AP Ch12 Kinetics.ppt.......................

CHEMICAL
CHEMICAL KINETICS
KINETICS:
:
THE RATES AND MECHANISMS
THE RATES AND MECHANISMS
OF CHEMICAL REACTIONS
OF CHEMICAL REACTIONS

Chemical kinetics is the study of
Chemical kinetics is the study of
the speed or rate of a reaction
the speed or rate of a reaction
under various conditions.
under various conditions.

Spontaneity is also important AND
Spontaneity is also important AND
a spontaneous reaction does NOT
a spontaneous reaction does NOT
imply a rapid reaction.
imply a rapid reaction.
The changing of diamond into
The changing of diamond into
graphite is spontaneous but so
graphite is spontaneous but so
slow that it is not detectable even
slow that it is not detectable even
in a lifetime.
in a lifetime.

A mechanism is a sequence of
A mechanism is a sequence of
events at the molecular level
events at the molecular level
that controls the speed and
that controls the speed and
outcome of the reaction.
outcome of the reaction.

FACTORS THAT AFFECT
FACTORS THAT AFFECT
REACTION RATES
REACTION RATES
The following conditions affect the
The following conditions affect the
speed
speed
of a chemical process:
of a chemical process:

1. Nature of the reactants
1. Nature of the reactants
Some reactant molecules react
Some reactant molecules react
in a hurry, others react very
in a hurry, others react very
slowly.
slowly.

Pointers…
Pointers…
Physical state
Physical state -
-
gasoline
gasoline (l)
(l) vs. gasoline
vs. gasoline (g)
(g)
K
K2
2SO
SO4(s)
4(s) + Ba(NO
+ Ba(NO3
3)
)2(s)
2(s) 
 no rxn.
no rxn.
**while both of these in the
**while both of these in the aqueous
aqueous
state do react.
state do react.

Chemical identity
Chemical identity
What is reacting?
What is reacting?
Usually, ions of opposite charge react
Usually, ions of opposite charge react
very rapidly.
very rapidly.
Usually, the more bonds between
Usually, the more bonds between
reacting atoms in a molecule, the
reacting atoms in a molecule, the
slower the reaction rate.
slower the reaction rate.
Substances with strong bonds (larger
Substances with strong bonds (larger
bond energies) will react much more
bond energies) will react much more
slowly.
slowly.

Examples:
Examples:
Metallic sodium reacts much faster with
Metallic sodium reacts much faster with
water than metallic calcium.
water than metallic calcium.
Oxidation of methane can be increased
Oxidation of methane can be increased
with an increase in temperature.
with an increase in temperature.
Photosynthesis is very slow and
Photosynthesis is very slow and
changes very little with an increase in
changes very little with an increase in
temperature.
temperature.

2. Concentration of reactants
2. Concentration of reactants
More molecules,
More molecules,
More collisions.
More collisions.

3. Temperature
3. Temperature
heat >em up & speed >em up
heat >em up & speed >em up
The faster they move, the more likely
The faster they move, the more likely
they are to collide.
they are to collide.

An increase in temperature
An increase in temperature
produces more successful
produces more successful
collisions that are able to
collisions that are able to
overcome the needed activation
overcome the needed activation
energy,
energy,
therefore,
therefore,
a general increase in reaction rate
a general increase in reaction rate
with increasing temperature.
with increasing temperature.

In fact,
In fact,
A general rule of thumb is that a
A general rule of thumb is that a
10°C increase in temperature will
10°C increase in temperature will
double the reaction rate.
double the reaction rate.
* This actually depends on the
* This actually depends on the
magnitude of the Ea* and the
magnitude of the Ea* and the
temperature range.
temperature range.

4. Catalysts
4. Catalysts
accelerate chemical reactions but are
accelerate chemical reactions but are
not themselves transformed.
not themselves transformed.
Biological catalysts are proteins called
Biological catalysts are proteins called
enzymes.
enzymes.

A catalyst is a substance that
A catalyst is a substance that
changes the rate of reaction by
changes the rate of reaction by
altering the reaction pathway.
altering the reaction pathway.
Most catalysts work by lowering
Most catalysts work by lowering
the activation energy needed for
the activation energy needed for
the reaction to proceed; therefore,
the reaction to proceed; therefore,
more collisions are successful and
more collisions are successful and
the reaction rate is increased.
the reaction rate is increased.

Remember!
Remember!
The catalyst is not part of the chemical
The catalyst is not part of the chemical
reaction and is not used up during the
reaction and is not used up during the
reaction.
reaction.
* (May be homogeneous or
* (May be homogeneous or
heterogeneous catalysts.)
heterogeneous catalysts.)

Example…
Example…
H
H2
2O
O2
2 decomposes relatively slowly into
decomposes relatively slowly into
H
H2
2O and O
O and O2
2
however…
however…
exposure to light accelerates this
exposure to light accelerates this
process AND with the help of MnO
process AND with the help of MnO2
2, it goes
, it goes
extremely FAST!!
extremely FAST!!

Note:
Note:
A catalyst lowers the activation energy
A catalyst lowers the activation energy
barrier. Therefore, the forward and
barrier. Therefore, the forward and
reverse reactions are both
reverse reactions are both
accelerated to the same degree.
accelerated to the same degree.
* (Some homogeneous catalysts
* (Some homogeneous catalysts
actually appear in the rate law
actually appear in the rate law
because their concentration affects
because their concentration affects
the reaction. Ex. NO catalyzing O
the reaction. Ex. NO catalyzing O3
3 )
)

5. Surface area of reactants
5. Surface area of reactants
exposed surfaces affect speed.
exposed surfaces affect speed.
Except for substances in the gaseous
Except for substances in the gaseous
state or solution, reactions occur at the
state or solution, reactions occur at the
boundary, or interface, between two
boundary, or interface, between two
phases.
phases.

The greater surface area exposed, the
The greater surface area exposed, the
greater chance of collisions between
greater chance of collisions between
particles, hence, the reaction should
particles, hence, the reaction should
proceed at a much faster rate.
proceed at a much faster rate.
Ex. coal dust is very explosive as
Ex. coal dust is very explosive as
opposed to a piece of charcoal.
opposed to a piece of charcoal.
Solutions are ultimate exposure!
Solutions are ultimate exposure!

THE COLLISION THEORY
THE COLLISION THEORY
OF REACTION RATES
OF REACTION RATES
Particles must collide.
Particles must collide.
Only two particles may collide at
Only two particles may collide at
one time.
one time.
Proper orientation of colliding
Proper orientation of colliding
molecules so that atoms can come
molecules so that atoms can come
in contact with each other to
in contact with each other to
become products.
become products.

The collision must occur with
The collision must occur with
enough energy to overcome the
enough energy to overcome the
electron/electron repulsion of the
electron/electron repulsion of the
valence shell electrons of the
valence shell electrons of the
reacting species and must have
reacting species and must have
enough energy to transform
enough energy to transform
translational energy into
translational energy into
vibrational energy in order to
vibrational energy in order to
penetrate into each other so that
penetrate into each other so that
the electrons can rearrange and
the electrons can rearrange and
form new bonds.
form new bonds.

