Transcript Chemical Kinetics - Bremerton School District

Chemical Kinetics
CHAPTER 12
Overview of Kinetics
 Macroscopic Study—




Rates of reaction, described by rate laws
Meaning of reaction rate—change of concentration of reactants
per unit of time.
How to determine rate from experimental data
How does temperature and concentration reaction affect rate
 Collision Theory—



Recall kinetic theory--particles that are in motion
Chemical reactions occur when particles collide with sufficient
energy and the correct orientation to cause a chemical reaction
Reaction mechanism—the detailed pathway taken by atoms and
molecules in the reaction process
12.1
What is kinetics?
 A study of reaction rates or speeds.
 Reaction Rates—a measure of how quickly a reaction
occurs. Is it spontaneous? If not how much energy
does it take?
How is it useful?
 Make sure the reaction proceeds at a fast enough rate
to be useful. (Is it practical? Show me the money!)
Solving Kinetic Problems
 You will either be given experimental data and need
to use mathematical relationships to solve for an
unknown .
 OR
 You will be given a graph, and need to understand
how to read the data to solve for the unknown. So
you need to understand the types of graphs involved
and the data they provide.
Graphing
How are Reaction Rates Measured?
 Reaction Rates are Measured in Terms of
Concentration
 Rate = final concentration – intial concentration
tfinal-tintial
 Rate = Δ [A]
Δt
--Where A is the reactant or product
being considered.
Consider the following reaction:
 2 N2O5(g)




4NO2(g) + O2 (g)
In terms of the reactant, the rate of reaction can be
written:
Rate = - [N2O5]final – [N2O5]intial = - Δ [N2O5]
tfinal – tintial
Δt
As the rate proceeds, N2O5 decomposes, so Δ [N2O5]
is negative or is decreasing.
Since rate must be positive, a negative sign must
be included for reaction rates of reactants.
Example Continued
 In terms of the products, the rate of reaction can be
written as:
 Rate = Δ [NO2]
or Rate = Δ [O2]
Δt
Δt
Where there is no negative sign since products are
made so both would be positive slopes already.
What is an instantaneous reaction rate?
 We can obtain the reaction rate at any given instant.
 The instantaneous reaction rate is the slope
of a line tangent to the curve of an instant in
time
 Instantaneous rate = - slope of tangent (for a
reactant)
Example of instantaneous rate, at t = 2.0 minutes,
concentration = .056M
Rate = -0.056 M = -.028M/min
2.o min
Reaction Rates and Stoichiometry
 Because the coefficients are different for each
reactant and product, these three rates will have
different values.
 Dividing by their Stoichiometric coefficients, we can
relate the three rates:
 Rate = Δ [N2O5] = Δ [NO2] = Δ [O2]
2Δt
4 Δt
Δt
 Now we can choose any of the reactants or products
to determine the rate.
Example of Instantaneous
Rate Reaction
Figure 12.5: A plot of [N O ] versus time for the
decomposition reaction of N O .
2
5
2
12a–13
5
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Company. All rights reserved.
Figure 12.1:
Starting with
a flask of
nitrogen dioxide
at 300°C, the
concentrations
of nitrogen
dioxide, nitric
oxide, and
oxygen are plotted
versus time.
12.2
Rate Law and Rate Constant
Experimental data, of the rate
of reaction versus [N2O5]:
4.5
 Using the general formula for a
3.5
The straight line through the origin
indicates that the rate is directly
proportional to [N2O5]: rate = k
[N2O5]
 In this case, k = the slope of the
line!
Rate (M/min)
line is y = mx + b
 Where m = slope, b = y-intercept
4
3
2.5
2
1.5
1
0.5
0
0
2
4
[N2O5]
6
What are Rate Laws?
 The reaction rate depends on the concentrations of




only the reactants. (see note below)
Rate = k[A]n
“k” the rate constant
n = the order of the reactant
Value of n, must be found experimentally, cannot be
found through stoichiometry!
Note:
• Chemical reactions are reversible
• When there are a lot of products and
little reactants the reaction can reverse,
so we usually measure rates soon after
mixing.
What affects reaction rates?
 Reactions occur as a result of collisions between
reactant molecules.
1.
2.
3.
4.
Concentration
Temperature
Nature of Reactants
Catalysts
What is a Reaction Order?
 The rate laws for most reactions have the general
form:
 Rate = k [reactant 1]m [reactant 2]n…
 The exponents m and n are called reaction orders,
and the sum of all the reaction orders (m + n) is the
overall reaction order.
 Infer the number of steps for the reaction!
Using Initial Rates to Determine Rate Laws
 Reaction order involving a single reactant
 For this type of reaction: A
Products
 the rate law has the form: rate = k [A]m where



