
C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq)
rate of consumption
of reactant =
rate of production
of product.
(IF 1:1 mol ratio)
Rate:
–[C4H9Cl] t =
[C4H9OH] t
↓ reactant = ↑ product

10. Reaction Rates: 2 NO2  2 NO + O2 change in conc. 1. slope =
change in time
1. slope =
2. consumption of reactants per time
–[NO2] t
concentration (M)
[NO2]
3. production of products per time
+[NO] t t
4. Are equalized to a
stoich. coefficient of 1
2 NO
–[NO2]
2 t =
[NO]
2 t =
[O2] t
1 O2
5. Slow down during rxn
(reactant collisions decrease)
time (t)

11. Rate Ratios = ? aA + bB cC + dD Rate 2 HI(g)  H2(g) + I2(g)
HW p. 619 #17,20
Rate Ratios
aA + bB
cC + dD
Rate 1 a
[A] t = b
[B] c
[C] d
[D] = − −
2 HI(g)  H2(g) + I2(g)
Initial rate of production of H2 is M∙s–1.
What is the rate of consumption of HI? = ? 2
mol HI
0.050 M∙s–1 H2 x =
0.10 M∙s–1 HI 1
mol H2
mol
L∙s
–0.10 M∙s–1 HI
mol ratio

12. Rate Laws rate = k[A]x[B]y[C]z
Rate equations (or rate laws) have the form:
rate = k[A]x[B]y[C]z
rate constant
order with respect to reactants
A, B, & C
…or…
number of each particle involved
in collision that affects the rate
overall order of reaction = x + y +…
Example:
overall order
= ___ order
(4 particles in collision)
rate = k[BrO3–][Br–][H+]2
4th

13. minimum E required to start reaction
Collision Model
reactant bonds break, then product bonds form
successful
reactants
Collision
products
Reaction rates depend on collisions
between reactant particles by:
collision frequency
collision energy
proper orientation
activation energy:
minimum E required to start reaction
(Eact )
unsuccessful

14. Reaction Coordinate Diagram
demo
transition state
…aka…
Energy
Profile
Eact
Potential Energy 
Demo: Transition State Ball Toss ∆E
Reaction progress 

15. Factors That Affect Reaction Rates
Concentration
Temperature
(exposed) Surface Area (particle size)
Catalyst
old foamy (elephant toothpaste)

16. Factors That Affect Reaction Rates
1) Concentration of Reactants
↑ concentration, ↑ collision frequency
increase pressure of gases
Fe(s) + O2(g)  Fe2O3(s)
20% of air is O2(g)
100% O2(g)

17. Factors That Affect Reaction Rates ↑ Temp, ↑ rate
2) Temperature
↑ Temp, ↑ rate
collisions of…
greater frequency
greater energy
Eact
at higher Temp more particles over Eact
unsuccessful collisions
(bounce off)
successful collisions
(react) 17

18. Temperature and k (rate constant)
↑ Temp, ↑ rate
rate = k [A]x
k is temp. dependent
(k changes with temp) 18

19. Factors That Affect Reaction Rates
3) Surface Area (particle size)
smaller pieces, more exposed surface area for collision.

20. Factors That Affect Reaction Rates
4) Catalyst
↑ rate by changing the reaction mechanism by…
demo
…lowering the Eact .
Uncatalyzed
Catalyzed
consumed, then produced (not used up)
potential energy
2 H2O2
Br2
Demo – Sudsy Kinetics (3 diff conc’s, volunteer choice for +10/0/-10 on test. Do 500 mL & glowing splint.
(intermediate)
+ 2 Br– + 2 H+
2 H2O + O2
Br–, H+
2 H2O H2O + O2
+ 2 Br– + 2 H+
reaction progress

21. Surface Catalysts
H2 + H2C=CH2
Catalysts can orient reactants to help bonds break and form.
H2 + H2C=CH2  H3C–CH3
CH3CH3 21

22. Enzymes biological catalysts in living systems.
A substrate fits into the active site of the enzyme much like a key fits into a lock.
(IMAFs work here)
HW p. 621
#50,51,64 22

23. Rate Laws rate = k[A]x[B]y[C]z recall…
Rate equations (or rate laws) have the form:
rate = k[A]x[B]y[C]z
rate constant
order with respect to reactants
A, B, & C
…or…
number of each particle involved
in collision that affects the rate
overall order of reaction = x + y +…
Example:
overall order
= ___ order
(4 particles in collision)
rate = k[BrO3–][Br–][H+]2
4th