This new collision
This new collision
product is at the
product is at the
peak of the
peak of the
activation energy
activation energy
hump and is called
hump and is called
the
the activated
activated
complex
complex or the
or the
transition state. At
transition state. At
this point, the
this point, the
activated complex
activated complex
can still either fall to
can still either fall to
reactants or to
reactants or to
products.
products.

With all of these criteria met,
With all of these criteria met,
the reaction may proceed in
the reaction may proceed in
the forward direction.
the forward direction.
Amazing that we have
Amazing that we have
reactions occurring at all!
reactions occurring at all!

12.1 CHEMICAL REACTION RATES
12.1 CHEMICAL REACTION RATES
The speed of a reaction is
The speed of a reaction is
expressed in terms of its “rate”,
expressed in terms of its “rate”,
some measurable quantity is
some measurable quantity is
changing with time.
changing with time.

The rate of a chemical reaction is
The rate of a chemical reaction is
measured by the decrease in
measured by the decrease in
concentration of a reactant or an
concentration of a reactant or an
increase in concentration of a
increase in concentration of a
product in a unit of time.
product in a unit of time.

Rate =
Rate =
change in concentration of a species
change in concentration of a species
time interval
time interval
When writing rate expressions, they
When writing rate expressions, they
can be written in terms of reactants
can be written in terms of reactants
disappearance or products
disappearance or products
appearance.
appearance.

Rate is not constant, it changes
Rate is not constant, it changes
with time. Graphing the data of
with time. Graphing the data of
an experiment will show an
an experiment will show an
average rate of reaction.
average rate of reaction.
You can find the instantaneous
You can find the instantaneous
rate by computing the slope of a
rate by computing the slope of a
straight line tangent to the curve
straight line tangent to the curve
at that time.
at that time.

Graph of Experimental Data
Graph of Experimental Data

Reaction Rate
Reaction Rate
Expressed as the Δ in concentration of a
Expressed as the Δ in concentration of a
reagent per unit time or Δ[A]/Δt
reagent per unit time or Δ[A]/Δt
Focus either on the disappearance of
Focus either on the disappearance of
reactants or the appearance of products
reactants or the appearance of products
 rate of Δ of a reactant is always
rate of Δ of a reactant is always negative
negative
 rate of Δ of a product is always
rate of Δ of a product is always positive
positive

Consider:
Consider:
2 NO
2 NO2(g)
2(g) →
→ O
O 2(g)
2(g)+ 2 NO
+ 2 NO(g)
(g)
 Oxygen can appear only half as rapidly
Oxygen can appear only half as rapidly
as the nitrogen dioxide disappears
as the nitrogen dioxide disappears
 NO appears twice as fast as oxygen
NO appears twice as fast as oxygen
appears.
appears.

Calculate the
Calculate the AVERAGE
AVERAGE rate at which
rate at which
[NO
[NO2
2] changes in the first 50.0 seconds:
] changes in the first 50.0 seconds:
RATE =
RATE = Δ [NO
Δ [NO2
2]
] =
= [.0079]-[0.0100]
[.0079]-[0.0100]
Δt
Δt 50.0 s
50.0 s
= -[-4.2 x 10
= -[-4.2 x 10-5
-5
mol/L • sec]
mol/L • sec]
= 4.2 x 10
= 4.2 x 10-5
-5
mol/L • sec
mol/L • sec
or, M • s-1
or, M • s-1

Note that the rate is
Note that the rate is NOT
NOT constant
constant
but decreases with time.
but decreases with time.
The rates given below are
The rates given below are average
average rates.
rates.

To find the value of the rate at a
To find the value of the rate at a
particular time, the
particular time, the
instantaneous rate
instantaneous rate, compute
, compute
the
the slope
slope of a line
of a line tangent to
tangent to
the curve
the curve at that point.
at that point.
Why the negative on NO
Why the negative on NO2
2?
?

RELATIVE RATES
RELATIVE RATES
We can consider the appearance of
We can consider the appearance of
products along with the
products along with the
disappearance of reactants.
disappearance of reactants.
The reactant’s concentration is
The reactant’s concentration is
declining, the product’s is
declining, the product’s is
increasing.
increasing.

Respect the algebraic sign AND
Respect the algebraic sign AND
respect the stoichiometry.
respect the stoichiometry.
Divide the rate of change in
Divide the rate of change in
concentration of each reactant
concentration of each reactant
by its stoichiometric coefficient
by its stoichiometric coefficient
in the balanced chem. eqn. and
in the balanced chem. eqn. and
this is foolproof and a breeze!
this is foolproof and a breeze!

Thus...
Thus...
Rate of Reaction =
Rate of Reaction =
-
- 1Δ[NO
1Δ[NO2
2]
] =
= 1 Δ[NO]
1 Δ[NO] =
= Δ[O2]
Δ[O2]
2 Δtime 2 Δtime
2 Δtime 2 Δtime Δtime
Δtime
Of course you can change these once
Of course you can change these once
the ratio is set. You might prefer:
the ratio is set. You might prefer:
-1 : +1 : +2
-1 : +1 : +2

Relative Rates from the
Relative Rates from the
balanced equation:
balanced equation:
Using the coefficients from the
Using the coefficients from the
balanced equation, students should
balanced equation, students should
be
be
able to give relative rates.
able to give relative rates.
For example:
For example:
4 PH
4 PH3 (g)
3 (g) 
 P
P4(g)
4(g) + 6 H
+ 6 H2(g)
2(g)

Initial rate rxn.
Initial rate rxn. =
=

Exercise
Exercise
What are the
What are the relative
relative rates of change
rates of change
in concentration of the products and
in concentration of the products and
reactant in the decomposition of
reactant in the decomposition of
Nitrosyl chloride, NOCl?
Nitrosyl chloride, NOCl?
2 NOCl
2 NOCl (g)
(g) →
→ 2 NO
2 NO(g)
(g) + Cl
+ Cl2(g)
2(g)

12.2 RATE LAWS:
12.2 RATE LAWS:
AN INTRODUCTION
AN INTRODUCTION
Reactions are reversible. So far,
Reactions are reversible. So far,
we’ve only considered the
we’ve only considered the
forward
forward
reaction. The reverse is equally
reaction. The reverse is equally
important.
important.

When the rate of the forward =
When the rate of the forward =
the rate of the reverse, we have
the rate of the reverse, we have
EQUILIBRIUM!
EQUILIBRIUM!
To avoid this complication we will
To avoid this complication we will
discuss reactions soon after
discuss reactions soon after
mixing--initial reaction rates, and
mixing--initial reaction rates, and
not worry
not worry
about the buildup of products and
about the buildup of products and
how that starts up the reverse
how that starts up the reverse
reaction.
reaction.

Initial Reaction Rates
Initial Reaction Rates
Begin with pure reactants, mix thoroughly, then
Begin with pure reactants, mix thoroughly, then
measure speed of rxn. Over time, the presence
measure speed of rxn. Over time, the presence
of products can alter results dramatically and lead
of products can alter results dramatically and lead
to confusing results.
to confusing results.
We’ll be talking initial reaction rates throughout
We’ll be talking initial reaction rates throughout
our discussions!
our discussions!

