Chapter 13: Chemical Kinetics - WordPress.com13.4: The Integrated Rate Law So far we have looked at...
Transcript of Chapter 13: Chemical Kinetics - WordPress.com13.4: The Integrated Rate Law So far we have looked at...
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Chapter 13: Chemical Kinetics
Mrs. Brayfield
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13.2: The Rate of a Chemical Reaction
When we measure rates (outside of chemistry) we may use something like miles per hour
But in chemistry, we are more interested in knowing the change in concentration of reactants over time
So our units are:
For example we can look at the rate for the following reaction:
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Rate of a Reaction
We can also express the rate as:
Or, if we look at the products we can express the rate as:
We use the ½ in front of the rate because for every 1 mol of
reactants we have ½ mol of products
Note that the concentrations of reactants decrease and the
concentration of the product increases
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Rate of a Reaction
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Expressing Rate
The instantaneous rate of the reaction is the slope of
the curve at any point (on the graph from previous slide)
However, we can generalize the rate by using a generic
chemical equation:
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Expressing Rate Example
For the balanced chemical reaction of:
Predict the rate of change in concentration of H2O2
during the time interval [from previous problem: in the
first 10.0 seconds of the reaction the concentration of I-
dropped from 1.000M to 0.868M]
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Rates of a Chemical Reaction
Homework Problems: #1, 2, 4, 6, 8
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13.3: The Rate Law
The rate of a chemical reaction typically depends on
more than one reactant
A rate law is the relationship between the rate of the
reaction and the concentration of the reactants:
For the reaction
Rate = k[A]n
Where: k = constant; n = a number called the reaction order
The reaction order is a number that determines how
the rate depends on the concentration of the reactants
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Rate Law
The value of n determines how the rate depends on
concentration of the reactant:
If n = 0, the reaction is zero order
This means the rate is independent of the concentration of A
If n = 1, the reaction is first order
This means that the rate is directly proportional to the concentration
of A
If n = 2, the reaction is second order
This means that the rate is proportional to the square of the
concentration of A
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Rate Law
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Rate Law
The order of a reaction can only be determined ONLY by
experiment!
When we have multiple reactants, simply add them to the
rate law:
For the reaction:
Where m is the order with respect to A and n is the order
with respect to B
HOWEVER the overall order is the sum of the exponents
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Rate Law Example
Consider the following reaction:
The initial rate of the reaction was measured at several
different concentrations of the reactants with the following
results:
From the data, determine:
a. The rate law for the reaction
b. The rate constant (k) for the reaction
[CHCl3] (M) [Cl2] (M) Initial Rate (M/s)
0.010 0.010 0.0035
0.020 0.010 0.0069
0.020 0.020 0.0098
0.040 0.040 0.027
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Rate Law Example, cont.
From 1 to 2: [A] doubles while [B] is constant
The rate also doubles
This tells us that the rate order for [A] is first order
From 2 to 3: [A] is constant while [B] doubles
The rate is 1.5 times the previous rate
This tells us that the rate order for [B] is ½ order
The rate law is: Rate = k[A][B]1/2
[CHCl3] (M) [Cl2] (M) Initial Rate (M/s)
1 0.010 0.010 0.0035
2 0.020 0.010 0.0069
3 0.020 0.020 0.0098
4 0.040 0.040 0.027
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Rate Law Example, cont.
To find k, pick one data set to plug into the rate equation:
Set #4:
Homework Problems #9, 10, 11, 12, 14, 16, 18
[CHCl3] (M) [Cl2] (M) Initial Rate (M/s)
1 0.010 0.010 0.0035
2 0.020 0.010 0.0069
3 0.020 0.020 0.0098
4 0.040 0.040 0.027
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13.4: The Integrated Rate Law
So far we have looked at the rate of a reaction and the
change in concentration of a reactant
But we really want to know the change in concentration
of a reactant over time
For this we need to find the integrated rate law
There are three integrated rate laws, one for each order
reaction
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First-Order Integrated Rate Law
We know that for a single reactant A:
And since
You can use calculus to integrate this rate law to find:
Where:
[A]t is the concentration of A at any time t
k is the rate constant
[A]0 is the initial concentration of A
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First-Order Integrated Rate Law
Notice that the equation looks like the equation for a
straight line:
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First-Order Integrated Rate Law Example
See page 484 in the book
In these problems you must first graph the data as shown
Excel is good
Use the graph to predict the concentration of SO2Cl2 at
1900s
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Second Example
Cyclopropane rearranges to form propene in the gas
phase according to the following reaction:
The reaction is first order in cyclopropane and has a
measured rate constant of 3.36x10-5s-1 at 720K. If the
initial cyclopropane concentration is 0.0445M, what will
be the cyclopropane concentration after 235.0 minutes?