24. Reaction Mechanisms
The sequence of molecular collisions and changes by which reactants become products is called the reaction mechanism.
Rxns may occur in separate elementary steps.
The overall reaction occurs only as fast as the slowest, rate-determining step. (RDS) 24

25. Slow First Step Rate law depends on slow step: (1st step)
NO2 (g) + CO (g)  NO (g) + CO2 (g)
A proposed mechanism for this reaction is:
Step 1: NO2 + NO2  NO3 + NO (slow)
Step 2: NO3 + CO  NO2 + CO2 (fast)
NO3 intermediate is produced then consumed.
Rate law depends on slow step: (1st step)
rate = k [NO2]2
CO is not involved in the slow RDS, so it does not appear in the rate law.
(OR…the order w.r.t. CO is ___)
0th 25

26. Slow Second Step Rate law depends on slow step: (2nd step)
2 NO(g) + Br2(g)  2 NOBr(g)
A proposed mechanism is:
Step 1: NO + Br2
NOBr (fast)
Step 2: NOBr2 + NO  2 NOBr (slow)
NOBr2 intermediate is produced then consumed.
Rate law depends on slow step: (2nd step)
rate = k2 [NOBr2] [NO]
But cannot include intermediate [NOBr2] (b/c it’s difficult to control conc.’s of intermediates). 26

27. Slow Second Step Step 1: NO + Br2 NOBr2 (fast)
Step 2: NOBr2 + NO  2 NOBr (slow)
From RDS Step 2: rate = k2 [NOBr2] [NO]
b/c step 1 is in equilibrium
From Step 1: rateforward = ratereverse
k1 [NO] [Br2] = k−1 [NOBr2]
solve for [NOBr2]
substitute for [NOBr2] k1
k−1
[NO] [Br2] = [NOBr2] 27

28. WS Reaction Mechanisms
Slow Second Step
Step 1: NO + Br2
NOBr (fast)
Step 2: NOBr2 + NO  2 NOBr (slow)
HW p. 625 #92,94
rate = k2 [NOBr2] [NO]
WS Reaction Mechanisms
k2k1
k−1
rate =
[NO] [Br2] [NO]
rate = k [NO]2 [Br2]
substitute for [NOBr2] k1
k−1
[NO] [Br2] = [NOBr2] 28

29. Rate Laws rate = k[A]x[B]y[C]z recall…
Rate equations (or rate laws) have the form:
rate = k[A]x[B]y[C]z
rate constant
order with respect to reactants
A, B, & C
…or…
number of each particle involved
in collision that affects the rate
overall order of reaction = x + y +…
Example:
overall order
= ___ order
(4 particles in collision)
rate = k[BrO3–][Br–][H+]2
4th

30. Orders from Experimental Data: Rate varies with Concentration
Initial Concentrations
Rate in M per unit time
Experiment
[BrO3–] , M
[Br–] , M
[H+] , M 1
0.0050
0.25
0.30 10 2
0.010 20 3
0.50 40 4
0.60
160

31. Orders from Experimental Data: Rate varies with Concentration
Initial Concentrations
Rate in M per unit time
Experiment
[BrO3–] , M
[Br–] , M
[H+] , M 1
0.0050
0.25
0.30 10 2
0.010 20 3
0.50 40 4
0.60
160
In experiments 1 & 2:
doubling [BrO3–] doubles rate
(other conc.’s kept constant)
rate is 1st order w.r.t. [BrO3–]
rate = k [BrO3–]x
if… 2 = [2]x
then… 1 = x
“w.r.t.” (with respect to)

32. Orders from Experimental Data: Rate varies with Concentration
Initial Concentrations
Rate in M per unit time
Experiment
[BrO3–] , M
[Br–] , M
[H+] , M 1
0.0050
0.25
0.30 10 2
0.010 20 3
0.50 40 4
0.60
160
In experiments 2 & 3:
doubling [Br–] doubles rate
(other conc.’s kept constant)
rate is 1st order w.r.t. [Br–]
rate = k [Br–]y
if… 2 = [2]y
then… 1 = y

33. Orders from Experimental Data: Rate varies with Concentration
Initial Concentrations
Rate in M per unit time
Experiment
[BrO3–] , M
[Br–] , M
[H+] , M 1
0.0050
0.25
0.30 10 2
0.010 20 3
0.50 40 4
0.60
160
In experiments 3 & 4:
doubling [H+] quadruples rate
(other conc.’s kept constant)
rate is 2nd order w.r.t. [H+]
rate = k [H+]z
if… 4 = [2]z
then… 2 = z

