1
Today…
• Turn in:
– Nothing
• Our Plan:
– Test Results
– Videos/Notes
– Investigation 10 Pre-Lab
• Homework (Write in Planner):
– Be prepared for Investigation 10 next class and
have report started.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
2
Introduction to Chemical Kinetics
• What is Kinetics? I’ll let Hank explain…
http://www.youtube.com/watch?v=7qOFtL3V
EBc
• Stop at 3:20
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
3
Kinetics in Action…
• Clock Reactions…
• http://www.youtube.com/watch?v=BqeWpy
wDuiY
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
4
Unit Preview…
Some Reactions
• Are fast:
• Are slow:
– Acid/base
neutralization
– Sodium + water
– PPT reactions
Prentice Hall © 2005
– Aluminum oxidation
– Iron oxidizing
– Plastic bottle
decomposing
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
5
And some depend…
• On temperature: fireflies
• On conditions: Iron oxide (humid or
desert)
• On enzymes: many processes in our body
• On a catalyst:
2CO + 2NO --> 2CO2 + N2 (automobile
pollution and catalytic converters)
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
6
Chemical Kinetics: A Preview
• Chemical kinetics is the study of:
– the rates of chemical reactions
– factors that affect these rates
– the mechanisms by which reactions occur
• Reaction rates vary greatly – some are very fast
(burning, precipitation) and some are very slow
(rusting, disintegration of a plastic bottle in
sunlight).
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
7
Variables in Reaction Rates
• Concentrations of reactants: Reaction rates generally
increase as the concentrations of the reactants are
increased.
• Temperature: Reaction rates generally increase rapidly as
the temperature is increased.
• Surface area: For reactions that occur on a surface rather
than in solution, the rate increases as the surface area is
increased.
• Catalysts: Catalysts speed up reactions and inhibitors slow
them down.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
8
Think of it like this…
• http://ed.ted.com/lessons/how-to-speed-upchemical-reactions-and-get-a-date
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
9
Theories of Chemical Kinetics:
Collision Theory
• Before atoms, molecules, or ions can react, they must first
collide.
• An effective collision between two molecules puts enough
energy into key bonds to break them.
• The activation energy (Ea) is the minimum energy that
must be supplied by collisions for a reaction to occur.
• A certain fraction of all molecules in a sample will have
the necessary activation energy to react; that fraction
increases with increasing temperature.
• The spatial orientations of the colliding species may also
determine whether a collision is effective.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
10
Distribution of Kinetic Energies
At higher temperature
(red), more molecules
have the necessary
activation energy.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
11
Key Point
Orientation of
molecules at the
time of their
collision will
determine whether
they react or not!
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
12
Analogy - Car Crash Example
Energy and orientation
of cars during a car
crash can establish the
change that occurs to the
cars.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
13
Importance of Orientation
One hydrogen atom
can approach another
from any direction …
Effective collision; the
I atom can bond to the
C atom to form CH3I
… and reaction will still occur; the
spherical symmetry of the atoms means
that orientation does not matter.
Prentice Hall © 2005
Ineffective collision;
orientation is important
in this reaction.
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
14
Transition State Theory
• The configuration of the atoms of the colliding species at
the time of the collision is called the transition state.
• The transitory species having this configuration is called
the activated complex.
• A reaction profile shows potential energy plotted as a
function of a parameter called the progress of the reaction.
• Reactant molecules must have enough energy to surmount
the energy “hill” separating products from reactants.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
15
Activated complex
• Species formed as a result of collisions
between energetic molecules that is an
intermediate between the reactants and
the products of a reaction. Once formed
the activated complex dissociates either
into products or back to the reactants
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
16
Reaction Profile
• A graphical representation of a
chemical reaction in terms of the
energies of the reactants,
activated complexes and
products
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
17
Reaction Profile
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
18
Exothermic
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
19
Endothermic
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
20
A Reaction Profile
CO(g) + NO2(g)
Prentice Hall © 2005

CO2(g) + NO(g)
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
21
An Analogy for Reaction Profiles
and Activation Energy
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
22
Stop!
• Investigation 10 - We’re going to test
the effect of different variables on the
rate of reaction next class.
