UNIT 3: Energy Changes and Rates of Reaction
Chapter 6: Rates of Reaction
UNIT 3 Chapter 6: Rates of Reaction
Chapter 6: Rates of Reaction
An important part of
studying chemical reactions
is to monitor the speed at
which they occur. Chemists
look at how quickly, or
slowly, reactions take place
and how these rates of
reaction are affected by
different factors.
The light produced by a firefly depends
on the speed of a particular chemical
reaction that occurs in its abdomen.
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UNIT 3 Chapter 6: Rates of Reaction
6.1 Chemical Reaction Rates
Chemical kinetics is the study of the rate at which
chemical reactions occur.
The term reaction rate, or rate of reaction refers to:
• the speed that a chemical reaction occurs at, or
• the change in amount of reactants consumed or
products formed over a specific time interval
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Section 6.1
UNIT 3 Chapter 6: Rates of Reaction
Section 6.1
Determining Reaction Rates
The reaction rate is often given in terms of the change in
concentration of a reactant or product per unit of time.
The change in concentration of
reactant A was monitored over time.
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UNIT 3 Chapter 6: Rates of Reaction
Section 6.1
Determining Reaction Rates
The change in concentration of reactant or product
over time is often graphed.
For the reaction A → B, over time, the
concentration of A decreases, and the
concentration of B increases.
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UNIT 3 Chapter 6: Rates of Reaction
Section 6.1
Average and Instantaneous Reaction Rates
Average rate of reaction:
• change in [reactant] or [product] over a given time
period (slope between two points)
Instantaneous rate of reaction:
• the rate of a reaction at a particular point in time
(slope of the tangent line)
Average rate of reaction
and instantaneous rate
of reaction can be
determined from a
graph of concentration
vs. time.
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UNIT 3 Chapter 6: Rates of Reaction
Section 6.1
Expressing Reaction Rates in Terms of
Reactants or Products
A known change in concentration of one reactant or
product and coefficients of a chemical equation allows
determination of changes in concentration of other
reactants or products.
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UNIT 3 Chapter 6: Rates of Reaction
Section 6.1
Express the rate of formation of
ammonia relative to hydrazine, for the
reaction on the previous slide.
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Answer on
the next slide
UNIT 3 Chapter 6: Rates of Reaction
The mole ratio of ammonia to
hydrazine is 4:3
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Section 6.1
UNIT 3 Chapter 6: Rates of Reaction
Section 6.1
Methods for Measuring Rates of Reaction
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UNIT 3 Chapter 6: Rates of Reaction
Section 6.1
Calculating Reaction Rates from
Experimental Data
The following data were collected in order to calculate the
rate of a reaction.
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UNIT 3 Chapter 6: Rates of Reaction
Calculating Reaction Rates from
Experimental Data
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Section 6.1
UNIT 3 Chapter 6: Rates of Reaction
Calculating Reaction Rates from
Experimental Data
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Section 6.1
UNIT 3 Chapter 6: Rates of Reaction
Section 6.1 Review
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Section 6.1
UNIT 3 Chapter 6: Rates of Reaction
Section 6.2
6.2 & 6.3: Collision Theory and Factors
Affecting Rates of Reaction
According to collision theory, a chemical reaction occurs
when the reacting particles collide with one another.
Only a fraction of collisions between particles result in a
chemical reaction because certain criteria must be met.
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UNIT 3 Chapter 6: Rates of Reaction
Effective Collision Criteria 1:
The Correct Orientation of Reactants
For a chemical reaction to occur, reactant molecules
must collide with the correct orientation relative to
each other (collision geometry).
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Five of many possible ways that NO(g) can collide with NO3(g) are shown.
Only one has the correct collision geometry for reaction to occur.
Section 6.2
UNIT 3 Chapter 6: Rates of Reaction
Effective Collision Criteria 2:
Sufficient Activation Energy
For a chemical reaction, reactant molecules must also
collide with sufficient energy.
Activation energy, Ea, is the minimum amount of
collision energy required to initiate a chemical reaction.
Collision energy depends on the kinetic energy of the
colliding particles.
The shaded part of the MaxwellBoltzmann distribution curve
represents the fraction of particles that
have enough collision energy for a
reaction (ie the energy is ≥ Ea).