Rate expression or rate law is
Rate expression or rate law is
the relation between
the relation between
reaction rate and the
reaction rate and the
concentrations of reactants
concentrations of reactants
given by a mathematical
given by a mathematical
equation.
equation.

CONCENTRATION AND
CONCENTRATION AND
REACTION RATE:
REACTION RATE:
THE RATE LAW OR RATE EXPRESSION:
THE RATE LAW OR RATE EXPRESSION:
Rates generally depend on
Rates generally depend on
reactant concentrations.
reactant concentrations.
To find the exact relation between rate
To find the exact relation between rate
and concentration, we must
and concentration, we must do some
do some
experiments
experiments and collect information.
and collect information.

Where C is a catalyst, the
Where C is a catalyst, the
rate expression will
rate expression will always
always
have the form:
have the form:
xX
C
bB
+
aA


Initial rxn rate =
Initial rxn rate = k
k[A]
[A]m
m
[B]
[B]n
n
[C]
[C]p
p
k
k = rate constant
= rate constant
[A] = concentration of reactant A
[A] = concentration of reactant A
[B] = concentration of reactant B
[B] = concentration of reactant B
[C] = concentration of the catalyst—won’t see
[C] = concentration of the catalyst—won’t see
this too often in AP
this too often in AP
m = order of reaction for reactant A
m = order of reaction for reactant A
n = order of reaction for reactant B
n = order of reaction for reactant B
p = order of reaction for the catalyst C
p = order of reaction for the catalyst C

Exponents can be zero, whole
Exponents can be zero, whole
numbers, or fractions --
numbers, or fractions --
AND MUST BE DETERMINED BY
AND MUST BE DETERMINED BY
EXPERIMENTATION!!
EXPERIMENTATION!!

THE RATE CONSTANT,
THE RATE CONSTANT, k
k
Is temperature dependent & must be
Is temperature dependent & must be
evaluated by experiment.
evaluated by experiment.
Example:
Example: rate =
rate = k
k[Pt(NH
[Pt(NH3
3)
)2
2Cl
Cl2
2]
]
k
k = 0.090/hr, therefore when [ion] = 0.018 mol/L
= 0.090/hr, therefore when [ion] = 0.018 mol/L
rate =
rate =
(.0090/hr)(0.018 mol/L) = 0.0016 mol/(L• hr)
(.0090/hr)(0.018 mol/L) = 0.0016 mol/(L• hr)

ORDER OF A REACTION
ORDER OF A REACTION
Order with respect to a certain reactant is
Order with respect to a certain reactant is
the
the exponent
exponent on its concentration term in
on its concentration term in
the rate expression.
the rate expression.
Order of the reaction is the sum of all the
Order of the reaction is the sum of all the
exponents on all the concentration terms
exponents on all the concentration terms
in
in
the expression.
the expression.

DETERMINATION OF THE
DETERMINATION OF THE
RATE EXPRESSION
RATE EXPRESSION
aA + bB
aA + bB →
→ xX
xX
initial rate =
initial rate = k
k[A]
[A]o
o
m
m
[B]
[B]o
o
n
n
the little subscript “o” means original.
the little subscript “o” means original.

Zero order
Zero order
 The change in concentration of
The change in concentration of
reactant has no effect on the rate.
reactant has no effect on the rate.
 These are not very common.
These are not very common.
 General form of rate equation:
General form of rate equation:
Rate = k
Rate = k

First order
First order
 Rate is directly proportional to the
Rate is directly proportional to the
reactants concentration; doubling
reactants concentration; doubling
[rxt], doubles rate. These are very
[rxt], doubles rate. These are very
common! Nuclear decay reactions
common! Nuclear decay reactions
usually fit into this category.
usually fit into this category.
Rate = k [A]
Rate = k [A]

Second order
Second order
 Rate is quadrupled when [rxt] is
Rate is quadrupled when [rxt] is
doubled and increases by a factor
doubled and increases by a factor
of 9 when [rxt] is tripled, etc.
of 9 when [rxt] is tripled, etc.
These are common, particularly in
These are common, particularly in
gas-phase reactions.
gas-phase reactions.
Rate = k [A]
Rate = k [A]2
2

Fractional orders are rare!
Fractional orders are rare!
Ex.
Ex. rate =
rate = k
k[A]
[A]o
o
m
m
[B]
[B]o
o
n
n
If m = 0 ; reaction is
If m = 0 ; reaction is zero order
zero order with respect to A
with respect to A
If m = 1 ; reaction is 1st order
1st order with respect to A
with respect to A
If m = 2 ; reaction is 2nd order
2nd order with respect to A
with respect to A
If n = 0 ; reaction is
If n = 0 ; reaction is zero order
zero order with respect to B
with respect to B
If n = 1 ; reaction is 1st order
1st order with respect to B
with respect to B
If n = 2 ; reaction is 2nd order
2nd order with respect to B
with respect to B

Adding the orders of each
Adding the orders of each
reactant gives the
reactant gives the overall order
overall order of
of
the reaction.
the reaction.

Since the rate stays the same
Since the rate stays the same
regardless of the concentration of
regardless of the concentration of
[A], it is zero order with respect to A.
[A], it is zero order with respect to A.
However, the rate doubles with a
However, the rate doubles with a
doubling of [B] and triples with a
doubling of [B] and triples with a
tripling of [B].
tripling of [B].
This indicates the rate is first order
This indicates the rate is first order
with respect to [B].
with respect to [B].

Summary
Summary
Initial reaction rate =
Initial reaction rate =
k
k[A]
[A]o
o
o
o
[B]
[B]o
o1 =
1 = k
k[B]
[B]o
o
1
1
The overall reaction rate =
The overall reaction rate =
1 + 0 = 1
1 + 0 = 1st
st
order overall.
order overall.

Now. . . . .
Now. . . . .
Use a set of the data to calculate
Use a set of the data to calculate k
k:
:
0.0050 mol/(L•hr) =
0.0050 mol/(L•hr) = k
k[0.20 mol/L]
[0.20 mol/L]1
1
k =
k = 2.5 x 10
2.5 x 10-2
-2
/hr
/hr
You should get the same value with any
You should get the same value with any
set of data!
set of data!

Ugly algebraic method is
Ugly algebraic method is
sometimes necessary
sometimes necessary
rate 1
rate 1 =
= k
k [reactant]
[reactant]m
m
[reactant]
[reactant]n
n
rate 2
rate 2 k
k [reactant]
[reactant]m
m
[reactant]
[reactant]n
n

Select a trial where one reactant
Select a trial where one reactant
concentration is held constant
concentration is held constant
SO THAT IT CANCELS;
SO THAT IT CANCELS;
the
the k
k’s will also cancel.
’s will also cancel.

Using trials 1 & 4:
Using trials 1 & 4:
0.50 x 10
0.50 x 10-
-2
2
=
= k
k [0.50]
[0.50]m
m
[0.20]
[0.20]n
n
1.00 x 10
1.00 x 10-2
-2
k
k [0.50]
[0.50]m
m
[0.40]
[0.40]n
n
so….
so…. ½ = [ ½ ]
½ = [ ½ ]n
n
and
and 
 n must be ONE to make that
n must be ONE to make that
true!
true!