34. Orders from Experimental Data: Rate varies with Concentration
Initial Concentrations
Rate in M per unit time
Experiment
[BrO3–] , M
[Br–] , M
[H+] , M 1
0.0050
0.25
0.30 10 2
0.010 20 3
0.50 40 4
0.60
160
Orders found by experiment are:
x = 1 , y = 1 , z = 2
Rate law is:
rate = k [BrO3–] [Br–] [H+]2

35. Orders in Rate Laws …only found experimentally (from data).
…do NOT come from the coefficients of reactants of an overall reaction.
…represent the number of reactant particles (coefficients) in the RDS of the mechanism.
…zero order reactants have no effect on rate b/c they do not appear in the RDS of the mechanism (coefficient of 0 in RDS).
…typically 0, 1, 2, but can be any # or fraction

36. Units of k (rate constant)
HW p. 619 #22,24,28
Units of k give info about order.
Rate is usually (M∙s–1), or (M∙min–1), etc.
(next slide)
Order
Rate Law
k Units
0th
1st
2nd
3rd
rate =
k[A]0
rate =
k[A]
rate =
k[A]2
rate =
k[A]2[B]
M = ? s
M = ?∙M s
M = ?∙M2 s
M = ?∙M3 s M s 1 s 1
M∙s 1
M2∙s
M∙s–1
s–1
M–1∙s–1
M–2∙s–1

37. Model for Solving Orders
HW p. 619 #28
Model for Solving Orders
Determine the rate law for the reaction (from experimental data).
general rate law
rate = k [ClO2]x [OH–]y
(a)
rate1
rate2
(0.0248)
k (0.060)x(0.030)y = =
( )
k (0.020)x(0.030)y
(0.0248) x
0.020 y
0.030 =
( )
9 =
(3)x
(1)y
9 = (3)x
x = 2 (2nd order)

38. rate = k [ClO2]2 [OH–]y rate = k [ClO2]2 [OH–]1 rate3 rate2 (0.00828)
HW p. 619 #28 (cont.)
Determine the rate law for the reaction (from experimental data).
rate = k [ClO2]2 [OH–]y
(a)
rate3
rate2
( )
k (0.020)2(0.090)y = =
( )
k (0.020)2(0.030)y
3 = (3)y
y = 1 (1st order)
rate = k [ClO2]2 [OH–]1
OR rate = k [ClO2]2 [OH–]

39. rate = k [ClO2]2 [OH–] Exp 1: (0.0248) = k (0.060)2(0.030) (0.0248)
HW p. 619 #28 (cont.)
Calculate the rate constant (with units).
rate = k [ClO2]2 [OH–]
(b)
Exp 1:
(0.0248)
= k (0.060)2(0.030)
(0.0248)
k =
(0.060)2(0.030)
k = 230
M–2∙s–1
M = ?∙M2∙M s

40. rate = k [ClO2]2 [OH–] rate = (230)(0.010)2(0.025) rate = 0.00058
HW p. 619 #28 (cont.)
Calculate the rate when [ClO2] = M and [OH–] = M.
rate = k [ClO2]2 [OH–]
(c)
rate
= (230)(0.010)2(0.025)
rate =
M∙s–1

41. Rate Laws
Differential rate laws (just called “rate law”) express the relationship between the concentration of reactants and
the rate of the reaction.
rate = k [A]
Integrated rate laws (on equation sheet) express the relationship between concentration of reactants and
the time of the reaction.

42. Integrated Rate Laws
Using calculus to integrate a first-order rate law gives us: ln
[A]t
[A]0
= −kt
[A]0 = initial conc. of A at t = 0 .
[A]t = conc. of A at any time, t .
given on exam
ln [A]t – ln [A]0 = –kt
ln [A]t = –kt + ln [A]0
y = mx + b 42

43. First-Order Processes
ln [A]t = –kt + ln [A]0
y = mx + b
first-order
If a reaction is first-order, a plot of ln [A] vs. t is a straight line, and the slope of the line will be –k.
m = –k
ln [A] t 43

44. First-Order Processes
CH3NC
CH3CN
at oC 44

45. First-Order Processes
When ln P is plotted as a function of time, a straight line results.
Therefore,
The process is first-order.
k is the negative slope: 5.1  10–5 s−1. 45

46. Second-Order Processes
Similarly, integrating the rate law for a process that is second-order in reactant A, we get…
given on exam 1
[A]t –
[A]0
= kt 1
[A]t
= kt +
[A]0
y = mx + b 46