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Section 6.2
UNIT 3 Chapter 6: Rates of Reaction
Section 6.2
Representing the Progress of a
Chemical Reaction
From left to right on a potential energy curve for a reaction:
• potential energy increases as reactants become closer
• when collision energy is ≥ maximum potential energy,
reactants will transform to a transition state
• products then form (or reactants re-form if ineffective)
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Exothermic
Endothermic
UNIT 3 Chapter 6: Rates of Reaction
Activation Energy and Enthalpy
The Ea for a reaction cannot be predicted from ∆H.
• ∆H is determined only by the difference in potential
energy between reactants and products.
• Ea is determined by analyzing rates of reaction at
differing temperatures.
• Reactions with low Ea occur quickly. Reactions with
high Ea occur slowly.
Potential energy diagram for the
combustion of octane.
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Section 6.2
UNIT 3 Chapter 6: Rates of Reaction
Section 6.2
Activation Energy for Reversible Reactions
Potential energy diagrams can represent both forward and
reverse reactions.
• follow left to right for the forward reaction
• follow right to left for the reverse reaction
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UNIT 3 Chapter 6: Rates of Reaction
Analyzing Reactions Using
Potential Energy Diagrams
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The BrCH3 molecule and OH- collide with the correct orientation and sufficient
energy and an activated complex forms. When chemical bonds reform, potential
energy decreases and kinetic energy increases as the particles move apart.
Section 6.2
UNIT 3 Chapter 6: Rates of Reaction
LEARNING CHECK
Describe the relative values of Ea(fwd)
and Ea(rev) for an exothermic reaction
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Answer on
the next slide
UNIT 3 Chapter 6: Rates of Reaction
LEARNING CHECK
Ea(rev) is greater than Ea(fwd)
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Section 6.2
UNIT 3 Chapter 6: Rates of Reaction
Section 6.2
Factors Affecting Reaction Rate
1. Nature of reactants
• reactions of ions tend to be faster than those of
molecules
2. Concentration
• a greater number of effective collisions are more
likely with a higher concentration of reactant
particles
3. Temperature
• with an increase in temperature, there are more
particles with sufficient energy needed for a reaction
(energy is ≥ Ea)
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UNIT 3 Chapter 6: Rates of Reaction
Section 6.2
Factors Affecting Reaction Rate
4. Pressure
• for gaseous reactants, the number of collisions in a
certain time interval increases with increased pressure
5. Surface area
• a greater exposed surface area of solid reactant means
a greater chance of effective collisions
6. Presence of a catalyst
• a catalyst is a substance that increases a reaction rate
without being consumed by the reaction
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UNIT 3 Chapter 6: Rates of Reaction
Section 6.2
A Catalyst Influences the Reaction Rate
A catalyst lowers the Ea of a reaction.
• this increases the fraction of reactants that have enough
kinetic energy to overcome the activation energy barrier
• a catalyzed reaction has the same reactants, products,
and enthalpy change as the uncatalyzed reaction
A catalyst decreases
both Ea(fwd) and Ea(rev).
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UNIT 3 Chapter 6: Rates of Reaction
Section 6.2
Catalysts in Industry
A metal catalyst is used for industrial-scale production of
ammonia from nitrogen and hydrogen.
Hydrogen and nitrogen molecules break apart when in contact with the
catalyst. These highly reactive species then recombine to form ammonia.
A catalyst (V2O5) is used for industrial-scale production
of sulfuric acid from sulfur.
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UNIT 3 Chapter 6: Rates of Reaction
Section 6.2
Catalysts in Industry
The Ostwald process uses a platinum-rhodium catalyst
for the industrial production of nitric acid.
Many industries use biological catalysts, called
enzymes, which are most often proteins.
For example: the use of enzymes decreases the amount
of bleach (an environmental hazard) needed to whiten
fibres used in paper production.
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UNIT 3 Chapter 6: Rates of Reaction
Section 6.2 Review
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Section 6.2
UNIT 3 Chapter 6: Rates of Reaction
Section 6.3
6.5: Rate Law
Initial rate is the rate of a chemical reaction at time zero.
• products of the reaction are not present, so the reverse
reaction cannot occur
• it is a more accurate method for studying the
relationship between concentration of reactant and
reaction rate
Initial rate is found by
determining the slope of a
line tangent to the curve at
time zero.