Exercise
Exercise
In the following reaction, a Co-Cl bond is
In the following reaction, a Co-Cl bond is
replaced by a Co-OH
replaced by a Co-OH2
2 bond.
bond.
[Co(NH
[Co(NH3
3)
)5
5Cl]
Cl]+2
+2
+ H
+ H2
2O
O →
→
[Co(NH
[Co(NH3
3)
)5
5H
H2
2O]
O]+3
+3
+ Cl
+ Cl
Initial rate =
Initial rate = k
k {[Co(NH
{[Co(NH3
3)
)5
5Cl]
Cl]+2
+2
}
}m
m

Using the data below, find the
Using the data below, find the
value of
value of m
m in the rate expression
in the rate expression
and calculate the value of
and calculate the value of k
k.
.
Exp. Initial Concentration Initial rate
Exp. Initial Concentration Initial rate
of [Co(NH
of [Co(NH3
3)
)5
5Cl]
Cl]+2
+2
mol/(L•min)
mol/(L•min)
(mol/L)
(mol/L)
1 1.0 x 10
1 1.0 x 10-3
-3
1.3 x 10
1.3 x 10-7
-7
2 2.0 x 10
2 2.0 x 10-3
-3
2.6 x 10
2.6 x 10-7
-7
3 3.0 x 10
3 3.0 x 10-3
-3
3.9 x 10
3.9 x 10-7
-7
4 1.0 x 10
4 1.0 x 10-3
-3
1.3 x 10
1.3 x 10-7
-7

Exercise 12.1
Exercise 12.1
The reaction between bromate ions
The reaction between bromate ions
and bromide ions in acidic aqueous
and bromide ions in acidic aqueous
Solution is given by the equation:
Solution is given by the equation:
BrO
BrO3
3
–
–
(aq) + 5 Br
(aq) + 5 Br –
–
(aq) + 6 H
(aq) + 6 H+
+
(aq)
(aq) 

3 Br
3 Br2
2 (
(l
l) + 3 H
) + 3 H2
2O (
O (l
l)
)

The table below gives the results of four
The table below gives the results of four
experiments
experiments. Using these data, determine
. Using these data, determine
the orders for all three reactants, the
the orders for all three reactants, the
overall reaction order, and the value of
overall reaction order, and the value of
the rate constant.
the rate constant.
What is the value of
What is the value of k
k? What are the units
? What are the units
of
of k
k?
?

TWO TYPES OF RATE LAW
TWO TYPES OF RATE LAW
 differential rate law
differential rate law--expresses
--expresses
how the rate depends on
how the rate depends on
concentration
concentration (most common &
(most common &
what we’ve been doing!)
what we’ve been doing!)
 integrated rate law
integrated rate law--expresses
--expresses
how the concentrations depend on
how the concentrations depend on
time
time

12.3 DETERMINING THE
12.3 DETERMINING THE
FORM OF THE RATE LAW
FORM OF THE RATE LAW
--experimental convenience
--experimental convenience
Note the shape of this curve! It will save
Note the shape of this curve! It will save
you time in the future!
you time in the future!

 Write the relative rate
Write the relative rate
expression:
expression:
 Write the differential rate law
Write the differential rate law
[expression]:
[expression]:

12.4 INTEGRATED RATE LAW -
12.4 INTEGRATED RATE LAW -
CONCENTRATION/TIME RELATIONSHIPS
CONCENTRATION/TIME RELATIONSHIPS
When we wish to know how long a
When we wish to know how long a
reaction must proceed to reach a
reaction must proceed to reach a
predetermined concentration of
predetermined concentration of
some reagent, we can construct
some reagent, we can construct
curves or derive an equation that
curves or derive an equation that
relates concentration and time.
relates concentration and time.

GRAPHICAL METHODS FOR
GRAPHICAL METHODS FOR
DISTINGUISHING FIRST AND
DISTINGUISHING FIRST AND
SECOND ORDER REACTIONS
SECOND ORDER REACTIONS
First order:
First order:
ln[A] = -
ln[A] = -k
kt + ln[A]
t + ln[A]o
o
y = ax + b
y = ax + b
Second order:
Second order:
1/[A] =
1/[A] = k
kt + 1/[A]
t + 1/[A]o
o
y = ax + b
y = ax + b

ln[reactant] vs. time
ln[reactant] vs. time 
 straight line for
straight line for
first order
first order in that reactant & since a = -
in that reactant & since a = -k
k
the slope of the line is
the slope of the line is negative
negative.
.
1/[reactant] vs. time
1/[reactant] vs. time 
 straight line for
straight line for
second order
second order in that reactant since a =
in that reactant since a = k
k
the slope is
the slope is positive
positive.
.

Using the graphing calculator
Using the graphing calculator
Set up your calculator so that
Set up your calculator so that
time is always in L1 and the y-
time is always in L1 and the y-
list
list
is alphabetical!
is alphabetical!

L1
L1 
 time (x variable throughout!)
time (x variable throughout!)
L2
L2 
 concentration:
concentration: [A]
[A]
straight line = zero order
straight line = zero order
L3
L3 
 ln concentration: ln [A]
ln concentration: ln [A]
straight line = first order
straight line = first order
L4
L4 
 reciprocal concentration: 1/[A]
reciprocal concentration: 1/[A]
straight line = second order
straight line = second order

Run 3 linear regressions – one
Run 3 linear regressions – one
each for:
each for:
L1, L2
L1, L2
L1, L3
L1, L3
L1, L4
L1, L4
and see which has the best “r”
and see which has the best “r”
[linear regression correlation
[linear regression correlation
coefficient in big people
coefficient in big people
language!]
language!]

Paste the best one into y= by
Paste the best one into y= by
hitting
hitting


to get the command back on
to get the command back on
the screen,
the screen,
then “fix” it to read
then “fix” it to read LinReg
LinReg {the
{the
combination that was the best
combination that was the best
regression}.
regression}.

Next, hit
Next, hit

 
 to Y-VARS
to Y-VARS
then
then


If you were successful, you’ll see
If you were successful, you’ll see
LinReg(ax +b) L
LinReg(ax +b) L1
1, L
, Lwhichever you chose
whichever you chose, Y
, Y1
1
displayed on your screen.
displayed on your screen.

The order of the reaction is
The order of the reaction is
0; 1; 2
0; 1; 2
respectively for each
respectively for each
combination.
combination.
|slope| = k
|slope| = k
Rate = k[rxt.]
Rate = k[rxt.]order
order

Next,
Next,
since linear,
since linear,
NEVER, EVER FORGET:
NEVER, EVER FORGET:
y = mx + b
y = mx + b
(TI uses an “a” instead of an “m”)
(TI uses an “a” instead of an “m”)

If L1, L3 was your best “r”, the
If L1, L3 was your best “r”, the
reaction is first order and
reaction is first order and
y = mx + b
y = mx + b
becomes
becomes
ln [conc.] = k
ln [conc.] = k
(DO use the proper sign for k here!)
(DO use the proper sign for k here!)
t + ln [conc.
t + ln [conc.o
o]
]

Do the same substitutions into
Do the same substitutions into
y = mx + b
y = mx + b
for the other formats!
for the other formats!

