Chapter 12
Chemical
Kinetics
Introduction
Consider the following:
diamond  graphite ΔG = -2.9 kJ
 Thoughts? Questions?
 If this is true, why don’t we see
diamonds in jewelry turning into
graphite?
 Hint: the reaction happens, yes, but at
what pace?
 This happens VERY slowly! Also ΔG is
very small, so just barely spontaneous.

Introduction
What have we learned about
thermochemical values and what they
do/don’t tell us?
 Reactions can be spontaneous but this
DOES NOT mean they occur fast.
 Chemical kinetics: branch of chemistry
that studies reaction rates.

What Affects Rates?
Brainstorm- what do you think?
Remember- in order to react, two particles
must collide!
(1) Identity of reactant- large & complex =
fewer chances that the collision will take
place at the correct location (rate slows).
(2) Temperature- greater KE means more
energetic and frequent collisions.
(3) Concentration- larger amounts mean
increased likelihood of collisions.

O
O
O
O
N
Br
N
Br
N
N
O N Br
Br Br
Br N O
Br Br
O N
Br N O
O N Br
Molecular Orientation
N O
No
Reaction
What Affects Rates?
(4) States of matter- reactants in the same
state are more able to collide & react; liquids
and gases are more able to collide due to
greater surface area.
(5) Adding a catalyst- adding this increases
the rate.
*Is recoverable at the end of the rxn.
*Reduces activation energy.
Unit Essential Question

What role do rates play in chemical
reactions?
Section 12.1
REACTION RATES
Lesson Essential Question: How is
the rate of a reaction determined?
Example
Using stoichiometry allows rates to be set equal to each
other! Thus the rate of disappearance of a reactant can be
said to equal the rate of appearance of a product.
Kinetics

2A + B  C + 3D (generic reaction)
Rate = D[substance]
Change in [ ] per unit time
Dt

Rate = -½D[A]/Dt = -D[B]/Dt = D[C]/Dt = 1/3 D[D]/Dt

Notice effect of coefficients!
Why are the first two negative?
Because reactant concentrations decrease!
This gives the initial reaction rate- how
quickly the reaction occurs as soon as the
reactants are mixed.




Practice
+ 3H2  2NH3
 What would the initial rate equal in terms
of each substance in the reaction?
 Rate = D[N2]
= -1/3D[H2] = ½ D[NH3]
Dt
Dt
Dt
 N2
Section 12.2
RATE LAWS: AN
INTRODUCTION
Rate Law Expression
2 NO2  2 NO + O2






Rate = k[NO2]n
This is called a rate law expression.
Only reactants are used! Why?
k is called the rate constant.
 k does not change for a given reaction
UNLESS T changes or a catalyst is added.
n is the order of the reactant(s) - usually a
positive integer.
Value of exponent can ONLY be determined
experimentally! Does NOT com from coefficients!
Understanding Rate Laws &
Orders: Rate = k[NO2]n




When n = 1: if reactant concentration is doubled,
the rate doubles; if reactant concentration is tripled,
the rate triples. First order with respect to this
substance.
When n = 2: if reactant concentration is doubled,
the rate increases four times (22 = 4). Second
order with respect to this substance.
When n = 0: reactant concentration has no effect
on the rate; it simply needs to be present in some
amount. Zero order with respect to this substance.
Can add orders together for all reactants to get the
overall order for the entire reaction.
Determining Rate Order & k
2NO (g) + O2 (g)  2NO2 (g)
Initial [NO] Initial [O2] Rate of NO2 formation (M/s)
0.01
0.01
0.05
0.02
0.01
0.20
0.01
0.02
0.10
Can determine rate orders by inspection or
mathematically.
Second order with respect to NO and first order with
respect to O2.
Determining Rate Order & k



Once you know the rate orders for each
reactant you can solve for K. How?
Choose any experiment and the corresponding
concentration values and rate. Plug them into
the rate equation and solve for k.
k = 5 x 104 /M2s
Sections 1&2 Homework

Pg. 567 #10, 13, 26, 28
AP Practice Question
What is the rate law for the following:
CO (g) + Cl2 (g)  COCl2 (g)
Experiment
1
2
3
Initial [CO] M Initial [Cl2] M
0.200
0.100
0.200
a) Rate = k[CO]
b) Rate = k[CO]2[Cl2]
0.100
0.200
0.200
Initial Rate
3.9 x 10-25
3.9 x 10-25
7.8 x 10-25
c) Rate = k[CO][Cl2]
d) Rate = k[CO][Cl2]2
AP Practice Question
The reaction (CH3)3CBr (aq) + H2O (l) 
(CH3)3COH (aq) + HBr (aq) has the following
rate law: Rate = k[(CH3)3CBr]. What will the
effect of decreasing the concentration of
(CH3)3CBr?
a) Reaction rate will increase.
b) More HBr will form.
c) Reaction rate will decrease.
d) Reaction will shift to the left.
Another Example

For the reaction:
BrO3- + 5 Br- + 6H+  3Br2 + 3 H2O

The general form of the rate law is:
Rate = k[BrO3-]a[Br-]b[H+]c

Use experimental data to determine the
values of a, b, and c.