• Complete steps 1 – 6 on the lab
handout and be prepared to conduct the
lab next class.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
23
Today…
• Turn in:
– Nothing
• Our Plan:
– Investigation 10
• Homework (Write in Planner):
– Lab Report Due Friday
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
24
Today…
• Turn in:
– Lab Report (rubric on top)
• Our Plan:
– Notes & Practice
• Homework (Write in Planner):
– Work on the Ch. 13 Homework
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
25
The Meaning of Rate
• The rate of a reaction is the change in
concentration of a product per unit of time
(rate of formation of product).
• Rate is also viewed as the negative of the
change in concentration of a reactant per
unit of time (rate of disappearance of
reactant).
• The rate of reaction often has the units of
moles per liter per unit time (mol∙L–1∙s–1 or
M∙s–1)
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
26
If the rate of consumption of H2O2 is 4.6 M/h, then …
… the rate of formation of H2O must also be 4.6 M/h, and …
… the rate of formation of O2 is 2.3 M/h
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
27
2 H2O2 --> 2 H2O + O2
1L
2.960 g O2 (0.09250
mole) produced in
60 s means …
Prentice Hall © 2005
… 0.1850 mol H2O2
reacted in 60 s.
0.1850 mol H2O2/L
Rate =
60 s
= 0.00131 M H2O2 s–1
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
28
Example 13.1
Consider the hypothetical reaction
A + 2B --> 3C + 2D
Suppose that at one point in the reaction, [A] = 0.4658 M
and 125 s later [A] = 0.4282 M. During this time period,
what is the average (a) rate of reaction expressed in M∙s–1
and (b) rate of formation of C, expressed in M∙min–1.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
29
Try It Out!
EX 13.1 A: Consider the hypothetical reaction
2A + B → 2C + D
Suppose that at some point during the reaction [D] =
0.2885 M and that 2.55 min (that is 2 min 33 sec)
later [D] = 0.3546 M.
a) What is the average rate of reaction during this
time period, expressed in M min-1?
b) What is the rate of formation of C expressed in
M s-1?
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
30
Rate of Reaction Expressed as the negative
of the Slope of a Tangent Line
• Average rate (green dotted
line)
• Initial Rate ( blue solid line)
• Instantaneous Rate (red line)
• Question: Over what time
interval are the instantaneous
rates greater than the average
rate measured for the 600 sec
period?
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
31
Average vs. Instantaneous Rate
Instantaneous rate is
the slope of the
tangent to the curve
at a particular time.
We often are interested
in the initial instantaneous rate; for the initial
concentrations of
reactants and products
are known at this time.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
32
Example 13.2
Use data from Table 13.1 and/or Figure 13.5 to
(a) determine the initial rate of reaction and
(b) calculate [H2O2] at t = 30 s.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
33
Try It Out
EX 13.2 A: From Figure 13.5,
a) Determine the instantaneous rate
of reaction at t = 300 s.
b) Use the result of a) to calculate a
value of [H2O2] at t = 310 s.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
34
The Rate Law of a Chemical Reaction
• The rate law for a chemical reaction relates the rate of
reaction to the concentrations of reactants.
aA + bB + cC …→ products
rate = k[A]n[B]m[C]p …
• The exponents (m, n, p…) are determined by experiment.
• Exponents are not derived from the coefficients in the
balanced chemical equation, though in some instances the
exponents and the coefficients may be the same.
• The value of an exponent in a rate law is the order of the
reaction with respect to the reactant in question.
• The proportionality constant, k, is the rate constant.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
35
Rate Law Examples
Rate = k[A]1 = k[A]
Reaction is first order in A
Rate = k[A]2
Reaction is second order in A
Rate = k[A]3
Reaction is third order in A
If we triple the
concentration of A in a
second-order reaction,
the rate increases by a
factor of ________.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
36
Effect of Concentration on Rate
Order of Rxn
Concentration
Change
Effect on Rate
0
Double/Triple…
Nothing (20)
1
Double
Double (21)
1
Triple
Triple (31)
2
Double
Quadruple (22)
2
Triple
8x (32)
2
Quadruple
16x (42)
3
Double
9x (23)
3
Triple
27x (33)
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
37
More About the Rate Constant k
• The rate of a reaction is the change in concentration with
time, whereas the rate constant is the proportionality
constant relating reaction rate to the concentrations of
reactants.
