CHAPTER 15Kinetics

1
CHAPTER 15 Kinetics
Concerns how quickly reactions occur and the
factors that affect reaction rates.
Different reactions can occur at vastly different
rates or speeds. Slow reactions rusting of iron,
spoiling of food Fast reactions light triggered
process such as photosynthesis ( gt ms (10-3 s) to
lt ps (10-12 s)
2
Speed of reactions linked to mechanisms
The rates or speeds of reactions are linked to
their mechanisms, the detailed way in which they
occur.
3
Speed of reactions linked to mechanisms (2)
Furthermore, the time it takes to get from one
place to another depends on the method of travel.
Regardless of the method, the distance between
the two places is always the same. The time it
takes to travel is the only thing affected by the
method.
4
Distinguish Between Kinetics and Energetics
Whether or not a chemical reaction will occur is
dependent on energy changes and other factors.
A decrease in energy (reactants versus products)
or an increase in disorder of the system will
determine whether or not a reaction will occur.
Knowing that a reaction will occur tells you
nothing about how quickly the reaction occurs.
Thus, we tend to distinguish between kinetics and
energetics.
Changes in energy are associated with state
functions (independent of path). Rates of
reactions are highly dependent upon the mechanism
by which a process takes place.
5
Chemical Equations and Kinetic Behavior
C4H9Cl(aq) H2O(l) ? C4H9OH(aq) HCl(aq)
REACTANTS ? PRODUCTS
BEFORE ? AFTER
6
Chemical Equations and Kinetic Behavior (2)
FLOUR EGGS WATER OIL
7
Stoichiometry versus Kinetics
Remember You cannot look at the balanced
equation and know how the concentration of a
particular reactant will affect the rate of a
reaction!
2 SLICES BREAD 1 SLICE CHEESE
1 SANDWICH
Remember about stoichiometry, limiting reactants
and yields?
8
Mechanisms may be simple or complex
Reactions may occur in one step e.g. A ?
Products or A B ? Products
Some chemical reactions involve multi-step
mechanisms e.g. A B ? Products (overall
reaction) mechanism A ? C C B
? Products
9
15.1 Rates of Chemical Reactions
Speed of an event is defined as the change that
occurs in a given interval of time. e.g. car
speed of 50 miles per hour
Reaction Rate for a chemical reaction, the
change in the concentration of reactants or
products per unit time (usually M/s or Ms-1)
10
Calculation of Average Rate
A ? B
Rate of reaction based on Rate of disappearance
of reactant A or Rate of appearance of product B
Use to designate concentration in moles per
liter
11
Decomposition of N2O5
N2O5(g) ? NO2(g) O2(g)
Figure 15.2 Reaction concentration versus
time. Average rate Initial rate
12
Reaction Rates and Stoichiometry
To get a consistent number for the rate of a
reaction, the rate of disappearance of reactant
or the rate of appearance of product can be
adjusted using the coefficients in the equation.
13
Example 15.1
Write the chemical equation (balanced) for which
the following rate expression would be
appropriate.
4PH3 ? P4 6H2
14
15.2 Factors that Affect Reaction Rates
1. The concentrations of reactants.
2. The temperature at which the reaction occurs.
3. The presence of a catalyst.
15
Exponents in Rate Laws and Reaction Orders
Rate kreactant1mreactant2n ...
Exponents m and n are called reaction orders.
Exponents can be 0, 1, 2 and (sometimes)
fractions.
Overall order is the sum of the exponents.
third
What is overall order for above?
16
Rate Laws versus Stoichiometry
Additional examples of equations and rate laws
2N2O5(g) ? 4NO2(g) O2(g) Rate kN2O5
CHCl3(g) Cl2(g) ? CCl4(g) HCl(g) Rate
kCHCl3Cl21/2
H2(g) I2(g) ? 2HI(g) Rate kH2I2
Note that the exponents in the rate law do not
always match the coefficients in the balanced
equation.
17
Units of the Rate Constant
The units of the rate constant, k, depend upon
the overall order of the reaction. All of the
units for the factors on the left side of the
rate equation need to cancel in such a way that
you get molarity per time (e.g. M/s) as the final
units.
First order process rate kA units of k
reciprocal time (s-1, min-1, etc.) Second order
process rate kA2 units of k M-1s-1
etc. Third order process rate kA2B units
of k M-2s-1 etc.
18
Determination of a Rate Law/Constant Using
Initial Rates
CH3CO2CH3 OH- ? products
Holding CH3CO2CH3 constant while doubling
OH-, leads to a doubling of the reaction rate.