Exercise 12.2
Exercise 12.2
The decomposition of N
The decomposition of N2
2O
O5
5 in the
in the
gas
gas
phase was studied at constant
phase was studied at constant
temperature.
temperature.
2 N
2 N2
2O
O5
5 (g)
(g) 
 4 NO
4 NO2
2 (g) + O
(g) + O2
2 (g)
(g)

The following results were collected:
The following results were collected:
[N
[N2
2O
O5
5]
] Time (s)
Time (s)
0.1000
0.1000 0
0
0.0707
0.0707 50
50
0.0500
0.0500 100
100
0.0250
0.0250 200
200
0.0125
0.0125 300
300
0.00625
0.00625 400
400
Determine the rate law and calculate
Determine the rate law and calculate
the value of
the value of k
k.
.

Once you have the CORRECT
Once you have the CORRECT
equation for the reaction’s rate
equation for the reaction’s rate
law in your calculator so that it can
law in your calculator so that it can
draw the CORRECT linear
draw the CORRECT linear
regression line…
regression line…
You can display the graph.
You can display the graph.

Make sure your plot 1 is ON and
Make sure your plot 1 is ON and
then set it up to read the
then set it up to read the
CORRECT axes.
CORRECT axes.
Check the max and min x-values
Check the max and min x-values
that zoom 9 assigned to the
that zoom 9 assigned to the
window.
window.

You can now solve for any
You can now solve for any
concentration EXACTLY between
concentration EXACTLY between
those max and min values.
those max and min values.
What if your window doesn’t have
What if your window doesn’t have
the proper time range?
the proper time range?
CHANGE IT!
CHANGE IT!

To solve,
To solve,
1.
1. Display your graph by hitting
Display your graph by hitting 
.
.
2.
2. Next hit
Next hit 
 to get to calculate then choose
to get to calculate then choose

 which is “value”.
which is “value”.
3.
3. Now your screen has the graph displayed
Now your screen has the graph displayed
AND in the lower left corner an x= with a
AND in the lower left corner an x= with a
flashing cursor.
flashing cursor.
4.
4. Just enter the time you want the
Just enter the time you want the
concentration for and voila!
concentration for and voila!

Exercise 12.3
Exercise 12.3
Using the data given in Ex. 12.2 (shown below),
Using the data given in Ex. 12.2 (shown below),
calculate
calculate
[N
[N2
2O
O5
5] at 150 s after the start of the reaction.
] at 150 s after the start of the reaction.
[N
[N2
2O
O5
5]
] Time (s)
Time (s)
0.1000
0.1000 0
0
0.0707
0.0707 50
50
0.0500
0.0500 100
100
0.0250
0.0250 200
200
0.0125
0.0125 300
300
0.00625
0.00625 400
400
Calculate the [N
Calculate the [N2
2O
O5
5] at the following times:
] at the following times:
200 s
200 s 400 s
400 s 600 s
600 s 1,000 s
1,000 s

HALF-LIFE AND REACTION RATE FOR
FIRST ORDER REACTIONS, t
FIRST ORDER REACTIONS, t1/2
1/2
The time required for one half of one
of the reactants to disappear.
[A] = ½[A]
[A] = ½[A]o
o or
or [A]
[A] = ½
= ½
[A]
[A]o
o
so... ln
so... ln [A]
[A] =
= k
k t
t½
½
[A]
[A]o
o/2
/2
and... ln 2 = t
and... ln 2 = t½
½

Rearrange, evaluate ln 2 and solve for
Rearrange, evaluate ln 2 and solve for
t
t½
½ and you get
and you get
t
t½
½ =
= 0.693
0.693
k
k
“
“Half life is INDEPENDENT OF
Half life is INDEPENDENT OF
ORIGINAL
ORIGINAL C
CONCENTRATION for 1
ONCENTRATION for 1st
st
order!!!”
order!!!”

Exercise 12.4
Exercise 12.4
A certain first-order reaction has a
A certain first-order reaction has a
half-life of 20.0 minutes.
half-life of 20.0 minutes.
a. Calculate the rate constant for this
a. Calculate the rate constant for this
reaction.
reaction.
b. How much time is required for this
b. How much time is required for this
reaction to be 75% complete?
reaction to be 75% complete?

SECOND ORDER REACTIONS, t1/2
SECOND ORDER REACTIONS, t1/2
[A] = ½[A]
[A] = ½[A]o
o or
or [A]
[A] = ½
= ½
[A]
[A]o
o
so...
so...
1
1 =
= k
k t
t½
½ +
+ 1
1
[A]
[A]o
o/2
/2 [A]
[A]o
o

Rearrange ,
Rearrange ,
2
2 -
- 1
1 =
= k
k t
t½
½
[A]
[A]o
o [A]
[A]o
o
k
k t
t½
½ =
= 1
1 solve for t
solve for t½
½,
,
[A]
[A]o
o
t
t½
½ =
= 1
1 for a 2
for a 2nd
nd
order rxn.
order rxn.
k
k[A]
[A]o
o

Exercise
Exercise
The rate constant for the first order
The rate constant for the first order
transformation of cyclopropane to
transformation of cyclopropane to
propene is 5.40 x 10
propene is 5.40 x 10-2
-2
/hr.
/hr.
-What is the half-life of this reaction?
-What is the half-life of this reaction?
-What fraction of the cyclopropane
-What fraction of the cyclopropane
remains after 51.2 hours?
remains after 51.2 hours?
-What fraction remains after 18.0 hours?
-What fraction remains after 18.0 hours?

Exercise
Exercise
For the reaction of (CH
For the reaction of (CH3
3)
)3
3CBr with OH
CBr with OH
-
-
,
,
(CH
(CH3
3)
)3
3CBr + OH
CBr + OH -
-
→
→ (CH
(CH3
3)
)3
3COH + Br
COH + Br -
-
The following data were obtained in
The following data were obtained in
the laboratory:
the laboratory:

TIME (s)
TIME (s) [(CH
[(CH3
3)
)3
3CBr]
CBr]
0
0 0.100
0.100
30
30 0.074
0.074
60
60 0.055
0.055
90
90 0.041
0.041
Plot these data as ln [(CH
Plot these data as ln [(CH3
3)
)3
3CBr]
CBr]
versus time. Sketch your graph.
versus time. Sketch your graph.

Is the reaction first order or
Is the reaction first order or
second
second
order?
order?
What is the value of the rate
What is the value of the rate
constant?
constant?

Exercise 12.5
Exercise 12.5
Butadiene reacts to form its
Butadiene reacts to form its
dimer
dimer
according to the equation:
according to the equation:
2 C
2 C4
4H
H6
6 (g)
(g) 
 C
C8
8H
H12
12 (g)
(g)

The following data were collected for
The following data were collected for
this reaction at a given temperature:
this reaction at a given temperature:
[C
[C4
4H
H6
6]
] Time (
Time (
 1 s)
1 s)
0.01000
0.01000 0
0
0.00625
0.00625 1000
1000
0.00476
0.00476 1800
1800
0.00370
0.00370 2800
2800
0.00313
0.00313 3600
3600
0.00270
0.00270 4400
4400
0.00241
0.00241 5200
5200
0.00208
0.00208 6200
6200
What is the order of this reaction?
What is the order of this reaction?
Explain.
Explain.
Sketch your graph as part of your
Sketch your graph as part of your

Continued:
Continued:
[C
[C4
4H
H6
6]
] Time (
Time (
 1 s)
1 s)
0.01000
0.01000 0
0
0.00625
0.00625 1000
1000
0.00476
0.00476 1800
1800
0.00370
0.00370 2800
2800
0.00313
0.00313 3600
3600
0.00270
0.00270 4400
4400
0.00241
0.00241 5200
5200
0.00208
0.00208 6200
6200
-What is the value of the rate constant for this reaction?
-What is the value of the rate constant for this reaction?
-What is the half-life for the reaction under the conditions
-What is the half-life for the reaction under the conditions
of this experiment?
of this experiment?