• The rate constant remains constant throughout a reaction,
regardless of the initial concentrations of the reactants.
• The rate and the rate constant have the same numerical
values and units only in zero-order reactions.
• For reaction orders other than zero, the rate and rate
constant are numerically equal only when the
concentrations of all reactants are 1 M. Even then, their
units are different.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
38
To find the overall order of a
reaction…
• Add the orders for each compound.
• Example:
– rate = [A]2[B]1 is 3rd order overall
– How about rate = [A]0[B]1[C]1?
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
39
Units of k (p. 536)
Overall Reaction
Order
Zero
First
Second
Third
Prentice Hall © 2005
Units of k
M s-1
s-1
M-1 s-1
M-2 s-1
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
40
Method of Initial Rates
• The method of initial rates is a method of establishing the
rate law for a reaction—finding the values of the exponents in
the rate law, and the value of k.
• A series of experiments is performed in which the initial
concentration of one reactant is varied. Concentrations of the
other reactants are held constant.
• When we double the concentration of a reactant A, if:
– there is no effect on the rate, the reaction is zero-order in A.
– the rate doubles, the reaction is first-order in A.
– the rate quadruples, the reaction is second-order in A.
– the rate increases eight times, the reaction is third-order
in A.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
41
The concentration of
NO was held the
same in Experiments
1 and 2 …
… while the
concentration of
Cl2 in Experiment
2 is twice that of
Experiment 1.
The rate in Experiment 2 is
twice that in Experiment 1,
so the reaction must be first
order in Cl2.
Which two experiments are
used to find the order of the
reaction in NO?
Prentice Hall © 2005
How do we find the value of k
after obtaining the order of the
reaction in NO and in Cl2?
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
42
Example 13.3
For the reaction 2 NO(g) + Cl2(g) → 2 NOCl(g)
described in the text and in Table 13.2, (a) what is the
initial rate for a hypothetical Experiment 4, which has
[NO] = 0.0500 M and [Cl2] = 0.0255 M? (b) What is the
value of k for the reaction?
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
So, when looking at data…..
43
• Zero order reaction: initial rate is
unaffected (20 )
• 1st order reaction: double the
concentration, doubles the rate (21 =2)
• 2nd Order Reaction: Initial rate
increases fourfold (22 =4)
• 3rd Order Reaction: Initial rate
increases eightfold (23 =8)
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
44
Try It Out
• Below is some rate data for the hypothetical
reaction, 2A + B --> C. What is the rate law
for this reaction?
Experiment
[A]0
[B]0
Rate (M/s)
1
2.0 M
1.0 M
0.100
2
2.0 M
2.0 M
0.400
3
4.0 M
1.0 M
0.100
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
45
Let’s Take a Break
• With your partner, Complete
Part 1 of the Partner Review
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
46
•Sum of exponents is equal
to zero
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
Zero Order
47
Rate of Reaction= k[A]0
• The concentration time
graph is a straight line
with a negative slope
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
48
Zero Order
• Rate of the reaction:
–IS equal to k
–remains constant throughout the
reaction
–is the negative of the slope of the
line when graph Molarity vs. time
(see pg. 544)
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
49
A Zero-Order Reaction
rate = k[A]0
=k
Rate is independent of
initial concentration
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
50
Zero Order
Integrated rate equation:
[A]t = -kt + [A]0
y = mx + b
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
51
Zero Order (cont.)
y = [A]t = conc of A at some
time
x = t = time
b = [A]0
m = -k (m, the slope of the
straight line)
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
52
First-Order Reactions
• In a first-order reaction, the exponent in the rate law is 1.
• Rate = k[A]1 = k[A]
• The integrated rate law describes the concentration of a
reactant as a function of time. For a first-order process:
ln
[A]t
[A]0
= –kt
ln [A]t – ln [A]0 = –kt
ln [A]t = –kt + ln [A]0
Look! It’s an
equation for a
straight line!