Rate is directly proportional to OH-.
Holding OH- constant while doubling
CH3CO2CH3, leads to a doubling of the reaction
rate. Rate is directly proportional to
CH3CO2CH3
Rate kCH3CO2CH3OH-
19
Determination of a Rate Law/Constant Using
Initial Rates
2NO(g) Cl2 (g) ? 2NOCl(g)
Holding Cl2 constant while doubling NO,
leads to a quadroupling of the reaction rate.
Rate is directly proportional to the square of
NO
Holding NO constant while doubling Cl2,
leads to a doubling of the reaction rate. Rate
is directly proportional to Cl2
Rate kNO2 Cl2
20
Rate law (2)
CONCLUSION since 2m 4, m 2
21
Rate law (3)
CONCLUSION since 2n 2, n 1
22
Rate Law (4)
Once the rate law is determined, the value of the
rate constant (k) can be determined by
rearranging the rate equation and plugging in
values for concentration and rate from any of the
experiments. It may be desirable to determine an
average from all of the experiments (see lab).
23
15.4 Change of Concentration with Time
First-Order Reactions
24
Figure 15.8 H2O2 Decomposition
25
Example 15.5
The first-order rate constant for the
decomposition of cyclopropane is 5.4 x 10-2 h-1.
If the initial concentration is 0.050 M, how long
will it take for the concentration to drop to
0.010 M?
26
Example 15.5 (2)
The first-order rate constant for the
decomposition of cyclopropane is 5.4 x 10-2 h-1.
If the initial concentration is 0.050 M, how long
will it take for the concentration to drop to
0.010 M?
27
Exercise 15.5
Here is a problem similar to 15.5 The
first-order rate constant for the decomposition
of N2O5, N2O5(g) ? 2NO2(g) O2(g), at 472 ?C is
6.82 x 10-3 s-1. If the initial concentration of
N2O5 is 0.0125 M, what is the concentration after
150 seconds?
Will use lnN2O5t -kt lnN2O50
28
Exercise 15.5 (cont)
lnN2O5150 -(6.82 x 10-3 s-1)(150 s)
ln(0.0125) -1.023 (-4.382) -5.405
N2O5150 e-5.405 4.49 x 10-3 M
Exercise 15.5 sucrose after 1 hr 0.0035 M
29
Second Order Reactions
Consider reactions that are second order in just
one reactant
Figure 15.9 Plots of kinetic data for
reaction 2NO2(g) ? 2NO(g) O2(g) at 300
?C
30
Half-Life
Half-life t1/2 , time required for the
concentration of a reactant to drop to one half
of its initial value
Figure 15.10 Successive half-lives
31
15.5 Particulate View of Reaction Rates
Collision Model based on the kinetic-molecular
theory and accounts for both concentration and
temperature effects on rates
1) molecules must collide to react
2) molecules must collide with sufficient energy
to react
3) molecules must collide with the appropriate
orientation
Figure 15.11 NO(g) O3(g)? NO2(g) O2(g),
Rate kNOO3
32
Temperature and Rate
Increasing the temperature causes the rates of
chemical reactions to increase. This is related
to the fact that as the temperature increases,
the number of collisions increases and the force
of the collisions increases. Figure 15.12 Kinetic
Energy Distribution
Molecules must posses a certain minimum energy to
react (activation energy, Ea).
Upon collision, kinetic energy of molecules used
to stretch, bend and break bonds leading to
chemical reactions.
33
Activation Energy
The rearrangement (isomerization) of methyl
isonitrile to form acetonitrile involves the
breaking of bonds to form a new compound.
34
Reaction Coordinate Diagrams
Figure 15.13 Reaction coordinate diagram for NO
CO2 ? NO2 CO These types of
diagrams map the progress of a reaction to the
potential energy (and thus the energy change). If
sufficient energy is supplied to the system the
transition state can form and the chemicals can
rearrange. They can then proceed to form
products or can return to the original state.
The products have a lower energy than the
reactants (exothermic reaction) or have a higher
energy (endothermic reaction).
35
Reaction Coordinate Diagrams (2)
Reactants/products Ea transition state ?E endo or
exothermic
Energy
Reaction pathway
Typical test question Draw the energy profile
diagram for a reaction that is exothermic (or
endothermic) and identify certain parts
36
Temperature and Orientation
Temperature increasing temperature increases the
fraction of molecules that have enough energy to
overcome the activation barrier.