ZERO-ORDER REACTIONS, t
ZERO-ORDER REACTIONS, t1/2
1/2
The time required for one half of one of
The time required for one half of one of
the reactants to disappear, BUT
the reactants to disappear, BUT
Rate =
Rate = k
k[A]
[A]0
0 =
= k
k (a big fat 1) =
(a big fat 1) = k
k
Integrated rate law is:
Integrated rate law is:
[A] = -
[A] = -k
kt + [A]
t + [A]o
o

[A] = ½[A]
[A] = ½[A]o
o or
or [A]
[A] = ½
= ½
[A]
[A]o
o
so...
so...
[A]
[A]o
o = - k t
= - k t½
½ + [A]
+ [A]o
o
2
2

K
K t
t½
½ =
= [A]
[A]o
o
2
2k
k
solve for t
solve for t½
½:
:
t
t½
½ =
= [A]
[A]o
o
2
2k
k
for a ZERO order rxn.
for a ZERO order rxn.

Zero-order reactions are most
Zero-order reactions are most
often encountered when a
often encountered when a
substance such as a metal surface
substance such as a metal surface
or an enzyme is required for the
or an enzyme is required for the
reaction to occur.
reaction to occur.
The enzyme or catalyst may
The enzyme or catalyst may
become saturated and therefore
become saturated and therefore
an increase in the
an increase in the
[reactant/substrate] has no effect
[reactant/substrate] has no effect

INTEGRATED RATE LAWS FOR
INTEGRATED RATE LAWS FOR
REACTIONS WITH MORE THAN ONE
REACTIONS WITH MORE THAN ONE
REACTANT
REACTANT
Must [still] be determined by
Must [still] be determined by
experiment! But we use a
experiment! But we use a
technique called “swamping”.
technique called “swamping”.

Flood the reaction vessel with
Flood the reaction vessel with
high concentrations of all but
high concentrations of all but
one reactant and perform the
one reactant and perform the
experiment.
experiment.
The reactants at high
The reactants at high
concentrations like say, 1.0 M
concentrations like say, 1.0 M
compared to the reactant with a
compared to the reactant with a
low concentration say, 1.0 x 10
low concentration say, 1.0 x 10-3
-3
M, stay the same.
M, stay the same.

“
“In English”—the rate is now
In English”—the rate is now
dependent on the concentration
dependent on the concentration
of the little guy since the big guy’s
of the little guy since the big guy’s
aren’t changing, therefore:
aren’t changing, therefore:
rate =
rate = k
k’ [little guy]
’ [little guy]
We now re-write the rate as a
We now re-write the rate as a
pseudo-rate-law
pseudo-rate-law and
and k
k’ is a
’ is a
pseudo-rate-constant
pseudo-rate-constant

This is what is happening in
This is what is happening in
the Crystal Violet lab!
the Crystal Violet lab!

12.6 REACTION
12.6 REACTION
MECHANISMS
MECHANISMS
The sequence of bond-making and
The sequence of bond-making and
bond-breaking steps that occurs during
bond-breaking steps that occurs during
the conversion of reactants to products.
the conversion of reactants to products.
Must be determined by experiment!
Must be determined by experiment!
Must agree with overall stoichiometry
Must agree with overall stoichiometry
AND the experimentally determined
AND the experimentally determined
rate law!
rate law!

ELEMENTARY STEPS
ELEMENTARY STEPS
Molecularity
Molecularity--number of molecules that
--number of molecules that
participate in an atomic rearrangement
participate in an atomic rearrangement
 unimolecular: involves one reactant
unimolecular: involves one reactant
molecule
molecule
 bimolecular: involves a collision
bimolecular: involves a collision
between two reactant molecules
between two reactant molecules
 termolecular: simultaneous collision
termolecular: simultaneous collision
between three reactant molecules
between three reactant molecules
[very rare!]*
[very rare!]*

RATE EXPRESSIONS FOR
RATE EXPRESSIONS FOR
ELEMENTARY STEPS
ELEMENTARY STEPS
The rate expression cannot be
The rate expression cannot be
predicted from overall stoichiometry.
predicted from overall stoichiometry.
The rate expression of an elementary
The rate expression of an elementary
step is given by the product of the
step is given by the product of the
rate constant and the concentrations
rate constant and the concentrations
of the reactants in the step.
of the reactants in the step.

THE PHYSICAL SIGNIFICANCE OF RAT
THE PHYSICAL SIGNIFICANCE OF RAT
EXPRESSIONS FOR ELEMENTARY
EXPRESSIONS FOR ELEMENTARY
STEPS
STEPS
the more molecules the more
the more molecules the more
collisions, the faster the rate
collisions, the faster the rate
the faster the molecules are
the faster the molecules are
moving, the more likely they will
moving, the more likely they will
collide, the faster the rate
collide, the faster the rate

MOLECULARITY AND ORDER
MOLECULARITY AND ORDER
an
an elementary step
elementary step is a reaction
is a reaction
whose rate law can be written
whose rate law can be written
from its molecularity
from its molecularity
NOT true of the overall reaction
NOT true of the overall reaction
order!
order!

Exercise
Exercise
Nitrogen oxide is reduced by
Nitrogen oxide is reduced by
hydrogen
hydrogen
to give water and nitrogen,
to give water and nitrogen,
2 H
2 H2(g)
2(g) + 2 NO
+ 2 NO(g)
(g) →
→ N
N2(g)
2(g) + 2 H
+ 2 H2
2O
O(g)
(g)

One possible mechanism to
One possible mechanism to
account for this reaction is:
account for this reaction is:
2 NO
2 NO(g)
(g) 
 N
N2
2O
O2(g)
2(g)
N
N2
2O
O2(g)
2(g) + H
+ H2(g)
2(g) →
→ N
N2
2O
O(g)
(g) + H
+ H2
2O
O(g)
(g)
N
N2
2O
O(g)
(g) + H
+ H2(g)
2(g) →
→ N
N2(g)
2(g) + H
+ H2
2O
O(g)
(g)

What is the molecularity of
What is the molecularity of
each of the three steps?
each of the three steps?
Show that the sum of these
Show that the sum of these
elementary steps is the net
elementary steps is the net
reaction.
reaction.

REACTION MECHANISMS
REACTION MECHANISMS
AND RATE EXPRESSIONS
AND RATE EXPRESSIONS
 determined by experiment
determined by experiment
 the rate of the overall reaction is
the rate of the overall reaction is
limited by, and is exactly equal
limited by, and is exactly equal
to, the combined rates of all
to, the combined rates of all
elementary steps
elementary steps up to and
up to and
including the slowest step in the
including the slowest step in the
mechanism
mechanism

 the slowest step is the
the slowest step is the rate
rate
determining step
determining step
 reaction intermediate
reaction intermediate--
--
produced in one step but
produced in one step but
consumed in another.
consumed in another.
 catalyst
catalyst--goes in, comes out
--goes in, comes out
unharmed and DOES NOT show
unharmed and DOES NOT show
up in the final rxn.
up in the final rxn.

Exercise 12.6
Exercise 12.6
The balanced equation for the reaction
The balanced equation for the reaction
of the gases nitrogen dioxide and
of the gases nitrogen dioxide and
fluorine is
fluorine is
2 NO
2 NO2
2 (g) + F
(g) + F2
2 (g)
(g) 
 2 NO
2 NO2
2F (g)
F (g)
The
The experimentally
experimentally determined rate
determined rate
law is
law is
Rate =
Rate = k
k [NO
[NO2
2][F
][F2
2]
]

Exercise 12.6, Continued
Exercise 12.6, Continued
A suggested mechanism for the reaction is
A suggested mechanism for the reaction is
NO
NO2
2 + F
+ F2
2 
 NO
NO2
2F + F Slow
F + F Slow
F + NO
F + NO2
2 
 NO
NO2
2F Fast
F Fast
Is this an acceptable mechanism? That is,
Is this an acceptable mechanism? That is,
does it satisfy the two requirements?
does it satisfy the two requirements?
Justify.
Justify.

12.7 A MODEL FOR KINETICS
12.7 A MODEL FOR KINETICS
Generally reactions occur more
Generally reactions occur more
rapidly at higher temperatures than
rapidly at higher temperatures than
at lower temperatures.
at lower temperatures.
The rate generally doubles for every
The rate generally doubles for every
10 K rise in temperature. It’s an
10 K rise in temperature. It’s an
exponential increase!
exponential increase!

TRANSITION STATE THEORY
TRANSITION STATE THEORY
 energy barrier must be overcome
energy barrier must be overcome
 reaction energy diagram
reaction energy diagram [humpy diagrams]
[humpy diagrams]
 transition state energy--max of rxn. E
transition state energy--max of rxn. E
diagram
diagram
 activated complex--deformed molecules in
activated complex--deformed molecules in
their transition state, formed at the E
their transition state, formed at the Ets
ts--
--
unstable, can go either way!
unstable, can go either way!
 Activation energy, E*, Ea--energy a reacting
Activation energy, E*, Ea--energy a reacting
molecule must absorb from its environment
molecule must absorb from its environment
in order to react.
in order to react.

Kinetics and net energy of reaction
Kinetics and net energy of reaction
Potential
Potential
Energy
Energy
-relationship between kinetics and thermodynamics
-relationship between kinetics and thermodynamics
-endo
-endothermic—products end up higher in energy than reactants
thermic—products end up higher in energy than reactants

 heat is absorbed or taken into or added to the system and given
heat is absorbed or taken into or added to the system and given
a positive sign.
a positive sign.
-exo
-exothermic—pictured above, products are of lower energy than
thermic—pictured above, products are of lower energy than
reactants
reactants 
 heat is lost to the surroundings and given a negative
heat is lost to the surroundings and given a negative
sign.
sign.
Reaction Coordinate
(time)
X2 + Y2
2 XY
Eactivation -- a kinetic quantity
E H
--a thermodynamic
quantity

COLLISION THEORY
COLLISION THEORY
Assumes molecules must collide in
Assumes molecules must collide in
order to react!
order to react!
Hindered by concentration,
Hindered by concentration,
temperature and geometry--
temperature and geometry--# of
# of
effective collisions
effective collisions

THE EFFECT OF TEMPERATURE OF REACTION
THE EFFECT OF TEMPERATURE OF REACTION
RATE: ARRHENIUS EQUATION
RATE: ARRHENIUS EQUATION
k
k = reaction rate constant
= reaction rate constant
= Ae
= Ae-E*/RT
-E*/RT
R is the “energy” R
R is the “energy” R
Or, 8.31 x 10
Or, 8.31 x 10-3
-3
kJ/K•mol
kJ/K•mol

A is the frequency factor units of
A is the frequency factor units of
L/(mol • s) & depends on the
L/(mol • s) & depends on the
frequency of collisions and the
frequency of collisions and the
fraction of these that have the
fraction of these that have the
correct geometry--
correct geometry--# of effective
# of effective
collisions
collisions
e
e-E*/RT
-E*/RT
is always less than 1 and is
is always less than 1 and is
the fraction of molecules having
the fraction of molecules having
the minimum energy required for
the minimum energy required for
reaction
reaction

*Notice in the equation: As "Ea"
*Notice in the equation: As "Ea"
increases, "k" gets smaller and
increases, "k" gets smaller and
thus, the rate would decrease.
thus, the rate would decrease.
Also, notice that as "T" goes up, "k"
Also, notice that as "T" goes up, "k"
increases and so the rate would
increases and so the rate would
also increase.
also increase.

Applying the laws of logarithms,
Applying the laws of logarithms,
taking the
taking the natural log of both
natural log of both
sides
sides, ln, we can rewrite the
, ln, we can rewrite the
equation:
equation:
ln k = ln A -
ln k = ln A - Ea
Ea
RT
RT

Rewrite in the form of an equation
Rewrite in the form of an equation
for
for
a
a straight line
straight line, we get:
, we get:
ln k = -
ln k = - Ea
Ea (
(1)
1) + ln [A]
+ ln [A]
R (T)
R (T)

Taking this equation, plot 1/T vs. ln [k],
Taking this equation, plot 1/T vs. ln [k],
and get a straight line.
and get a straight line.
From the straight line, find the
From the straight line, find the slope
slope
and then find the
and then find the activation energy.
activation energy.
slope = -
slope = - Ea
Ea
R
R
so ... Ea = - (R) (slope)
so ... Ea = - (R) (slope)

OR,
OR, find Ea
find Ea from data given
from data given
mathematically:
mathematically:
ln =
ln =

Used to calculate
Used to calculate
-value of activation energy from
-value of activation energy from
temperature dependence of the
temperature dependence of the
rate constant
rate constant
-rate constant for a given temp -
-rate constant for a given temp -
if the E* [also known as E
if the E* [also known as Ea
a] and A
] and A
factor are known.
factor are known.

Example - Arrhenius
Example - Arrhenius
Calculate the activation energy for
Calculate the activation energy for
the following set of data:
the following set of data:
T (°C)
T (°C) k (l/mol- s)
k (l/mol- s)
3
3 1.4 x 10
1.4 x 10-3
-3
13
13 2.9 x 10
2.9 x 10-3
-3
24
24 6.2 x 10
6.2 x 10-3
-3
33
33 1.2 x 10
1.2 x 10-2
-2

Points to remember!!
Points to remember!!
1. Ea is smaller; k is greater; the
1. Ea is smaller; k is greater; the
reaction is faster.
reaction is faster.
2. Ea is greater; k is smaller; the
2. Ea is greater; k is smaller; the
reaction is slower.
reaction is slower.