• At times, it is convenient to replace molarities in an
integrated rate law by quantities that are proportional to
concentration.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
53
Rate of reaction=
1
k[A]
Integrated rate equation
ln[A]t = -kt + ln[A]0
y
Prentice Hall © 2005
= mx +
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
b
Chapter Thirteen
54
st
1
Order
• Easy test for a first order
reaction is to plot the natural log
of reactant conc. vs. time and
see if the graph is linear.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
55
Decomposition of H2O2
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
56
Example 13.4
For the first-order decomposition of H2O2(aq), given k =
3.66 x 10–3 s–1 and [H2O2]0 = 0.882 M, determine (a) the
time at which [H2O2] = 0.600 M and (b) [H2O2] after 225 s.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
57
Another Example
H202 initially at a conc of 2.32 M, is allowed to decompose.
What will the [H202] be 1200 s later? Use k = 7.3 x 10-4 s-1
for this first order decomposition.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
58
Try It Out
EX 13.4 A: The decomposition of nitramide,
NH2NO2, is a first order reaction:
NH2NO2 (aq) → H2O (l) + N2O (g)
The rate law is rate = k[NH2NO2], with k =
5.62 x 10-3 min-1 at 15ᵒC. Starting with 0.105
M NH2NO2,
a) At what time will [NH2NO2] = 0.0250 M
b) What is [NH2NO2] after 6.00 hours?
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
59
Half-life
Time required for ½ of the
reactant to be consumed
Equation:
t1/2 = ln2
k
Prentice Hall © 2005
= 0.693
k
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
60
Example 13.5
• Use data from Figure 13.7 (p. 542) to
evaluate the
a) half-life and
b) Rate constant for the first order
decomposition of N2O5 at 67 ᵒC:
N2O5 (g) → 2NO2 (g) + ½ O2 (g)
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
61
Example 13.5 A
Use the result of Example 13.5 to determine
a) The time required to reduce the quantity of
N2O5 to 1/16 of its initial value and
b) The mass of N2O5 remaining after a 4.80 g
sample of N2O5 has decomposed for 10.0
min.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
62
Sum of exponents = 2
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
63
Second-Order Reactions
• A reaction that is second order in a reactant has a rate law in
which the exponent for that reactant is 2.
Rate = k[A]2
What do we plot
• The integrated rate law has the form:
vs. time to get a
straight line?
1
1
–––– = kt + ––––
[A]t
[A]0
• The half-life of a second-order reaction depends on the initial
concentration as well as on the rate constant k:
1
t½ = –––––
k[A]0
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
64
Example 13.7
The second-order decomposition of
HI(g) at 700 K is represented in
Figure 13.9.
HI(g) → ½ H2(g) + ½ I2(g)
Rate = k[HI]2
What are the: (a) rate constant and
(b) half-life of the decomposition
of 1.00 M HI(g) at 700 K?
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
65
Try It Out
EX 13.7 A: If, in the second order reaction A →
products, it takes 55 s for the concentration of
reactant A to fall to 0.40 M from an initial
concentration of 0.80 M, what is the rate constant
k for the reaction?
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
66
Summary of Kinetic Data
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
67
Short Cut!!
• [A] vs time (straight line = Zero
order)
• ln [A] vs. time ( straight line = first
order)
• 1/[A] vs. time (straight line =
second order)
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
68
Try it in your note packet!
Time, min
0
5
10
15
25
Prentice Hall © 2005
[A]
1.0
0.63
0.46
0.36
0.25
ln[A]
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
1/[A]
Chapter Thirteen
69
Crash Course Review
• http://www.youtube.com/watch?v=7qOFtL
3VEBc
• 3:10 – 5:30
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
70
Let’s Take a Break
• Complete Part 2 of the
Partner Review.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
71
Today…
• Turn in:
– Get out a piece of notebook paper
• Our Plan:
– Scavenger Hunt Review
– Investigation 11 Pre-Lab
• Homework (Write in Planner):
– Prepare Lab Report
–
Homework
Problems
(Due
2/12)
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
72
Pre Lab
• Complete steps 1 – 5 on the
handout and have your formal lab
report started.
• Be prepared to experiment next
class.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
73
Today…
• Turn in:
– Nothing
• Our Plan:
– Investigation 11
• Homework (Write in Planner):
– Lab Report due Next Class (2/10)
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
74
Today…
• Turn in:
– Lab Report (rubric on top)
• Our Plan:
–
–
–
–
Review Activity
Elephant Toothpaste Demo
Notes – Catalysts & Rate Determining Steps
Finish Homework Problems
• Homework (Write in Planner):
– Homework Problems
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
75
Effect of Temperature on
the Rates of Reactions
• In 1889, Svante Arrhenius proposed the following
expression for the effect of temperature on the rate
constant, k:
k = Ae–Ea/RT
• The constant A, called the frequency factor, is an
expression of collision frequency and orientation; it
represents the number of collisions per unit time that are
capable of leading to reaction.