Orientation Factor molecules must be oriented in
a certain way during collisions
Figure 15.13 Effective and Ineffective
collisions NO(g) O3(g)? NO2(g) O2(g), (NO
and O3 must approach each other in proper
orientation and with sufficient energy)
37
Arrhenius Equation
These dependence of rate on these factors is
summarized by the Arrhenius equation.
k Ae-Ea/RT k rate constant A frequency
factor (relates to number of collisions and the
geometry) e-Ea/RT fraction of molecules with
minimum energy Ea activation energy R gas
constant T absolute temperature
38
Arrhenius Equation (3)
39
Catalysts and Catalysis
Catalyst substance that changes the speed of a
chemical reaction without undergoing a permanent
change itself in the process
2KClO3(s) ? KCl(s) 3O2(g) slow reaction
without MnO2 catalyst
Catalysts are important in our daily lives e.g.
catalytic converters in automobiles industrial
manufacturing enzymes within our bodies
40
Homogeneous Catalysis
Homogeneous Catalyst a catalyst that is present
in the same phase as the reacting molecules
2H2O2(aq) ? 2H2O(l) O2(g) The reaction is
slow without a catalyst such as bromide ion.
2Br-(aq) H2O2(aq) 2H(aq) ? Br2(aq)
2H2O(l) Br2(aq) H2O2(aq) ? 2Br-(aq)
2H(aq) O2(g)
41
Homogeneous Catalysis (2)
Figure 15.16 The conversion of cis-2-butene to
trans-2-butene using I2 as a catalysts. Note that
the I2 dissociates to form I atoms that can help
to break the double bond of the butene and
provide an alternate mechanism for the reaction.
Figure 15.17 Reaction coordinate diagram for the
conversion of cis-2-butene to trans-2-butene. By
altering the mechanism of the reaction, a
catalyst provides a low energy pathway and
decreases the activation energy for the process.
42
15.6 Reaction Mechanisms
Balanced equation tells us what we started with
(reactants) and what was generated during the
reaction (products)
Reaction mechanism tells us about how the
process (reaction) occurs
Reactions that occur in 1 elementary
step H3C?N?C ? H3C?C?N NO(g) O3(g) ?
NO2(g) O2(g)
Molecularity unimolecular (1), bimolecular (2),
termolecular (3)
43
Multistep Mechanisms
Below 225 ?C, reaction involves 2 elementary
steps NO2(g) CO(g) ? NO(g) CO2(g)
1 NO2(g) NO2(g) ? NO3(g) NO(g) 2
NO3(g) CO(g) ? NO2(g) CO2(g)
_______________________________________ overall
NO2(g) CO(g) ? NO(g) CO2(g)
NO3 is an intermediate formed in 1 step,
consumed in another step
44
Conversion of Ozone to Oxygen
It has been proposed that the conversion of ozone
into O2 proceeds via two elementary steps.
Identify the molecularity of each step, write the
overall reaction and identify any
intermediates. O3(g) ? O2(g) O(g) O3(g)
O(g) ? 2O2(g)
unimolecular
bimolecular
O is an intermediate
45
Rate Equations for Elementary Steps
Rate laws are linked to molecularity If we know
a reaction is an elementary step, then we know
rate law.
A ? products based on collision theory and
statistics Rate kA
A B ? products Rate kAB Table
Elementary steps and their rate laws
46
Example 15.12
Example 15.12 The self oxidation of ClO? to give
ClO3? and Cl?
If the following reaction occurs in a single
elementary step, predict the rate law H2(g)
Br2(g) ? 2HBr(g)
Rate kH2Br2
Actual (or Experimental) Rate Law Rate
kH2Br21/2 Conclusion?
47
Rate Laws for Multistep Mechanisms
Below 225 ?C, reaction involves 2 elementary
steps Experimental rate law Rate kNO22
Based on molecularity Rate k1NO22
48
Mechanisms with an Initial Fast Step
2 NO(g) Br2(g) ? 2NOBr(g)
Experimental rate law Rate kNO2Br2
Rule out termolecular process. Why?
49
Mechanisms with an Initial Fast Step (2)
Rate k2NOBr2NO (based on step 2)
50
Mechanisms with an Initial Fast Step (2)
Equilibrium rate of forward rxn rate of
reverse rxn k1NO Br2 k-1NOBr2
NOBr2 (k1/ k-1)NO Br2 Rate
k2NOBr2NO Rate k2 (k1/ k-1)NO Br2
NO kNO2 Br2