Chemical Kinetics
How are chemical reaction rates
determined and what factors
affect these rates?
Objectives





To understand rates of reaction and the
conditions affecting rates.
To derive the rate equation, rate constant,
and reaction order from experimental data.
To use integrated rate laws.
To understand collision theory of reaction
rates and the role of activation energy.
To relate reaction mechanisms and rate
laws.
Chemical Kinetics
Ch 14 hmwk problems:
11 – 15, 20, 21, 25, 29, 33, 38, 45, 47,
50, 55, 58, 59, 62, 66, 69, 87, 98
Why are Kinetics
Important?



Medicinal: How quickly or slowly will a
medication work?
Environmental: Is the rate at which
ozone depletion occurs equal to the
rate at which it is formed?
Industrial: How long will it take to
produce a chemical? Is there a way to
change this?
What are Kinetics Anyway?



Chemical Kinetics is the study of
chemical reaction rates, or speeds.
Can you think of an example of a
chemical reaction with a fast rate?
combustion of gasoline
Can you think of an example of a
chemical reaction with a slow rate?
formation of a diamond
What Makes a Reactions
Occur?


If a reaction is to occur between two or
more reactants, the reactants must come in
contact with each other, and have enough
energy to both beak existing bonds and
form new bonds.
The rate at which these reactants come in
contact with each other affects the rate of
the overall chemical reaction.
Factors Affecting
Reaction Rates

The Physical State of the Reactants:
Reactions are limited by the area of
contact between two reactants.
Therefore, the greater the surface
area of a solid reactant, the faster a
reaction can proceed.
Factors Affecting
Reaction Rates

The Concentration of Reactants: As
the concentration of one or more
reactants increases, the more
frequently the reactants can collide,
leading to increased reaction rates.
Factors Affecting
Reaction Rates

The temperature at which the reaction
occurs: The rate of a chemical reaction
occurs increases with increasing
temperature. As the molecules move
more rapidly, they collide more
frequently and with greater energy,
increasing the reaction rate.
Factors Affecting
Reaction Rates

The presence of a catalyst: Catalysts
are reagents that increase the speed
of a reaction without being consumed
in the reaction by affecting how the
reactants collide together. We will
discuss this in greater depth later.
Reaction Rates


Since a rate is expressed as change over
time, a reaction rate is expressed as the
change in the concentration of reactants
over time.
For a simple reaction, A —› B, the rate
would be can be described as the rate of
appearance of B or the rate of
disappearance of A, since one mole of A will
produce to one mole of B.
Reaction Rates


For a simple reaction, A —› B, the rate
would be expressed as either Δ[B]/ Δt
or -Δ[A]/ Δt
Rates are always expressed as positive
quantities, therefore, a negative sign
must be used because the [A]
decreases as the reaction progresses
A reaction vessel initially contains 1.00
M of reactant A. Twenty seconds later,
the reaction vessel contains only 0.54 M
of reactant A and 0.46 M of product B.
Calculate the average rate of the
disappearance of reactant A.




Recall, A —› B, and rate = -Δ[A]/ Δt .
Rate = -[0.54M – 1.00M]/(20s – 0s)
Rate = 0.46M/20s
Rate = 2.3 x 10-2 M/s
Reaction Rates and
Stoichiometry



What if the reaction stoichiometry is
more complicated than our simple 1:1
ratio?
Let’s look at 2HI —› H2 + I2
Rate = -½ Δ[HI]/Δt = Δ[H2]/ Δt
= [I2]/ Δt
Reaction Rates and
Stoichiometry

In general, for the reaction
aA + bB —› cC + dD
the rate is given by
Rate=
-1/aΔ[A]/Δt = -1/bΔ[B]/Δt
= 1/cΔ[C]/ Δt = 1/dΔ[D]/ Δt
How is the rate at which ozone disappears related to
the rate at which oxygen appears in the reaction
2O3 —› 3O2?
Rate = -1/2Δ[O3]/Δt = 1/3Δ[O2]/Δt
If the rate at which O2 appears is 6.0 x 10-5 M/s at a
particular instant, at what rate is O3 disappearing at
the same time?
Rate = -1/2Δ[O3]/Δt = 1/3Δ[O2]/Δt
Rate = -1/2Δ[O3]/Δt = 1/3(6.0 x 10-5 M/s)
Rate = -Δ[O3]/Δt = 2/3(6.0 x 10-5 M/s)
Rate = -Δ[O3]/Δt = 4.0 x 10-5 M/s
You should now be able to
complete # 11 & 12
Rate Laws