47. Second-Order Processes1
[A]t
= kt +
[A]0
y = mx + b
second-order
If a reaction is second-order, a plot of 1/[A] vs. t is a straight line, and the slope of the line is k.
m = k 1
[A] t 47

48. Second-Order Processes
The decomposition of NO2 at 300°C is described by the equation
NO2 (g)
NO (g) + 1/2 O2 (g)
and yields the following data:
Time (s)
[NO2], M
0.0
50.0
100.0
200.0
300.0 48

49. Second-Order Processes
Graphing ln [NO2] vs. t yields:
The plot is NOT a straight line, so the process is NOT first-order in [A].
Time (s)
[NO2], M
ln [NO2]
0.0
−4.610
50.0
−4.845
100.0
−5.038
200.0
−5.337
300.0
−5.573 49

50. Second-Order Processes
Graphing 1/[NO2] vs. t, however, gives this plot.
Because this IS a straight line, the process is second-order in [A].
Time (s)
[NO2], M
1/[NO2]
0.0
100
50.0
127
100.0
154
200.0
208
300.0
263 50

51. Zero-Order Processes
If a reaction is zero-order, a plot of [A] vs. t is a straight line, with a slope = rate.
zero-order
[A] t 51

52. zero-order Summary of Integrated Rate Laws and Linear Graphs [A] t
WS Kinetics III #14
HW p. 620 #33, 38 t
first-order
second-order
m = k
m = –k 1
[A]
ln [A] t t 1
[A]t
= kt +
[A]0
ln [A]t = –kt + ln [A]0

53. Half-life (t1/2): time at which half of initial amount remains.
[A]t = 0.5 [A]0
initial concentration at time, t = 0
concentration at time, t
Half-life (t1/2) is constant
for 1st order only. 53

54. Half-life (t1/2) depends on k:
For a 1st order process:
[A]t1/2 = 0.5 [A]0
ln [A]t – ln [A]0 = –kt
ln 0.5 [A]0 – ln [A]0 = –kt1/2
0.5 [A]0
[A]0 ln
= −kt1/2
ln 0.5 = −kt1/2
−0.693 = −kt1/2
given on exam
= t1/2
0.693 k 54

55. Half-life & Radiometric Dating
A wooden object from an archeological site is subjected to radiometric dating by carbon-14.
The activity of the sample due to 14C is measured to be 11.6 disintegrations per second (current amount of C-14).
The activity of a carbon sample of equal mass from fresh wood is 15.2 disintegrations per second (assumed as original amount of C-14).
The half-life of 14C is 5715 yr.
What is the age of the archeological sample?

56. Half-life & Radiometric Dating
A wooden object from an archeological site is subjected to radiometric dating by carbon-14.
The activity of the sample due to 14C is measured to be 11.6 disintegrations per second (current amount of C-14).
The activity of a carbon sample of equal mass from fresh wood is 15.2 disintegrations per second (assumed as original amount of C-14).
The half-life of 14C is 5715 yr.
What is the age of the archeological sample?
[A]t = 11.6 current dis/s at time, t
[A]0 = 15.2 initial dis/s at time, t0 = 0 s
t1/2 = 5715 yr
t = ? yr

57. Half-life & Radiometric Dating
First, determine the rate constant, k.
ln Nt − ln N0 = –kt
easy way
(given: t1/2 = 5715 yr)
= t1/2
0.693 k
ln 0.5N0 − ln N0 = –k(5715)
0.5 N0 N0 ln
= −k(5715)
= 5715 yr
0.693 k
0.5 ln
= −k(5715)
= k
0.693
5715 yr
–0.693
= −k(5715)
–0.693
–5715
= k
k = 1.21  10−4 yr−1

58. Half-life & Radiometric Dating
Now t can be determined:
ln Nt − ln N0 = –kt
= −(1.21  10−4) t
ln(11.6) – ln(15.2)
= −(1.21  10−4) t
–0.270
= t
2230 yr

59. M∙s–1 s–1 M–1∙s–1 Summary 0th Order 1st Order 2nd Order [A] ln[A] 1
Rate Law
Rate = k
Rate = k[A]
Rate = k[A]2
Integrated Rate Law
[A] = –kt + [A]0
ln[A] = –kt + ln[A]0
Linear plot
[A]
ln[A]
k & slope of line
Slope = rate
Slope = –k
Slope = k
M∙s–1
s–1
M–1∙s–1
Half-Life
depends on [A]0
y = mx + b
y = mx + b 1
[A] t t t
M = ? s
M = ?∙M s
M = ?∙M2 s
Units of k
HW p. 620 #34, 36, 40