Then determine the reaction order for
each reactant and the overall reaction.
Initial concentrations (M)
BrO30.10
0.20
0.20
0.10
Br0.10
0.10
0.20
0.10
H+
0.10
0.10
0.10
0.20
Rate (M/s)
8.0 x 10-4
1.6 x 10-3
3.2 x 10-3
3.2 x 10-3
Initial concentrations (M)
Rate (M/s)
BrO3BrH+
0.10
0.10
0.10
8.0 x 10-4
0.20
0.10
0.10
1.6 x 10-3
0.20
0.20
0.10
3.2 x 10-3
0.10
0.10
0.20
3.2 x 10-3
Now we have to see how the rates
change with concentration!
Answers to Example

The powers of the reactants are as
follows: a = 1; b = 1; c = 2, so:
Rate = k[BrO3-] [Br-] [H+]2
 The reaction is first order with respect to
BrO3- and Br-, second order with respect
to H+, and fourth order overall.
 How would you calculate the value of
the rate constant? k = 8.0 L3/mol3∙s
Section 12.4
THE INTEGRATED
RATE LAW
Integrated Rate Law



Allows us to determine information about
concentrations over time, not just at a
particular instant during the reaction.
For example, with the integrated rate law we
can determine how long it will take to use up
a reactant, form a product, etc.
Integrated rate laws come from setting the
rate law and rate expression equal to each
other for each order, and integrating over
time.
First Order Integrated Rate Law







ln[A]t – ln[A]0 = -kt
[A]t = concentration of reactant at a given
time, t
[A]0 = initial concentration of reactant
Can be written as: ln[A]t = -kt + ln[A]0
Has the format of a line!
Note: slope = -k so adjust signs accordingly
when solving for k!
Graphing ln[A] vs. time gives a straight line for
first order!
Second Order Integrated Rate Law





1/[A]t – 1/[A]0 = kt
Can be written as: 1/[A]t = kt + 1/[A]0
This also has the format of a line!
Note: slope = k so no need to adjust signs!
Graphing 1/[A] vs. time gives a straight line
for second order!
Practice
HI decomposes in a second order
reaction. The value of k = 2.40x10-21/(Ms)
at 25°C. How long will it take the
concentration of HI to decrease from
0.200M to 0.190M at 25°C?
1.10 x 1020 s
Half Life
• Half life: time required for a reactant to
reach half of its original concentration.
• Does not depend on concentrations for
first order; only depends on k: t1/2 = ln2/k
= 0.693/k
• Does depend on concentrations for
second order, and also k: t1/2 = 1/k[A]0
• Note: radioactive decay is first order!
AP Practice Question
The rate constant, k, for radioactive
beryllium-11 is 0.049/s. What mass of a
0.500mg sample of beryllium-11 remains
after 28 seconds? Find how many half lives make
a)
b)
c)
d)
0.250mg
0.125mg
0.0625mg
0.375mg
up the time given. Then halve
the initial amount that many
times.
t1/2 = 14s. 14 goes into 28 two
times, so Be goes through
two half lives; divide 0.500 by
two twice.
Practice

The rate constant for the radioactive
decay of thorium-232 is 5.0x10-11/year.
What is the half life of thorium?
1.386 x 1010 years
Practice

HI decomposes in a second order
process to its elements. The value of
the rate constant is 2.40 x 10-21/Ms at
25°C. What is the half life for the
decomposition of 0.200M HI at 25°C?

2.08 x 1021 seconds
Determining Rate Order Example




If you are given data on a reaction
(concentrations of reactant and times), plot the
data to determine the order.
Plot ln[A] vs. time (first order) and 1/[A] vs. time
(second order).
 Whichever gives a straight line is the order of
the reaction!
To find k, determine the slope of the line
(remember, m = –k for first order and m = k for
second order).
Use appropriate half life equation to calculate t1/2.
Determining Order Example


The following data was collected for the reaction:
2C4H6(g)  C8H12(g)
[C4H6] (mol/L)
Time (±1s)
0.01000
0
0.00625
1000
0.00476
1800
0.00370
2800
0.00313
3600
0.00270
4400
0.00241
5200
0.00208
6200
What is: the reaction order, the rate constant
value, and the half life of the reaction?
Determining Order Example

[C4H6] (mol/L)
Time (±1s)
0.01000
0
0.00625
1000
0.00476
1800
0.00370
2800
0.00313
3600
0.00270
4400
0.00241
5200
0.00208
6200
CAN’T determine order by inspection because
the time at which [ ]’s are measured are not the
same! MUST use integrated rate laws/graphing.
Zero Order Rate Law
Rate = k[A]0 = k
 Rate does not change with
concentration.
 Integrated: [A] = -kt + [A]0
 Half life: t1/2 = [A]0/2k

When Do Zero Order Rate Laws Occur?
Typically happen on a surface because
the surface area stays constant.
 Rate of contact to surface impacts
rate of reaction, not the concentration
of the substance on the surface.
 Also applies to enzyme chemistry.
 Rate depends on how quickly enzyme
works.