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UNIT 3 Chapter 6: Rates of Reaction
The Rate Law
The rate law shows the relationship between reaction
rates and concentration of reactants for the overall
reaction.
rate = k[A]m[B]n
m: order of the reaction for reactant A
n: order of the reaction for reactant B
k: rate constant
m + n: order of the overall reaction
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Section 6.3
UNIT 3 Chapter 6: Rates of Reaction
Section 6.3
Graphing Reaction Rate in Terms of
Concentration
To study the effects of concentration on reaction rate:
• different starting concentrations of reactant are used
• initial rates are calculated using the slopes of the tangent
lines from concentration vs time curves
• initial rates are plotted against starting
concentration
Initial rates are determined
(A) and these are plotted
against concentration (B).
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UNIT 3 Chapter 6: Rates of Reaction
Section 6.3
First-order Reactions
The initial rate vs starting concentration graph on the
previous slide is a straight line.
• the equation of the line can be expressed as:
rate = k[A]
• This represents a first-order reaction
For reactions with more than one reactant (e.g. A and B):
• if experiments for each reactant produce straight lines,
the rate is “first order with respect to reactant A and first
order with respect to reactant B.”
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Example1:
The empirically determined rate law equation for the reaction
between nitrogen dioxide and fluorine gas :
2NO2 + F2 → 2NO2F
r = K[NO2]1[F2]1
The order of reaction with respect to NO2, F2 is 1 respectively.
the overall order of reaction is (1 + 1) = 2 ( The reaction is second
order overall)
Example 2:
The discomposition of dinitrogen pentoxide is first order with respect
to N2O5. If the initial rate of consumption is 2.1 x 10-4 mol/(L*s)
when the initial concentration of N2O5 is 0.40 mol/L. Predict what
the rate would be if another experiment were performed in which
the initial concentration of N2O5 were 0.80 mol/L
2N2O5
→
2NO2
+O2
Solution:
First: write the rate law equation for the system:
r = K[N2O5]1 (first order reaction, so exponent is 1)
r = 0.8 mol/L = x
0.4 mol/L 2.1 x 10-4mol/(L*s)
Since it is first order, any change in [N2O5] will have the same effect
on the rate. The [N2O5] is doubled from 0.4 to 0.8 mol/L, so the rate
of consumption will double from 2x (2.1 x 10-4) to 4.2 x 10-4
mol/(L*s)
UNIT 3 Chapter 6: Rates of Reaction
Second-order Reactions
For chlorine dioxide in this reaction:
• the initial rate vs concentration curve is parabolic
• the reaction is proportional to the square of [ClO2]
• it is a second order reaction
with respect to this reactant
rate = k[A]2
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Section 6.3
Example 3:
The dimerization reaction of 1,3-butadiene (C4H6) is a second
order with respect to C4H6. If the initial rate of reaction were
32 mmol/(L*min) C4H6 at a given initial concentration of C4H6,.
What would be the initial rate of reaction if the initial
concentration of C4H6 were doubled
2C4H6
→ C8H12
Solution:
r = K[C4H6]2 (let x be the initial concentration)
x = 32
2x2
If the initial concentration is doubled (multiplied by 2), the initial
rate will be multiplied by 22, or 4. The new rate is 4 x 32 mmol
C4H6/(L*min) or 0.13 mol /(L*min) C4H6
The rate equation is
DETERMINATION OF EXPONENT OF THE RATE
LAW
The exponents of the Rate Law must be determined
experimentally. To find the exponent of the rate law, we must
study how changes in concentration affect the rate of reaction.
Example: [A]+ B → products
Rate = k[A]m[B]n
DETERMINATION OF EXPONENT OF THE RATE LAW
The exponents of the Rate
Law must be determined
experimentally by study
how changes in
Concentration affect the
Rate of reaction.