The rate law is an equation that relates the
rate of a reaction to the concentration of
reactants (and catalysts) raised to various
powers
The rate law for our last example,
aA + bB —› cC + dD, would be expressed
as
Rate = k[A]m[B]n, where k is the rate
constant and m and n are typically
small, whole numbers.
Rate Laws


k in the rate law is the rate constant
defined as a proportionality constant
in the relationship between rate and
concentrations.
The rate constant is fixed at a
particular temperature (but will
change as the temperature changes)
Rate Laws
We can classify reactions by their orders
– The reaction order, with respect to a given
reactant species, equals the exponent of the
concentration of that species in the rate law, as
determined experimentally
– The overall order of the reaction equals the
sum of the orders of all reactant species in the
rate law
– In the case Rate = k[A]m[B]n, the overall order
of this reaction is given by m+n.
Rate Laws

For the following reaction, the rate law can be
written as is indicated below:
NH4+ + NO2- —› N2 + 2H2O
Rate = k[NH4+][NO2-]


Because the exponent of [NH4+] is one, the rate in
NH4+ is first order. The rate in [NO2-] is also first
order. The overall reaction order is second order.
The exponents in a rate law indicate how the rate is
affected by the concentration of each reactant.
For the following reaction, the rate law can be written
as is indicated below:
NH4+ + NO2- —› N2 + 2H2O
Rate = k[NH4+][NO2-],
What is the effect on the rate if [NO2-] doubles?
Because the exponent of [NO2-] in the rate law is one,
doubling its concentration will double the rate.
What is the effect on the rate if [NH4+] triples?
Because the exponent of [NH4+] in the rate law is one,
tripling its concentration will triple the rate.
For a given reaction, the rate law is given by
Rate = k[NO2]2
What is the order of the reaction with respect to NO2?
Because the exponent of NO2 in the equation is 2, the
reaction is second order with respect to NO2.
Because the concentration of only one reactant is
expressed in the rate law, the overall order is also
second.
What is the effect on the rate if [NO2] triples?
If the concentration triples, then the rate increases by
9 since [3]2.
Rate Laws: Additional
Examples
2N2O5 —› 4NO2 + O2
CHCl3 + Cl2 —› CCl4 + HCl
H2 + I2 —› 2HI
Rate = k[N2O5]
Rate = k[CHCl3][Cl2]½
Rate = k[H2][I2]
Note: While exponents in rate laws are often the same
as the coefficients in a balanced equation, this is
not a rule! Exponents in a rate law can only
be determined experimentally!
Determining a Rate Law
Experimentally – Initial Rate
Method
Trial
[NO]
[H2]
R0 (M/s)
1
0.10
0.10
1.23 x 10-3
2
0.10
0.20
2.46 x 10-3
3
0.20
0.10
4.92 x 10-3
Using the data above, determine the rate law,
calculate the value of k and determine the rate when
[NO] = 0.050M and [H2] = 0.150 M.
Determining a Rate Law
Experimentally – Initial Rate
Method
A. Determine the rate law
Recall, the Rate = k[NO]x[H2]y. We can
determine the values of x and y by
using the experimental data and the
equation and solving for the unknown
variables.
Determining a Rate Law
Experimentally – Initial Rate
Method
Let’s compare trial two to trial one:
Rate 2 = [NO]x[H2]y = [0.10]x[.20]y
Rate 1 = [NO]x[H2]y = [0.10]x[.10]y
We can simplify our expression and substitute
the values for Rate 1 and Rate 2
Rate 2 = 2.46 x 10-3 = [.20]y
Rate 1 = 1.23 x 10-3 = [.10]y
Notice, we are only left with one unknown
variable to solve for.
Determining a Rate Law
Experimentally – Initial Rate
Method
Rate 2 = 2.46 x 10-3 = [.20]y
Rate 1 = 1.23 x 10-3 = [.10]y
2y = 2
y must equal 1.
So far we have determined that
Rate = k[NO]x[H2]1. We must repeat the
process to solve for x by comparing trial 3
to trial 1.
Determining a Rate Law
Experimentally – Initial Rate
Method
Rate 3 = [NO]x[H2]y = [0.20]x[.10]y
Rate 1 = [NO]x[H2]y = [0.10]x[.10]y
Rate 3 = 4.92 x 10-3 = [.20]x
Rate 1 = 1.23 x 10-3 = [.10]x
2x = 4
X must equal 2. Therefore, the experimentally
determined rate law is Rate = k[NO]2[H2]1
Determining a Rate Law
Experimentally – Initial Rate
Method
B. Calculate the value of k
If Rate = k[NO]2[H2]1, then simple
substitution from the data table will
allow us to calculate k.
Rate = k[NO]2[H2]1 (use trial 1)
1.23 x 10-3M/s = k[0.10M]2[0.10M]1
1.23 x 10-3M/s = k(0.0010M3)
1.2 M-2·s-1 = k
Determining a Rate Law
Experimentally – Initial Rate
Method
C. Determine the rate when [NO] =
0.050M and [H2] = 0.150 M
Since Rate = 1.2M-2·s-1[NO]2[H2]1
Rate = 1.2M-2·s-1[0.050M]2[0.150M]1
Rate = 4.5 x 10-4 M/s
You should now be able to
complete # 13, 14, 15, 20,
21, 25
Instantaneous Rates