m = order of reaction
If m = 0, the reaction is “zero-order”
If m = 1, the reaction is “first-order”, etc.
 The order of the reaction cannot be obtained from
the chemical equation, it must be determined
from experimental data.
Graphs represent concentration versus time!
To determine the order…
 You need the rates and concentrations for two
different instances:
rate 1 = k [A]1m
rate 2 = k [A]2m
We can solve for m by dividing the second rate by the
first:
rate 2 = [A]2m = [A]2 m
rate 1 = [A]1m
[A]1
Example Problem
 Given the following data, determine the reaction
order, m, the decomposition of N2O5:
[N2O5]
Rate (M/s)
0.90 M
5.4 x 10-4
0.45 M
2.7 x 10-4
2 N2O5 (g)
4 NO2 (g) + O2 (g)
Reaction Order with More than One Reactant
 Most reactions are of the form: A + B
products
 And the general form of the rate law is:
 Rate = k [A]m [B]n
 And the overall reaction order: order = m + n
 For example, if the rate law for a reaction is:
 Rate = k [NO]2 [O2], the reaction is second-order in NO, firstorder in O2, and third-order overall.
Given the following data,
Experiment
[O2]
[NO]
Rate (M/s)
1
0.0110M
0.0130 M
3.21 x 10-3
2
0.0220 M
0.0130 M
6.40 x 10-3
3
0.0110 M
0.0260 M
12.8 x 10-3
Determine m and n, and thus, the reaction order and the rate law for the
reaction: O2 (g) + 2NO (g)
2 NO2 (g)
a. First, use data from two experiments where the [NO] remains
constant (e.g. Experiments #1 and # 2) to get m.
b. Next use data from two experiments were the [O2] remains constant
(e.g. Experiments #1 and #3) to get n.
c. Finally write the rate law for the reaction.
Integrated Rate Equations
 Using calculus, we can develop integrated rate




equations to relate reactant concentration to time.
Zero-order reaction of this type A
products,
we get: rate = -k
Thus, the rate for a zero-order reaction is constant
and independent of the concentration of the
reactants. It does not matter how much you have!!!
Plotting [A] versus time should give us a straight
line.
[A]
Figure 12.7:
A plot of
[A] versus t
for a zeroorder
reaction.
First-Order Reactions
 For a first-order reaction of this type:
A
Products, we get:
rate = - Δ [A]t = k [A]t where [A]t is the [A] at time t
Δt
Rearranging the equation above and integrating gives us
the relationship between concentration and time:
ln [A]t = -kt
[A]o
or
ln [A]t - ln [A]o = -kt
Figure 12.4: A plot of ln[N2O ] versus time.
5
Only for first-order reactions does one get a straight line
when plotting ln [A] versus time!
12a–28
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Company. All rights reserved.
Half-life (t1/2)
 The time required for one half of a sample to decompose.
For first-order reactions, half-life is a fixed value,
independent of concentration.
We can solve for the half-life:
ln [A]t = ln ½[A]o = ln 1 = -0.693 = -kt1/2
[A]o
[A]o
2
So t1/2 = 0.693
k
Second-Order Reactions
 For a second-order reaction of this type:
A
products
 We get:
rate = - Δ[A]t = k [A]2
Δt
Rearranging the equation above and integrating gives us
the relationship between concentration and time:
1 - 1 = kt
or 1 = kt + 1
[A]t [A]o
[A]t
[A]o
Figure 12.6: (a) A plot of ln[C H ] versus t. (b) A
plot of 1y[C H ] versus t.
4
4
6
Only for
second-order
reactions
does one get
a straight
line when
plotting
1/[A]t versus
time!
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rights reserved.
12a–
31
6
To determine the order of reaction given experimental
data, plot the data and determine which set of data gives a
linear plot
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rights reserved.
12b–
32
The Collision Model
 Main Idea: Molecules must collide to react
 Three Factors:
1. Kinetic Energy—for a reaction to occur, molecules
have to be moving quickly enough that they can break
and reform bonds when they collide, if moving too
slowly they will merely bounce off each other.