• The term e–Ea/RT represents the fraction of molecular
collisions sufficiently energetic to produce a reaction.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
76
I like this equation better…
• Look at Eq. 13.16 on p. 551 of the
text!
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
77
Example 13.9
Estimate a value of k at 375 K for the decomposition of
dinitrogen pentoxide illustrated in Figure 13.15, given that
k = 2.5 x 10–3 s–1 at 332 K.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
78
Try it Out
EX 13.9 B: Di-tert-butyl peroxide (DTBP) is
used as a catalyst in the manufacture of
polymers. In the gaseous state, DTBP
decomposes to acetone and ethane by a firstorder reaction.
(C4H9)2O2 (g) → 2(CH3)2CO (g) + C2H6 (g)
The half-life of DTBP is 17.5 h at 125ᵒC and
1.67 h at 145ᵒC. What is the activation energy,
Ea, of the decomposition reaction?
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
79
Reaction Mechanisms
• Analogy: a banana split is made by steps in sequence:
slice banana; three scoops ice cream; chocolate sauce;
strawberries; pineapple; whipped cream; end with
cherry.
• A chemical reaction occurs according to a reaction
mechanism—a series of collisions or dissociations—
that lead from initial reactants to the final products.
• Like making a banana split, three molecules will not
collide simultaneously very often, so steps of a
reaction mechanism involve only one or two reactants
at a time.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
80
Reaction Mechanisms
• An elementary reaction represents, at the
molecular level, a single step in the progress of the
overall reaction.
• A proposed mechanism must:
– account for the experimentally determined rate
law.
– be consistent with the stoichiometry of the
overall or net reaction.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
81
Molecularity
The molecularity of an elementary reaction refers to the
number of free atoms, ions, or molecules that collide or
dissociate in that step.
Termolecular processes
are unusual, for the
same reason that three
basketballs shot at the
same time are unlikely
to collide at the same
instant …
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
82
The Rate-Determining Step
• The rate-determining step is the crucial step in
establishing the rate of the overall reaction. It is usually the
slowest step.
• Some two-step mechanisms have a slow first step followed
by a fast second step, while others have a fast reversible
first step followed by a slow second step.
Fast
Mechanism for
2 NO + O2 --> 2 NO2
Slow
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
83
An Example
• Given the reaction:
2A + 2B → C + D
rate = k[A]2[B]
Could take place by the following three-step
mechanism:
I. A + A ↔ X (fast)
II. X + B → C + Y (slow)
III. Y + B → D (fast)
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
84
Intermediates
• X & Y are called intermediates
because they appear in the mechanism,
but they cancel out of the balanced
equation.
• They are products from one reaction
and then a reactant in the next.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
85
An Example
• The steps of a reaction mechanism must add
up to equal the balanced equation with all
intermediates cancelling out. Let’s try it
with our example.
I. A + A ↔ X (fast)
II. X + B → C + Y (slow)
III. Y + B → D (fast)
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
86
Rate-determining Step
• As in any process where many steps are involved,
the speed of the whole process can’t go faster than
the speed of the slowest step in the process.
• The slowest step of a reaction is the ratedetermining step.
• Because the slowest step is the most important
step in determining the rate of a reaction, the
slowest step and the steps leading up to it are used
to see if the mechanism is consistent with the rate
law for the overall reaction.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
87
An Example
• Let’s look at our reaction again and show
that it is consistent with the rate law:
2A + 2B → C + D
rate = k[A]2[B]
I. A + A ↔ X (fast)
II. X + B → C + Y (slow)
III. Y + B → D (fast)
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
88
An Example (p. 199 Cracking AP
Chemistry Exam)
1. Write rate law for slow step.
2. X is an intermediate, we need to eliminate
it from the rate law.