Calculated from a
graph of
experimental data
that represents
concentration vs.
time.
Instantaneous Rates



Recall, the slope of a graph is
Δy/Δx. In a concentration vs.
time graph, the Δy = Δ[reactant]
and Δx = Δtime.
Therefore, the slope of the graph
= Δ[reactant]/ Δtime and will
yield the instantaneous rate of
reaction progression at any given
point in time.
Average reaction rates can be
determined from the average of
several calculated instantaneous
rates.
Rate Orders
Recall, the overall order of a reaction is
equal to the sum of the exponents in a
rate law
Rate = k[A]0 or Rate = k is zero order
Rate = k[A]1 is first order
Rate = k[A]2 or Rate = k[A][B] is second order
Rate = k[A]3 or Rate = k[A]2[B] is third order
A closer look at Zeroorder Reactions
Rate = k[A]0 or Rate = k is zero order
This is referred to as the differential rate law.
If we integrate the above equation, we find a
very useful equation:
[A] = -kt + [A0]
This will allow us to solve for unknown
concentration values of zero-order reactions
A closer look at Zeroorder Reactions
A zero-order reaction begins with [X] =
0.50 M. How long does it take to
react all but 0.10M if k=0.0410
M/min? What is the half-life? What is
the [X] after 5 min?
A closer look at Zeroorder Reactions
A. [X] = 0.10 M, [X0] = 0.50 M and
k=0.0410 M/min
[A] = -kt + [A0]
[0.10M] = -(0.0410M/min)t + [0.50M]
-0.40M = -(0.0410M/min)t
10. min = t
A closer look at Zeroorder Reactions
B. What is the half-life?
Recall, a half-life is the time it takes for half a
reactant to be consumed.
So, if [X0] = 0.50 M, then [X] must equal 0.25
M
If [A] = -kt + [A0], then
[0.25M] = -(0.0410M/min)t½ + [0.50M]
-0.25M = -(0.0410M/min)t½
6.1 min = t ½
A closer look at Zeroorder Reactions
C. What is the [X] after 5 min?
If [A] = -kt + [A0], then
[X] = -(0.0410M/min)5min + [0.50M]
[X] = -.205M + 0.50M
[X] = 0.30 M
A closer look at Zeroorder Reactions
Notice, [A] = -kt + [A0] closely mimics
y = mx + b
Therefore, a graph of change in concentration
vs. time will be linear for a zero-order
reaction.
A closer look at Firstorder Reactions
Rate = k[A]1 is first order
This is referred to as the differential rate law.
If we integrate the above equation, we find a the
integrated rate law for first order equations:
ln[At] - ln[A0] = -kt
Or
ln[At] = -kt + ln[A0]
(y) = (mx) + b
This will allow us to solve for unknown concentration
values of first-order reactions
A closer look at Secondorder Reactions
Rate = k[A]2 or is second order
This is referred to as the differential rate law.
If we integrate the above equation, we find a the
integrated rate law for second order equations:
1/[At] = kt + 1/[A0]
(y) = (mx) + (b)
This will allow us to solve for unknown concentration
values of second-order reactions
Determining Order From
Graphical Data
Time
(min)
0
200
400
600
800
1000
1200
1600
[X]
0.0200
0.0160
0.0131
0.0106
0.0086
0.0069
0.0056
0.0037
An experiment was
conducted were a reaction
was allowed to proceed
and the reactant
concentration was
measured over a period of
time. What is the order of
this reaction?
Determining Order From
Graphical Data
Recall, the integrated rate law of zero, first
and second order reactions can be
rearranged in the form of y=mx + b
Zero-order: [A] = -kt + [A0]
First-order: ln[At] = -kt + ln[A0]
Second-order: 1/[At] = kt + 1/[A0]
If we create three plots of our data, we can
determine order: [X] vs. t; ln[X] vs. t; 1/[X]
vs. t. The plot that yield a linear graph will
determine its order.
Determining Order From Graphical Data
Notice, only one graph yields
a straight line: The plot of
ln[X] vs. t. Therefore, the
reaction must be first order.
You should now be able to
complete # 29, 33, 38
A Microscopic View of
Reaction Rates
Consider the reaction
NO + O3 —› NO2 + O2
where Rate = k[NO][O3]
Imagine these reactants in rapid, random
motion inside a flask. They strike the walls
of the reaction vessel and collide with other
molecules.
Will all collisions result in a chemical reaction?
What conditions must be met for these
reactants to react together?
How a Chemical Reaction
Takes Place