AP Practice Question
A reaction follows the rate law:
Rate = k[A]2. Which of the plots below will
give a straight line?
a)
b)
c)
d)
1/[A] vs. 1/time
[A]2 vs. time
1/[A] vs. time
ln[A] vs. time
AP Practice Question
For the following reaction:
NO2 (g) + CO (g)  NO (g) + CO2 (g)
the rate law is: Rate = k[NO2]. If a little CO is
added to a mixture that had 0.10M NO2 and
0.20M CO, which statement is true?
a) Both k and the rxn. rate remain the same.
b) Both k and the rxn. rate increase.
c) Both k and the rxn. rate decrease.
d) Only k increases; rxn. rate stays the same.
AP Practice Question
The decomposition of ammonia to the
elements is a first-order reaction with t1/2 =
200s at a certain T. How long will it take
the partial pressure of NH3 to decrease
from 0.100atm to 0.00625atm?
Remember- P is directly linked to # of
a) 200s
moles for gases! Treat pressures like
b) 400s
amounts.
How much pressure would one half
c) 800s
life give? Is this the final pressure
d) 1,000s
being asked for?
Section 4 Homework/Examples

Pg. 569 # 31, 39, 43, 46
Section 12.5
RATE LAWS: A SUMMARY
Important Points
1.
2.
Forward reaction is important- rate
laws will only contain reactants.
Two types of rate laws:
• Rate Law
•
Shows how rate depends on concentration.
• Integrated Rate Law
•
Shows how concentration depends on time.
Important Points cont.
3.
4.
5.
Type of law depends on type of data
collected.
• Can write the other if we have one.
Initial rate method used if several
experiments are done.
• Order of reactant can be determined
Can graph experimental data to
determine rate law (order).
• K can obtained from the slope
Summary of Rate Laws
Further Look at Activation Energy

Activation energy (Ea): the minimum
energy required to start a reaction.
 Need E to break bonds!
 KE when molecules collide can supply
the energy needed to break bonds and
start the reaction.
 This is why higher T increases rates!
 E is then released when new bonds
form.
Activation Energy Graph
(activated complex)
Note the difference!
The activated complex/transition state is the orientation of
the reactants as bonds are breaking and forming.
Arrhenius Equation

Relationship between T and k can be used to
find the activation energy: k = Ae-Ea/RT

k = rate constant; A = frequency factor; Ea =
activation E; R = ideal gas constant; T =
temperature in K

Take ln of both sides: ln k = -(Ea/R)(1/T) + ln A

ln k vs 1/T is a straight line: slope = -Ea/R.
Since R is known, Ea can be calculated.

A could also be calculated: y-intercept.
AP Practice Question
The energy difference between the
reactants and the transition state is:
a)
b)
c)
d)
the free energy.
the heat of reaction.
the activation energy.
the kinetic energy.
AP Practice Question
The purpose for striking a match against
the side of a box is to:
a)
b)
c)
d)
supply the free energy for the rxn.
supply the activation energy for the rxn.
supply the heat of reaction.
supply the kinetic energy for the rxn.
Section 12.6
REACTION MECHANISMS
Lesson Essential Question:
1) How do the steps of a reaction
impact the rate of the reaction?
Reaction Mechanisms
The reaction mechanism is a series of
steps that occur in a chemical reaction.
 Reactions do not typically occur in one
step. Why might this be?
 Think about orientations and energy!
 Particles need to collide in the correct
location AND with enough energy.