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Second-Order Integrated Rate Law
We know that for a single reactant A:
And since
You can use calculus to integrate this rate law to find:
Where:
[A]t is the concentration of A at any time t
k is the rate constant
[A]0 is the initial concentration of A
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Second-Order Integrated Rate Law
Notice that the equation looks like the equation for a
straight line:
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Zero-Order Integrated Rate Law
We know that for a single reactant A:
And since
You can use calculus to integrate this rate law to find:
Where:
[A]t is the concentration of A at any time t
k is the rate constant
[A]0 is the initial concentration of A
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Zero-Order Integrated Rate Law
Notice that the equation looks like the equation for a
straight line:
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Second-Order Integrated Rate Law Example
Use the following data to show that the reaction is NOT
first order: [see excel sheet]
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Finding the Rate Order Example
Using the following data find the overall rate order for
the reaction of a single reactant:
It is second order!
Time (s) [A] (M)
0 0.1000
5 0.0141
10 0.0078
15 0.0053
20 0.0040
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Half-Life
The half-life (t½) of a reaction is the time required for the
concentration of a reactant to fall to one-half of its initial
value
For first-order reactions:
At a time equal to the half-life (t = t½):
Note that for a first-order reaction t½ does not depend on conc.
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Half-Life
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Half-Life Example
A first-order reaction has a half-life of 26.4 seconds. How
long will it take for the concentration of the reactant to
fall to one-eighth of its initial value?
1st half-life: 1/2 initial
2nd half-life: 1/4 initial
3rd half-life: 1/8 initial
3 half-lives x 26.4 seconds = 79.2 seconds
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Half-Life
For a second-order reaction, the half-life equation is:
For a zero-order reaction, the half-life equation is:
See your book for the proof
On page 490 there is a nice summary chart of the three
rate laws
Homework Problems: #20, 22, 24, 26, 28
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13.5: The Affect of Temperature on Rate
Changing the temperature affects the rate of a reaction
The rate constant, k, is only constant when the temperature
remains constant
Chemist Svante Arrhenius investigated this relationship:
Where A is the frequency factor
R is the ideal gas constant
T is the temperature
Ea is the activation energy
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Activation Energy
The activation energy is the energy barrier or hump that
must be surmounted for the reactants to be turned into
products
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Activation Energy
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Activation Energy
The higher the Ea, the slower the reaction rate
The higher the hill the harder it is to go over it
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Collision Theory
In the collision model a chemical reaction occurs after a
sufficiently energetic collision between two reactant
molecules
http://phet.colorado.edu/en/simulation/reactions-and-
rates
Homework Problems #31, 43
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13.6: Reaction Mechanisms
Most chemical reactions do NOT take place in just one step alone (they take multiple steps)
For example:
Does not just take one step to complete
This does not show the intermediate steps that it took to complete
The full reaction is:
This is called a reaction mechanism
Or a series of individual steps by which an overall reaction occurs
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Reaction Mechanisms
Each step in a reaction mechanism is called an elementary
step
Which cannot be broken into smaller steps
In any reaction mechanism, the elementary steps must
add to the overall reaction:
In this case HI is the reaction intermediate
This is not found in the overall reaction but plays a key role in
the mechanism
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Reaction Mechanisms
If we look at the reaction:
And if the experimentally determined rate is:
We can say the possible mechanism is:
The slow step is called the rate determining step
We can look at the proposed energy diagram
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Reaction Mechanisms
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Reaction Mechanisms
The mechanism doesn’t have to be proven to be valid
Simply because other mechanism may exist and can be proven
The rate determining step is also the step that the rate
law is written for the overall reaction
So what is the rate law for the following reaction:
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Reaction Mechanisms
If the overall reaction is:
And the experimental rate law is:
Given the proposed mechanism, determine if the
mechanism is valid:
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Reaction Mechanisms
So the rate law for the rate determining step is:
However N2O2 is an intermediate and rate laws cannot
be expressed in terms of the intermediate
So we need to find a way to substitute the [N2O2]…
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Reaction Mechanisms Example
Predict the overall reaction and rate law that would result
from the following two-step reaction:
Homework Problems: #45, 47, 48
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13.7 Catalysis
A catalyst is a substance that increases the rate of a
chemical reaction but is not consumed by the reaction
For example:
This reaction by itself is very slow
However when you introduce a Cl atom (from
chlorofluorocarbons) this happens very quickly:
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Catalysis
Homework Problem #50
Review Problems #54, 56, 61, 62, 71