Only a tiny fraction of reactant
collisions leads to reaction. Why?
– Reactant molecules must collide in the
proper orientation.
– They must collide with the minimum
energy required to initiate a chemical
reaction. This is called activation
energy (Ea) and the value varies from
reaction to reaction.
Activation Energy
The diagrams shows the change in potential energy of the molecules
during the reaction. Energy must be supplied to the reactants to stretch
and break reactant bonds into an intermediate form (represented at the
peak of the graph) before proceeding to the final product. The energy
difference between the starting molecule and the highest energy along the
path represent the activation energy. The difference in the potential
energies of the reactants and products indicate whether the reaction was
endo- or exothermic.
The Arrhenius Equation
Most reaction rate data is based on three factors
1. The fraction of molecules possessing the
minimum Ea or greater
2. The number of collisions occurring per second
3. The fraction of collisions that have the
appropriate orientation
Arrhenius related these factors mathematically:
k = Ae-Ea/RT
The frequency factor, A, is considered constant.
The Arrhenius Equation
The Arrhenius can be written in several
forms:
k = Ae-Ea/RT
lnk = -Ea/RT + lnA
ln(k1/k2) = Ea/R (1/T2 – 1/T1)
(notice this last formula closely mimics y=mx + b)
Using the Arrhenius
Equation
Use the data to determine the activation energy for
the zero-order decomposition of HI on a platinum
surface.
Temp (K)
Rate Constant (M/s)
573
2.91 x 10-6
673
8.38 x 10-4
773
7.65 x 10-2
We can determine the activation energy for a
reaction from a plot of the lnk vs. 1/T. According
to ln(k1/k2) = Ea/R (1/T2 – 1/T1), the slope of
this graph will equal Ea/R.
Using the Arrhenius
Equation
Temp (K)
Rate Constant
(M/s)
lnk
1/T (K-1)
573
2.91 x 10-6
-12.75
0.00175
673
8.38 x 10-4
-7.08
0.00149
773
7.65 x 10-2
-2.57
0.00129
ln k vs. 1/T
0
0
0.0005
0.001
0.0015
-2
ln k
-4
-6
y = -22115x + 25.926
-8
-10
-12
-14
1/T (K-1)
0.002
Using the Arrhenius
Equation
The graph yields a straight line with a
slope of -2.2x104K. According to the
equation, ln(k1/k2) = Ea/R (1/T2 –
1/T1), this equals Ea/R.
So, -2.2x104 K= Ea/R
-2.2x104 K= Ea/8.314 J/K·mol
Ea= 1.8 x 102 kJ/mol
You should now be able to
complete # 45, 47, 50
Reaction Mechanisms



We know that reactants must change form
(often forming an intermediate state) in
order to proceed to becoming products.
This process can be described in much
greater detail if we look at a chemical
reaction as occurring over a series of steps.
This is referred to as a reaction mechanism.
Reaction Mechanisms


Reaction mechanisms can be very
simple and involve only one step, or
very complicated, involving many
steps.
Reaction mechanisms are often
speculated. It is very difficult to
scientifically prove that a chemical
reaction happens in a specific way.
Reaction Mechanisms
The steps of a reaction mechanism can
be described by the number of
molecules that participate as reactants
in the step. This is called the
molecularity of the reaction.
H3CNC —> H3CCN
unimolecular
NO + O2 —> NO2 + O2
bimolecular
And so on
Reaction Mechanisms


Reactions involving only one step are
referred to as elementary reactions
Reaction involving several steps are
referred to as multistep reactions.
This often involves the formation and
consumption of an intermediate.
Multistep Mechanisms
Let’s consider the reaction
NO2 + CO —> NO + CO2
Studies of the reaction have shown that it most likely
occurs in two steps,
NO2 + NO2 —> NO3 + NO
NO3 + CO —> NO2 + CO2
Thus, we say the reaction occurs by a two-step
mechanism.
Each step is represented by an elementary reaction
and the rate law can be written directly from its
molecularity.
Multistep Mechanisms
The chemical equations for the elementary reactions
in a multistep mechanism must always add to give
the chemical equation of the overall process.
NO2 + NO2 —> NO3 + NO
NO3 + CO —> NO2 + CO2
2 NO2 + NO3 + CO —> NO2 + NO3 + NO + CO2
This equation can be simplified by eliminating
substances that appear on both sides to yield:
NO2 + CO —> NO + CO2
Notice, NO3 appears in the reaction mechanism, but it
neither a reactant nor a product. It is formed in
one elementary step and consumed in the next.
NO3 is an intermediate.
Multistep Mechanisms


Each step of a mechanism has its own rate law.
Each step is represented by an elementary reaction
and the rate law can be written directly from its
molecularity.
Step 1: NO2 + NO2 —> NO3 + NO Rate1 = k1[NO2]2
Step 2: NO3 + CO —> NO2 + CO2 Rate2 = k2[NO3][CO]
 It is important to remember that the overall rate
law of a chemical reaction does not necessarily
follow the molecularity of the balanced chemical
equation! It must be determined experimentally.
Multistep Mechanisms



Most chemical reactions occur by
mechanisms involving two or more
steps.
These steps often occur at different
rates.
The overall rate of a chemical reaction
is determined by the slowest step of
its mechanism. This is referred to as
the rate-determining step.
Multistep Mechanisms –
Slow Initial Step
Let’s take another look at our example.
Step 1: NO2 + NO2 —> NO3 + NO
Step 2: NO3 + CO —> NO2 + CO2
Overall: NO2 + CO —> NO + CO2
If we know that step 1 is much, much slower
than step two, step 1 must be ratedetermining. Therefore, the overall rate law
for this reaction must be Rate = k[NO2]2
Multistep Mechanisms –
Fast Initial Step
If the initial step is fast, and therefore not
rate-determining, it is likely that the ratedetermining step involves an intermediate
as a reactant.
Let’s look at another example:
Step 1: NO + Br2 <—> NOBr2 (fast)
Step 2: NOBr2 + NO —> 2NOBr (slow)
Overall: 2NO + Br2 —> 2NOBr
How would you write a rate law for the overall
reaction?
Multistep Mechanisms –
Fast Initial Step
You may have been tempted to say
Rate = k[NOBr2][NO]
However, notice that an intermediate appears
in the rate law. The concentrations of
intermediates are usually unknown. Thus,
our rate law depends on an unknown
concentration. We must rewrite our rate
law to include only known quantities.
But how???
Multistep Mechanisms –
Fast Initial Step
Let’s go back to the mechanism
Step 1: NO + Br2 <—> NOBr2 (fast)
Step 2: NOBr2 + NO —> 2NOBr (slow)
Overall: 2NO + Br2 —> 2NOBr
Write a rate law the first step
Step 1: NO + Br2 <—> NOBr2 (fast)
Rate1 = k1[NO][Br2]
But notice the equilibrium? The rate above is for the forward
reaction, what about the reverse?
Rate-1 = k-1[NOBr2]
Multistep Mechanisms –
Fast Initial Step
Write a rate law the second step
Step 2: NOBr2 + NO —> 2NOBr (slow)
Rate2 = k2[NOBr2][NO]
Recall, the original overall rate law was written
as Rate = k[NOBr2][NO]. We must rewrite
the rate law without [NOBr2].
The three elementary rate laws written from
the mechanism should give us plenty of
substitution to choose from.
Multistep Mechanisms –
Fast Initial Step
We must rewrite Rate = k[NOBr2][NO] using
the elementary rate laws shown below.
Rate1 = k1[NO][Br2]
Rate-1 = k-1[NOBr2]
Rate2 = k2[NOBr2][NO]
Recall that because of equilibrium, Rate1 = Rate-1.
So, k1[NO][Br2] = k-1[NOBr2]
Notice, we can now solve for NOBr2.
(k1/k-1) [NO][Br2] = [NOBr2]
Multistep Mechanisms –
Fast Initial Step
Let’s make the substitution.
Rate = k[NOBr2][NO]
Rate = k(k1/k-1) [NO][Br2][NO]
Since k’s are constant, let’s rename k(k1/k-1) as K.
Rate = K [NO][Br2][NO]
And now simplify
Rate = K [NO]2[Br2] where K = k(k1/k-1).
This is a viable rate law since the concentrations of all
species in the rate law are known.
You should now be able to
complete # 55, 58, 59, 62
Catalysis



A catalyst is a substance that changes the
speed of a chemical reaction without
undergoing a permanent chemical change.
A catalyst can be homogeneous (in the
same phase as the reacting molecules) or
heterogeneous (in a different phase as the
reacting molecules)
Enzymes in the human body are examples
of organic catalysts.
Homogeneous Catalysis
H2O2(aq) —> H2O(l) + O2(g)
The decomposition of hydrogen peroxide can
be catalyzed by I- (given by the dissociation
of KI) as shown.
Step 1: 2I-(aq) + 2K+(aq) + H2O2(aq) —> I2(aq) + 2H2O(g)
Step 2: I2(aq) + H2O2(aq) —> 2I-(aq) + 2K+(aq) + O2(g)
Notice, adding steps 1 & 2 yields
2H2O2(aq) —> 2H2O(l) + O2(g)
How does this work?
Homogeneous Catalysis
As a general rule, a catalyst lowers the overall
activation energy for a chemical reaction
Heterogeneous Catalysis



The initial set usually begins with adsorption
of reactants onto the catalyst surface.
Atoms/Ions on the surface of a solid are
very reactive because of the unused
bonding capacity.
Be sure to read this section of your text
book carefully so you understand how
heterogeneous catalysis works!!!
SAMPLE INTEGRATIVE EXERCISE Putting Concepts Together
Formic acid (HCOOH) decomposes in the gas phase at elevated temperatures
as follows:
The decomposition reaction is determined to be first order. A graph of the
partial pressure of HCOOH versus time for decomposition at 838 K is shown as
the red curve in Figure 14.28. When a small amount of solid ZnO is added to
the reaction chamber, the partial pressure of acid versus time varies as shown
by the blue curve in Figure 14.28.
Figure 14.28 Variation
in pressure of
HCOOH(g) as a
function of time at 838
K. The red line
corresponds to
decomposition when only
gaseous HCOOH is
present. The blue line
corresponds to
decomposition in the
presence of added ZnO(s).
SAMPLE INTEGRATIVE EXERCISE continued
(a) Estimate the half-life and first-order rate constant for formic acid
decomposition.
(b) What can you conclude from the effect of added ZnO on the
decomposition of formic acid?
(c) The progress of the reaction was followed by measuring the partial
pressure of formic acid vapor at selected times. Suppose that, instead,
we had plotted the concentration of formic acid in units of mol/L. What
effect would this have had on the calculated value of k?
(d) The pressure of formic acid vapor at the start of the reaction is 3.00
 10 2 torr. Assuming constant temperature and ideal-gas behavior, what
is the pressure in the system at the end of the reaction? If the volume of
the reaction chamber is 436 cm3, how many moles of gas occupy the
reaction chamber at the end of the reaction?
(e) The standard heat of formation of formic acid vapor is ΔHf° = 378.6 kJ/mol.
Calculate ΔHº for the overall reaction. Assuming that the activation
energy (Ea) for the reaction is 184 kJ/mol, sketch an approximate
energy profile for the reaction, and label Ea, ΔHº, and the transition
state.
You should now be able to
complete # 66, 87, 88, 98
You Should Now Be Able To…
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
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Identify factors which affect reaction rates.
Calculate the rate of production of a product or
consumption of a reactant using mole ratios and
the given rate.
Determine the rate law for a reaction from given
data, overall order and value of the rate constant
(including units!)
Determine the instantaneous rate of a reaction.
You Should Now Be Able To…
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


Use integrated rate laws to determine
concentrations at a certain time, t, and create
graphs to determine the order of a reaction. Also,
determine the half-life reaction.
Determine the activation energy for the reaction
using the Arrhenius equation.
Graphically determine the activation energy using
the Arrhenius equation.
Write the rate law from a given mechanism and
identify catalysts and intermediates present.