3. Solve for [X] in terms of [A].
4. Substitute for [X] in our second step.
5. Now we have the rate law.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
89
Example 13.10 For the reaction H2(g) + I2(g) --> 2 HI(g), a proposed
mechanism is below. What is the net equation for the overall reaction, and what
is the order of the reaction according to this mechanism?
Fast step: I2
k1
2I
k–1
Slow step: 2 I + H2
Prentice Hall © 2005
k2
2 HI
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
90
EX 13.10 A
The decomposition of nitrosyl choride,
2NOCl (g) → 2NO (g) + Cl2 (g)
Is a first-order reaction. Propose a mechanism
for this reaction consisting of one fast step and
one slow step.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
91
Catalyst
• A catalyst provides an alternative reaction
pathway of lower activation energy
• Participates in a chemical reaction w/o
undergoing permanent change
• Speeds up without being consumed in the
reaction. It is neither a reactant nor product
in a reaction.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
92
Catalysis – how does it work
• In general, a catalyst works by changing the
mechanism of a chemical reaction.
• Often the catalyst is consumed in one step of
the mechanism, but is regenerated in another
step.
• The pathway of a catalyzed reaction has a
lower activation energy than that of an
uncatalyzed reaction, so more molecules at a
fixed temperature have the necessary
activation energy.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
93
Effect of Catalyst on Reaction
Profile and Activation Energy
A catalyst lowers the
activation energy, making it
easier for the reactants to
“climb the energy hill” and
form the products.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
94
A Kinetics Pick Up Line…
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
95
Homogeneous Catalysis
Ozone decomposition catalyzed
by chlorine atoms has a much
lower activation energy and
proceeds much more rapidly
than the uncatalyzed reaction
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
96
Heterogeneous Catalysis
• Many reactions are catalyzed by the surfaces of
appropriate solids.
• A good catalyst provides a higher frequency of effective
collisions.
• Four steps in heterogeneous catalysis:
– Reactant molecules are adsorbed.
– Reactant molecules diffuse along the surface.
– Reactant molecules react to form product molecules.
– Product molecules are desorbed (released from the
surface).
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
97
Heterogeneous Catalysis
Hydrogen is adsorbed
onto the surface of a
nickel catalyst. A C=C
approaches …
… and is adsorbed.
Hydrogen atoms
attach to the carbon
atoms, and the
molecule is desorbed.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
98
A Surface-Catalyzed Reaction
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
99
Enzyme Catalysis
• Enzymes are high-molecular-mass proteins that usually
catalyze one specific reaction, or a set of similar reactions.
• The reactant substance, called the substrate (S), attaches
itself to an area on the enzyme (E) called the active site, to
form an enzyme-substrate complex (ES).
• The enzyme–substrate complex decomposes to form
products (P), and the enzyme is regenerated.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
100
Factors Influencing Enzyme Activity
The rates of enzyme-catalyzed reactions are influenced by
factors such as concentration of the substrate, concentration
of the enzyme, acidity of the medium, and temperature.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
101
Mercury Poisoning: An Example of
Enzyme Inhibition
When Hg reacts
with an enzyme …
… the Hg binds to
sulfur atoms …
… changing the shape
of the active site, so
that it no longer “fits”
the substrate.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
102
Homogeneous Catalysis
Step 1:
A + catalyst → intermediate + C
Step 2: B + intermediate → D + catalyst
Net equation:
A+B→C+D
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
103
Identifying Catalysts and
intermediate species
O3
ClO·
+ Cl∙ --> ClO∙ + O2
+ O∙ --> Cl∙ + O2
Catalyst?
Intermediate?
Net Equation?
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
104
Crash Course Review
• http://www.youtube.com/watch?v=7qOFtL
3VEBc
• 7:15 - end
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
105
Finish the HW
All homework
problems are due on
Wednesday!
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
106
Today…
• Before Class:
– Mark Homework Questions on the Board
• Our Plan:
– Homework Questions/Check HW
– Worksheet Race
– Study Guide
• Homework (Write in Planner):
– Test Next Class
– Breakfast Club 6 am on Friday
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
107
Study Guide Changes
• 7 – do 39 a only
• 11 – do 25 a and b only
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
108
Today…
• Turn in:
– Nothing
• Our Plan:
– Study Guide Questions
– Test
• Homework (Write in Planner):
– Complete the POGIL (Day 1 ONLY)
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen