Chemical Kinetics
Kinetics
Chemical kinetics is the branch of
chemistry concerned with the rate of
chemical reactions and the mechanism
by which chemical reactions occur.
Presented by Mark Langella,
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Collision Theory
 In
order for two particles to react
chemically, they must collide. Not only
must they collide, but it must be an
“effective collision.” That is, they must
have the correct amount of energy and
collide with the proper orientation in space.
 Any factor which increases the likelihood
that they will collide will increase the rate
of the chemical reaction.
Presented by Mark Langella,
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FACTORS WHICH AFFECT THE
RATE OF A CHEMICAL
REACTION (Bonds Must Break)
 Surface
Area/ Contact Area (Opportunity
for collisions)
 Concentration ( Increase Frequency)
 Temperature ( Increase Frequency)
 Catalyst ( Effective Collisions)
 Nature of Reactants ( Effective collisions)
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Surface Area /Contact Area
 Sugar
Cube vs. Packet of Sugar
 Steel vs Steel Wool
 Two Sided Candle
 Coffee Creamer
 Alka-tablets
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Concentration
 Sudsy
Kinetics
 Magnesium in Various Concentrated Acids
 Which Reactant has the greater
concentration effect ( that’s another story)
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Which of the following hydrogen peroxide
and sodium iodide mixtures will give the
greatest rate of reaction?
30%
10%
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3%
Temperature
 Spoiling
of Fruit
 Alka Seltzer Hot and Cold
 Light Stick Kinetics
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Temp vs. Reaction rate
 The
effect of temperature on the rate of a
chemical reaction can best be explained in
terms of the kinetic molecular theory. The
higher the temperature, the faster
molecules move. The faster they more, the
faster they collide and therefore, react at a
faster rate.
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Every ten degrees rate doubles
T0
T0 + 10
Energy, E
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Arrhenius Equation
-
Collisions must have enough energy to
produce the reaction (must equal or
exceed the activation energy).
-
Orientation of reactants must allow
formation of new bonds.
Presented by Mark Langella,
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Temperature
T1
Activation Energy
T2
Energy
Presented by Mark Langella,
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12_300
T1
T2 > T1
T2
0
0
Ea
Energy
Plot showing the number of collisions with a particular
energy at T1& T2, where T2 > T1 -- Boltzman Distribution.
Presented by Mark Langella,
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Arrhenius Equation
(continued)
k  Ae Ea / RT





k = rate constant
A = frequency factor
Ea = activation energy
T = temperature
R = gas constant
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Arrhenius Equation
 If
the natural logarithm of each side of the
Arrhenius Equation is taken, the following
equation results:
 ln(k) = -Ea/R (1/T) + ln(A)
 y = mx + b

 m = -Ea/R when ln(k) is plotted versus 1/T.
Presented by Mark Langella,
PWISTA.com
Presented by Mark Langella,
PWISTA.com
THE RELATIONSHIP BETWEEN k
AND T
k = A -
Ea
RT
e
or ln k = ln A - E a / RT
Presented by Mark Langella,
PWISTA.com
Since the second form of the equation follows the form, y = mx + b,
then a graph of ln k vs 1/T should give a straight line.
The activation energy of the reaction may be calculated from the slope
of the ln k vs 1/T plot.
Slope = - Ea / R
Therefore, Ea = - R (slope)
Presented by Mark Langella,
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If you have data of two points
k1
ln
  Ea / R(1/ T 1  1/ T 2)
k2
Presented by Mark Langella,
PWISTA.com
Presented by Mark Langella,
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Presented by Mark Langella,
PWISTA.com
Presented by Mark Langella,
PWISTA.com
Presented by Mark Langella,
PWISTA.com
Final Notes on Kinetics
•
•
•
As T increases, so does k.
Ea & E are independent of T.
A catalyst lowers Ea and increases
the rate of both kf & kr.
Presented by Mark Langella,
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Light Stick Kinetics
 Bend
a light stick to break the glass tube
inside mixing the hydrogen peroxide with
the phenyl oxalate ester. Repeat this
several times to eliminate any large
sections of glass and expedite mixing.
Place the light stick aside for two to three
hours. Your instructor may have activated
and aged the light sticks for you.
Presented by Mark Langella,
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 Prepare
the data collection device by
slipping the one-hole #00 stopper
prepared for this experiment over the
temperature probe. It might be helpful to
lubricate the hole in the stopper with
silicone oil or spray. Cut the top from an
aged and activated light stick with a sharp
knife or scissors. Divide its contents
equally with another lab group into two
small test tubes.
Presented by Mark Langella,
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
Place the temperature probe into the mixture.
Adjust the position of the temperature probe so
that it is centered in the solution but just visible
when you sight into the light sensor hole. It must
not extend too deep as to interfere with the light
sensor readings. Clamp the foam block onto the
ring stand with a large clamp. Gently force the
light sensor into the hole prepared for it on the
side of the foam block. It should be deep enough
to fit securely but not touch the test tube.
Presented by Mark Langella,
PWISTA.com

Place the switch on the light sensor to the 600
lux sensitivity. Plug the light sensor into Channel
one (CH1) of a powered Lab Pro and the
temperature probe into Channel two (CH2).
 Connect the LabPro to your computer by means
of the appropriate cable. Double click the Logger
Pro icon on your computer. It should auto-ID
both sensors. If not, check your connections and
try again. If your instructor has a file setup for
this experiment, open it now.
Presented by Mark Langella,
PWISTA.com

Remove the test tube and temperature probe
from the foam block, placing it into a beaker of
hot (45 - 50°C) water. Using the temperature
probe as a handle, stir the water with the test
tube observing the temperature until it remains
more or less constant. Remove the test tube
from the water, dry the outside, and insert it into
the foam block.
 When everything is ready, click on the green
“COLLECT” icon on the icon toolbar, and
observe the data. The experiment will run for
fifteen minutes.
Presented by Mark Langella,
PWISTA.com
 When
the measurement is complete,
disassemble the apparatus, returning all
items to their designated areas. Dispose of
the spent light stick solution as indicated
by your instructor.
 Clean both the temperature probe and the
test tube with a brush using soap.
Presented by Mark Langella,
PWISTA.com
1
a) Click on the label, Lumination (lux) on
the vertical axis of your graph. Select the
natural log of the light intensity, “ln
Illumination,” for the vertical axis. Click on
the label, “Time (s),” on the horizontal axis
of your graph. Select “Inverse
Temperature.” Auto scale your graph.
Your graph now displays the ln intensity vs
1/temperature data necessary to find the
activation energy.
Presented by Mark Langella,
PWISTA.com

1 (b) Drag your curser across the most linear
portion of your log I vs. 1/T plot. The linear
section should be highlighted by a dark gray
rectangle. Select the linear fit icon in the icon
toolbar. The best fit solution will be displayed.
Record these values.

1 (c) Calculate the activation energy for the light
stick reaction from the slope found in Question
#1b above. Express your answer in kilojoules
per mole (kJ/mol).
Presented by Mark Langella,
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Catalysis
Catalyst:
A substance that speeds up a reaction
without being consumed
Enzyme:
A large molecule (usually a protein)
that catalyzes biological reactions.
Homogeneous
catalyst: Present in the same
phase as the reacting molecules.
Heterogeneous
catalyst: Present in a different
phase than the reacting molecules.
Presented by Mark Langella,
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12_303
Uncatalyzed
pathway
Energy
Catalyzed
pathway
Products
E
Reactants
Reaction progress
Energy plots for a catalyzed and an uncatalyzed pathway
for an endothermic reaction.
Presented by Mark Langella,
PWISTA.com
12_304
Number of collisions
with a given energy
Number of collisions
with a given energy
Effective
collisions
(uncatalyzed)
Effective
collisions
(catalyzed)
Ea (catalyzed )
E a (uncatalyzed )
Energy
Energy
(a)
(b)
Effect of a catalyst on the number of reaction-producing
collisions. A greater fraction of collisions are effective
Presented by Mark Langella,
for the catalyzed reaction.
PWISTA.com
Heterogeneous Catalysis
Steps:

1.

2. Migration of the adsorbed reactants on the
surface.

3.
Reaction of the adsorbed substances.

4.
Escape, or desorption, of the products.
Adsorption and activation of the reactants.
Oscillating Flask Demo
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Homogeneous Catalysis
 Catalyst
is in the same phase as the
reacting molecules.
 NO(g) + 1/2O2(g) ----> NO2(g)

NO2(g) ----> NO(g) + O(g)

O2(g) + O(g) ----> O3(g)

3/2 O2(g) ----> O3(g)
 What is the catalyst in this reaction?
 What are the intermediates?
Presented by Mark Langella,
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Nature of Reactant
 Whoosh
Bottle
 Closed vials with 2 cm of


White phenolphthalein in NaCl added to
calcium Hydroxide
Ferric Chloride added to Sodium Acetate
Presented by Mark Langella,
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Spontaneity
•
•
•
tendency for a reaction to occur.
does not mean that the reaction will be
fast!
Diamonds will spontaneously change into
graphite, but the process is so slow that it
is not detectable.
Presented by Mark Langella,
PWISTA.com
Reaction Mechanism
•
•
•
the steps by which a chemical process
occurs.
allows us to find ways to facilitate
reactions.
can be changed by the use of a catalyst.
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Reaction Rate
Change
in concentration (conc) of a
reactant or product per unit time.
conc of A at time t2  conc of A at time t1
Rate =
t2  t1
 A 

t
Presented by Mark Langella,
PWISTA.com
Presented by Mark Langella,
PWISTA.com
Reaction Rate
 It
is customary to work with positive
reaction rates, so a negative sign is used
in some cases to make the rate positive.
 Rates determined over a period of time
are called average rates.
 Instantaneous rate equals the negative
slope of the tangent line.
Presented by Mark Langella,
PWISTA.com
12_291
0.0100
NO2
Concentrations (mol/L)
0.0075
0.0026
[NO2 ]
0.0006
70s
t
110 s
0.005
NO
0.0003
70s
0.0025
O2
50
100
150
200
250
300
Time (s)
The concentrations of nitrogen dioxide, nitric oxide,
oxygen plotted versus time.
Presented by Mark Langella,
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350
400
12_1575
Time
(a)
Time
(b)
(c)
Representation of the reaction of 2 NO2(g) ---> 2 NO(g) + O2(g).
a) t = 0
b) & c) with increased time, NO2 is changed into
NO and O2.
Presented by Mark Langella,
PWISTA.com
Rate Laws
(differential)
Rate
= k[NO2]n
k
= rate constant
n = rate order
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Types of Rate Laws
Differential
Rate Law: expresses how
rate depends on concentration.
Integrated
Rate Law: expresses how
concentration depends on time.
Presented by Mark Langella,
PWISTA.com
Rate Laws Summary
•
•
•
•
•
•
Differential rate law -- rate of a reaction depends on
concentration.
Integrated rate law -- concentration depends on time.
Rate laws normally only involve concentrations of
reactants.
Experimental determination of either rate law is
sufficient.
Experimental convenience dictates which rate law is
determined experimentally.
Rate law for a reaction often indicates reaction
mechanism.
Presented by Mark Langella,
PWISTA.com
12_292
1.00
Rate = 5.4 x 10-4 mol/L.s
[N2O5] (mol/L)
.80
Rate = 2.7 x 10-4 mol/L.s
.60
.40
.20
400
800
1200
1600
2000
Time (s)
Plot of the concentration of N2O5 as a function of time for the
reaction 2N2O5(soln) ---> 4NO2(soln) + O2(g). Rate at 0.90 M is
twice the rate at 0.45 M.
Presented by Mark Langella,
PWISTA.com
Method of Initial Rates
Rate: the “instantaneous rate”
just after the reaction begins.
Initial
The
initial rate is determined in
several experiments using different
initial concentrations.
Presented by Mark Langella,
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Overall Reaction Order
Sum
of the order of each component in
the rate law.
rate
The
= k[H2SeO3][H+]2[I]3
overall reaction order is 1 + 2 + 3
= 6.
Presented by Mark Langella,
PWISTA.com
First-Order Rate Law
For aA  Products in a 1st-order reaction,
 A
Rate =
k A
t
 Integrated
first-order rate law is
 ln[A]
y

= kt + ln[A]o
= mx + b
If a reaction is first-order, a plot of ln[A] versus time
is a straight line.
Presented by Mark Langella,
PWISTA.com
Half-Life of a First-Order
Reaction
t1/2
0.693

k
t1/2
= half-life of the reaction
k = rate constant
For
a first-order reaction, the half-life
does not depend on concentration.
Presented by Mark Langella,
PWISTA.com
12_294
[N2O5]0
0.1000
0.0900
[N2O5] (mol/L)
0.0800
0.0700
0.0600
[N2O5]0
2
[N2O5]0
0.0500
0.0400
0.0300
4
0.0200
[N2O5]0
0.0100
8
50
t1/2
100
150
200
t1/2
250
t1/2
Time (s)
Plot of [N2O5] versus time for the decomposition reaction
of N2O5. Note that the half-life for a 1st order reaction is
constant.
Presented by Mark Langella,
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300
350
400
Second-Order Rate Law
aA  products in a second-order
reaction,
 For
 A
Rate =
k A
t
 Integrated
rate law is
2
1
1
 kt +
 A
 A o
y = mx + b
If a reaction is second-order, a plot of 1/[A]
versus time is a straight line.
Presented by Mark Langella,
PWISTA.com
Half-Life of a Second-Order
Reaction
t1/2
1

kA
o
t1/2
= half-life of the reaction
k = rate constant
Ao = initial concentration of A
The
half-life is dependent upon the
initial concentration.
Presented by Mark Langella,
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Zero-Order Rate Law

For aA---> products in zero-order reaction,
 Rate

= k[A]o = k
The integrated rate law is
 [A]
= -kt + [A]0
 y = mx + b


If a reaction is zero-order, the plot of [A] versus time is a straight
line.
Example -- surface of a solid catalyst cannot hold a greater
concentration of reactant.
Presented by Mark Langella,
PWISTA.com
Table 12.6 Summary of the Kinetics for Reactions of the Type aA  Products That Are
Zero, First, or Second Order in [A]
Order
Zero
First
Second
Rate = k
Rate = k[A]
[A] = -kt + [A]0
ln[A] = -kt + ln[A]0
Plot needed to give a straight line
[A] versus t
ln[A] versus t
Rate = k[A]2
1
1
= kt +
[A]
[A]0
1
versus t
[A]
Relationship of rate constant
to the slope of straight line
Slope = -k
Slope = -k
Rate law
Integrated rate law
Half-life
t1/2 =
[A]0
2k
KNOW THIS TABLE!!!!
Presented by Mark Langella,
PWISTA.com
t1/2 =
0.693
k
Slope = k
t1/2 =
1
k[A]0
A Summary

1. Simplification: Conditions are set such
that only forward reaction is important.


2. Two types: differential rate law
integrated rate law

3. Which type? Depends on the type of
data collected - differential and integrated
forms can be interconverted.
Presented by Mark Langella,
PWISTA.com
A Summary
(continued)

4. Most common: method of initial rates.

5. Concentration v. time: used to determine
integrated rate law, often graphically.

6. For several reactants: choose conditions
under which only one reactant varies
significantly (pseudo first-order conditions).
Presented by Mark Langella,
PWISTA.com
Reaction Mechanism
-
The series of steps by which a
chemical reaction occurs.
-
A chemical equation does not tell us
how reactants become products - it is
a summary of the overall process.
Presented by Mark Langella,
PWISTA.com
Reaction Mechanism
(continued)
 The
reaction
6CO2  6H2O

light
C6 H12O6  6O2
has many steps in the reaction
mechanism.
Presented by Mark Langella,
PWISTA.com
Often Used Terms
Intermediate:
formed in one step and
used up in a subsequent step and so is
never seen as a product.
Molecularity:
the number of species that
must collide to produce the reaction
indicated by that step.
Elementary
Step: A reaction whose rate
law can be written from its molecularity.
uni, bi and termolecular
Presented by Mark Langella,
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Presented by Mark Langella,
PWISTA.com
Reaction Mechanism
Requirements
 1.
The sum of the elementary steps must
give the overall balanced equation for the
reaction.
 2. The mechanism must agree with the
experimentally determined rate law.
Presented by Mark Langella,
PWISTA.com
Rate-Determining Step
In
a multistep reaction, it is
the slowest step. It therefore
determines the rate of
reaction.
Presented by Mark Langella,
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Fast Equilibrium Reaction

When the 1st step is not the slow step, but a
fast equilibrium, the rate can be determined as
follows:

2A + B ---> C 2nd order in B, 1st order in A.
Step 1 A + B <----> D
(Fast equilibrium)
Step 2 D + B ----> E
Slow
Step 3 E + A ----> C + B Fast
A + B <---> D
D + B ---> E
E + A ---> C + B
2 A + B ---> C







1st requirement that elementary steps
equal overall reaction is met.
Presented by Mark Langella,
PWISTA.com
Fast Equilibrium Reaction
(continued)







kf[A][B] = kr[D] (fast)
[D] = kf/kr([A][B])
Rate = k2[D][B] (slow)
Substitute fast equation in terms of [D] into
slow reaction.
Rate = k2kf/kr([A][B][B])
Rate = k[A][B]2
2nd requirement is also met, this is, then, a
possible mechanism.
Presented by Mark Langella,
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