Exercise
Exercise
The colorless gas dinitrogen tetroxide
The colorless gas dinitrogen tetroxide
decomposes to the brown gas NO
decomposes to the brown gas NO2
2 in a
in a
first order reaction with a value of:
first order reaction with a value of:
k
k = 4.5 x 10
= 4.5 x 103
3
/s at 274K.
/s at 274K.
If
If k
k is 1.00 x 10
is 1.00 x 104
4
/s at 283K, what is the
/s at 283K, what is the
energy of activation?
energy of activation?

Exercise 12.7
Exercise 12.7
The reaction: 2N
The reaction: 2N2
2O
O3
3(g)
(g) →
→ 4 NO
4 NO2
2(g) + O
(g) + O2
2(g)
(g)
Was studied at several temperatures and
Was studied at several temperatures and
the following values of k were obtained:
the following values of k were obtained:
Calculate the value of Ea for this reaction.
Calculate the value of Ea for this reaction.
Sketch your graph.
Sketch your graph.

Exercise 12.8
Exercise 12.8
The gas-phase reaction between
The gas-phase reaction between
methane and diatomic sulfur is
methane and diatomic sulfur is
given
given
by the equation:
by the equation:
CH
CH4
4 (g) + 2S
(g) + 2S2
2 (g)
(g) 

CS
CS2
2 (g) + 2H
(g) + 2H2
2S (g)
S (g)

At 550° C the rate constant for this
At 550° C the rate constant for this
reaction is 1.1 M
reaction is 1.1 M-1
-1

 s
s-1
-1
and at 625° C
and at 625° C
the rate constant is 6.4 M
the rate constant is 6.4 M-1
-1

 s
s-1
-1
.
.
Using these values, calculate E
Using these values, calculate Ea
a for
for
this reaction.
this reaction.

12.8 CATALYSIS
12.8 CATALYSIS
Alter the mechanism so
Alter the mechanism so
the activation energy
the activation energy
barrier can be lowered.
barrier can be lowered.
Catalysts are not
Catalysts are not
altered during the
altered during the
reaction--they serve to
reaction--they serve to
lower the activation
lower the activation
energy and speed up
energy and speed up
the reaction by offering
the reaction by offering
a different pathway for
a different pathway for
the reaction
the reaction


E is NOT changed for the process
E is NOT changed for the process
biological catalysts are enzymes--
biological catalysts are enzymes--
proteins w/ specific shapes
proteins w/ specific shapes
ATP synthetase is the most
ATP synthetase is the most
important enzyme in the human
important enzyme in the human
body!
body!

HETEROGENEOUS CATALYST
HETEROGENEOUS CATALYST
different phase
different phase
than reactants
than reactants
usually involves
usually involves
gaseous
gaseous
reactants
reactants
adsorbed on the
adsorbed on the
surface of a solid
surface of a solid

 adsorption
adsorption—refers to the
—refers to the
collection of one substance on the
collection of one substance on the
surface of another
surface of another
 absorption
absorption—refers to the
—refers to the
penetration of one substance into
penetration of one substance into
another; water is absorbed by a
another; water is absorbed by a
sponge
sponge

 hydrogenation of unsaturated
hydrogenation of unsaturated
hydrocarbons—especially important
hydrocarbons—especially important
in converting unsaturated fats [oils]
in converting unsaturated fats [oils]
into saturated fats [solids like Crisco]
into saturated fats [solids like Crisco]
 C=C bonds are converted into C-C
C=C bonds are converted into C-C
bonds by adding a pair of hydrogens
bonds by adding a pair of hydrogens
“across the double bond”
“across the double bond”

H H H H
C=C + H2 (g)  H C C H
H H H H
ethene (ethylene) ethane
A simple example of
A simple example of
hydrogenation involves ethylene:
hydrogenation involves ethylene:
This reaction uses a solid catalyst in
This reaction uses a solid catalyst in
the form of Pt, Pd, or Ni. The
the form of Pt, Pd, or Ni. The
hydrogen and ethylene adsorb on
hydrogen and ethylene adsorb on
the catalyst surface where the
the catalyst surface where the
reaction occurs. The catalyst
reaction occurs. The catalyst
allows for metal-hydrogen
allows for metal-hydrogen
interactions that weaken the strong
interactions that weaken the strong
H-H bonds and facilitate the
H-H bonds and facilitate the
reaction.
reaction.

Typically involves 4 steps:
Typically involves 4 steps:
1.
1. Adsorption and activation of the
Adsorption and activation of the
reactants
reactants
2.
2. Migration of the adsorbed
Migration of the adsorbed
reactants on the surface
reactants on the surface
3.
3. Reaction of the adsorbed
Reaction of the adsorbed
substances
substances
4.
4. Escape, or desorption, of the
Escape, or desorption, of the
products
products

Catalytic converters are also
Catalytic converters are also
heterogeneous catalysts.
heterogeneous catalysts.
They have been
They have been
placed in
placed in
automobiles since
automobiles since
1974. [I know!
1974. [I know!
You weren’t born
You weren’t born
yet! Don’t rub it
yet! Don’t rub it
in.] Gasoline
in.] Gasoline
containing lead
containing lead
RUINS the
RUINS the
catalytic converter
catalytic converter
in your car!
in your car!

HOMOGENEOUS CATALYST
HOMOGENEOUS CATALYST
exists in the same phase as the
exists in the same phase as the
reacting molecules.
reacting molecules.
Freons or chlorofluorocarbons
Freons or chlorofluorocarbons
[CFC’s] were used until recently as
[CFC’s] were used until recently as
refrigerants and as propellants in
refrigerants and as propellants in
aerosol cans.
aerosol cans.

Freon-12 (CCl
Freon-12 (CCl2
2F
F2
2) is relatively inert
) is relatively inert
and thus remains in the
and thus remains in the
environment for a long time.
environment for a long time.
Eventually they migrate upward
Eventually they migrate upward
into the upper atmosphere and
into the upper atmosphere and
are decomposed by high-energy
are decomposed by high-energy
light.
light.

Among the decomposition
Among the decomposition
products are chlorine atoms:
products are chlorine atoms:
CCl
CCl2
2F
F2
2 (g)
(g) →
→ CClF
CClF2
2 (g)
(g) + Cl
+ Cl (g)
(g)

These chlorine atoms can catalyze
These chlorine atoms can catalyze
the decomposition of ozone:
the decomposition of ozone:
Cl
Cl (g)
(g) + O
+ O3 (g)
3 (g) 
 ClO
ClO (g)
(g) + O
+ O2 (g)
2 (g)
O
O(g)
(g) +
+ ClO
ClO(g)
(g) 
 Cl
Cl(g)
(g) + O
+ O2 (g)
2 (g)
O
O(g)
(g) + O
+ O3(g)
3(g) 
 2 O
2 O2(g)
2(g)

READ THE LAST COUPLE OF PAGES
READ THE LAST COUPLE OF PAGES
OF THIS CHAPTER FOR SOME GREAT
OF THIS CHAPTER FOR SOME GREAT
DESCRIPTIVE STUFF!
DESCRIPTIVE STUFF!
(That translates into multiple choice
(That translates into multiple choice
points!)
points!)

AP Ch12 Kinetics.ppt.......................

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