Reaction Mechanisms





Kinetics can tell us about the mechanism;
balanced equations can’t.
Ex: NO2 (g) + CO (g)  NO (g) + CO2 (g)
The proposed mechanism is:
NO2 (g) + NO2 (g)  NO3 (g) + NO (g)
NO3 (g) + CO (g)  NO2 (g) + CO2 (g)
How do people know that the overall
reaction doesn’t happen in one step?
Experimental data suggests the following
rate law: Rate = k[NO2]2. No CO is present!
Reaction Mechanisms
A mechanism should always end up giving
the overall reaction being considered.
 Think of this part like Hess’ Law- cancel
out species that appear on opposite sides
to arrive at the overall reaction.
 Examine the previous mechanism:
NO2 (g) + NO2 (g)  NO3 (g) + NO (g)
+
NO3 (g) + CO (g)  NO2 (g) + CO2 (g)
NO2 (g) + CO (g)  NO (g) + CO2 (g)

Reaction Intermediates
Example continued:
NO2 (g) + NO2 (g)  NO3 (g) + NO (g)
+
NO3 (g) + CO (g)  NO2 (g) + CO2 (g)
NO2 (g) + CO (g)  NO (g) + CO2 (g)
 Reaction intermediates are substances that
are produced and consumed.
 What is/are the intermediate(s) above?
 Only NO3 because it is first produced, and
later consumed. NO2 is there to start with!

Rate-Determining Step



Each step in a mechanism can be called
an elementary step.
One elementary step is always slower than
the rest; it’s called the rate-determining
step (RDS).
This step is essentially the only one
affecting the rate of the overall reaction
since the other steps, in comparison, are
much faster.
Rate-Determining Step



What does this mean?
Rate laws CAN be written directly from the
RDS!
We CANNOT do this with the overall
reaction being examined (as we saw in
previous sections) because it is not the
RDS!
Acceptable Reaction Mechanisms

Two requirements for acceptable
mechanisms:
 Using a process like Hess’ Law, the
proposed mechanism must equal the
overall reaction.
 The RDS must equal the
experimentally determined rate law.
 Rate cannot be faster than the
slowest step!
Rate-Determining Step
Example continued:
NO2 (g) + NO2 (g)  NO3 (g) + NO (g) slow
NO3 (g) + CO (g)  NO2 (g) + CO2 (g) fast
 Which is the RDS?
 The first one!
 The rate law for the overall reaction is:
Rate = k[NO2][NO2]  Rate = k[NO2]2

Reaction Mechanism Practice
2NO2 + F2  2NO2F
 Proposed mechanism:
NO2 + F2  NO2F + F
F + NO2  NO2F

slow
fast
What is the RDS? first elementary step
 What is the rate law for this reaction?
Rate = k[NO2][F2]
 What is/are the intermediate(s)? F

Molecularity






Molecularity tells you the number of
molecules colliding.
Unimolecular- one molecule breaks down
or rearranges.
Bimolecular- two molecules collide.
Termolecular- three molecules collide.
Which of these would you expect to be
least likely to occur?
Termolecular; the odds of three molecules
colliding in exactly the correct orientation is
highly unlikely.
Example
2NO2(g) + F2(g)  2NO2F(g)
 Rate = k[NO2][F2]
 Suggested mechanism:
k1
NO2 + F2
NO2F + F (slow)
k2
F + NO2
NO2F
(fast)
 Is this an acceptable mechanism (does
it satisfy the two requirements)? YES
 Being acceptable does not mean it is
the correct mechanism!

Example
Another possible mechanism:
NO2 + F2  NOF2 + O
(slow)
NO2 + O  NO3
(fast)
NOF2 + NO2  NO2F + NOF (fast)
NO3 + NOF  NO2F + NO2 (fast)
 Perform additional reactions to decide on
the most likely mechanism.
 You can determine the rate law for a
reaction given a proposed mechanism.
 Use RDS information!

Section 12.8
CATALYSTS
Catalysts



Speeds up a reaction without being
consumed; the catalyst will still be there in
the end although it may change along the
way.
An iron ion will speed up the
decomposition of hydrogen peroxide:
2Fe+3 + H2O2  2Fe+2 + O2 + 2H+
2Fe+2 + H2O2 + 2H+  2Fe+3 + 2H2O
Notice the iron ends up exactly how it
started!
Homogeneous Catalysts




Homogeneous catalysts (same state as
reactants) allow reactions to proceed by a
different mechanism - a new pathway.
New pathway has a lower activation
energy.
More molecules will be able to overcome
this activation energy.
Does not change overall DE.
 Starting and ending energies of
reactants and products are not altered.
Homogeneous Catalysts

Two additional examples:
 Enzymes are biological catalysts.
 Chlorofluorocarbons (CFC’s) catalyze
the decomposition of ozone.
Heterogeneous Catalysts



Heterogeneous catalysts (different state
from reactants) allow more surface for the
reaction to occur AND allow for better
orientations of reactants.
Typically the catalyst is a solid for liquid or
gaseous reactants.
Ex: making ammonia (Haber process)
requires an iron catalyst for nitrogen and
hydrogen gases to react.
Additional Info.

For reactions in equilibrium (when both
forward and reverse reactions are
considered), BOTH are affected by adding
a catalyst.
 The rates for both increase since they
share the same Ea.
Sections 7&8 Homework

Pg. 571 #49, 50, 51, 53, 55, 65