Activation Energy (Ea): the minimum amount of energy required to
initiate a chemical reaction
2. Concentration—The higher the concentration, the
more molecules present, higher probability of
collisions, thus a higher reaction rate!
3. Orientation of molecules (or steric factor)—
molecules have to be in the correct orientation for a
reaction to occur (see figure 12.13)
Importance of the steric factor
Figure 12.13: Several possible orientations for a
collision between two BrNO molecules.
Orientations (a) and (b) can lead to a reaction,
but orientation (c) cannot.
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rights reserved.
12b–
35
Kinetic Molecular Theory
 Recall with the kinetic molecular theory of gases, the
higher the temperature, the higher the kinetic
energy of molecules.
 By increasing temperature, more molecules have the
required activation energy for a reaction.
Activation Energy
Figure 12.12: Plot showing the number of
collisions with a particular energy at T1 and
This figure shows the
T2, where T2 > T1.
energy of molecules at
two temperatures.
At higher temperature
T2 more molecules
have higher energy
The shaded region
indicates molecules
with the activation
energy (Ea)
At the higher
temperature, more
molecules have Ea.
Transition-State Model: Energy Profiles
 We can show a reaction in terms of an Energy
Profile.
 The transition state (also called the activated
complex) is the arrangement of atoms at the peak
of the energy profile.
 The activation energy, Ea, is the difference in
energy between the reactants and the transition
state.
 The difference in energy between reactants and
products is ΔHrxn. Note the next slide for a diagram.
Figure 12.11: (a) The change in potential energy as a function of
reaction progress for the reaction 2BrNO
2NO + Br2. The activation
energy Ea represents the energy needed to disrupt the BrNO molecules
so that they can form products. The quantity DE represents the net
change in energy in going from reactant to products. (b) A molecular
representation of the reaction.
The Arrhenius Equation
 Svante Arrhenius noted that for most reactions, the
temperature dependence for the rate constant is
given by:
k = A e-Ea/RT
Where A = frequency factor which accounts for the
frequency of collision and the probability that the
molecules are in the correct orientation.
Ea = activation energy
R = 8.314 J/mol *K
T = Temperature in Kelvins
The Arrhenius Equation continued…
 If we take the natural log of both sides and
rearranging the equation we get:
ln k = - Ea + ln A
RT
From this equation plotting ln k versus 1/T should give
a straight line;
Thus, we can determine the activation energy for a
given reaction if we have experimental data on its
rate constant at different temperatures.
Figure 12.14: Plot
of ln(k) versus 1/T
for the reaction
2N2O5(g)
4NO2(g) + O2(g).
The value of the
activation energy
for this reaction
can be obtained
from the slope of
the line, which
equals -Ea/R.
“Two-Point” Equation Relating k and T
 We can also determine the activation energy for a
given reaction if we have experimental data for the
reaction at two different temperatures:
ln k2 = Ea 1 - 1
k1
R T1
T2
This equation will also allow us to determine the rate
constant at a different temperature if we have the
activation energy.
Example
 The rate constant for a first-order reaction is
0.346 s-1 at 298K. What is the rate constant at 355 K
if the activation energy for the reaction is 50.2
kJ/mol?
Reaction Mechanisms
 A reaction mechanism is a sequence of steps
by which a reaction occurs at the molecular level.
 Elementary Steps: The individual steps that make
up a reaction mechanism
 The molecularity of a reaction describes the
number of molecules reacting in an elementary step.



Unimolecular: only one reactant
Bimolecular: two reactants
Termolecular: three reactants (less common as very low
probability of 3 molecules colliding in the correct orientation
for a reaction )
The types of elementary steps and their
corresponding rates are given below:
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Catalysts
 Substances that accelerate a chemical reaction but
are not themselves transformed into a
product of the reaction. If the catalyst is the same
state as the reactant(s) it is a homogeneous catalysts
and if it is not it is a heterogeneous catalysts (usually
a solid).
How to tell if it is a intermediate or a catalyst
 If it is a catalyst, it will appear first on the
reactant side.
 If it is an intermediate, it will first appear on
the product side.
Important Equations
 Arrhenius Equation
 k = A e-Ea/RT
Note:
From this equation plotting ln k versus
1/T should give a straight line;
 ln k = - Ea + ln A
Thus, we can determine the activation
RT
energy for a given reaction if we have
 A = frequency factor which
experimental data on its rate constant
accounts for the frequency of
collision and the probability that at different temperatures.
the molecules are in the correct
orientation.
 Ea = activation energy
 R = 8.314 J/mol *K
 T = Temp in Kelvins
Two different temperature problems:
ln k2 = Ea 1 - 1
ln k1
R T1
T2