Example: A + B → products
Rate = k[A]m[B]n
Example: Experimental Data
Initial Concentration
[A]
1. 0.10
2. 0.20
3. 0.30
4. 0.30
5. 0.30
[B]
0.10
0.10
0.10
0.20
0.30
Initial rate of formation
of products
0.20
0.40
0.60
2.40
5.40
To determine the value of m, we use the
results from Exp.# 1 and 2 or 3, in which
only [A] changes:
rate 2 = 0.4 mol/L.s =k(0.2 mol/L)m
rate 1 0.2 mol/L.s k(0.1mol/L)m
2.0 = (2.0)m
Meaning, if [A] is doubled the rate doubles (#1 & 2). Also, if
[A] is tripled (#1 &3), the rate is tripled .Therefore, the
exponent of A must be 1 ie [A]1
To determine n, we use the results from Exp. #3 and 4 or 5,
in which only [B] changes: This time it is the concentration of
B that affects the rate
rate 4 = 2.4 mol/L.s = k(0.20 mol/L)n
rate 3 0.6 mol/L.s K(0.10 mol/L)n
n
4
= 2
When [B] is doubled the rate increases by factor of 4 which
equals 22
[A]
1) 0.10
2) 0.20
3) 0.30
4) 0.30
5) 0.30
[B]
0.10
0.10
0.10
0.20
0.30
Rates
0.20
0.40
0.60
2.40
5.40
rate 5 = 5.4 mol/L.s = k(0.30 mol/L)n
rate 3 0.60 mol/L.s k(0.10 mol/L)n
9
= 3
Calculating Constant, K, from the Rate Law:
Using the data from set # 1, K can be calculated from
the
r= k [A]1 [B]2
[A]
[B]
K=Rate
[A][B]
0.10
0.20
0.30
0.30
0.30
K =0.20 mol L-1s-1
(0.10 mol L-1)(0.10 mol L-1)
K=0.2 mol L-1s-1
0.010 mol2 L-2
K=2.0 x 101 L mol-1 s-1
Let’s check other sample on P.378-379
0.10
0.10
0.10
0.20
0.30
Rates
0.20
0.40
0.60
2.40
5.40
Try this Example: Determine the exponents of the Rate Law when the following
data were measured for the reduction of Nitric Oxide with
Hydrogen 2NO(g)+2H2(g)→N2(g)+2H2O(g)
Initial Concentrations
Initial Rate of Disappearance of NO
(mol L-1s-1)
Solution:
1. Find the order of the reaction
-3
1.37 X 10 2. Solve for K value and unit
2.75 X 10-3 3. Then write the rate law
Equation
[NO]
[H2]
1) 0.10
0.10
2) 0.10
0.20
3) 0.20
0.10
5.47 X 10-3
1. What is the rate law for this reaction?
2. What is the unit of the constant?
Solution:
The general form of the rate law for this reaction:Rate = K[NO]m[H2]n
To determine m, use result where only NO
changes:
Rate 3 = 5.47x10-3 mol/L*s= 0.20 mol/L
Rate 1 1.37 x 10-3mol/L
0.10 mol/L
4 = [2]n, ie [2]2 , Thus m = 2
To determine n, use result where only [H2] changes:
Rate 2 = 2.75x10-3 mol/L*s= 0.20 mol/L
Rate 1 1.37 x 10-3mol/L
0.10 mol/L
2.0 = [2]n, ie [2]1 , Thus m = 1
Therefore, r = k[NO]2[H2]1
Using trial # 1, find k
0.00137 mol/L.s = k x 0.100 mol/L x (0.100 mol/L)2
k = 0.00137 mol/L.s
=
1.37 L2/mol2.s
0.100 mol/L x (0.100 mol/L)2
r = 1.37 L2/(mol2.s)[NO]2[H2]
Note: Monitor the Unit of the rate constant, K, closely
as it varies with different order of reactions
UNIT 3 Chapter 6: Rates of Reaction
Section 6.3
6.6: Reaction Mechanisms
A reaction mechanism is the series of elementary steps
that occur as reactants are converted to products.
For example, oxygen and nitrogen are not formed directly
from the decomposition of nitrogen dioxide:
It occurs in two elementary steps:
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A balanced equation does not tell us how the reactants
become product, it only tells us the reactant, the product, and
the stoichiometry, but gives no information about the reaction
mechanism.
O is an intermediate, it is neither a reactant nor a
product, but formed and consumed during the reaction
sequence.
Each of the two reactions is called an elementary step, a
reaction whose rate law can be written from its molecules.
A reaction with one molecule is called a unimolecular step,
two or three molecules are called bimolecular and
termolecular steps respectively
A unimolecular step is always first order,bimolecular, second
order Termolecular steps are uncommon
Elementary step
Molecularity
Rate law
A → Product
A + A → Products
A + B → Products
Unimolecular
Bimolecular
Bimolecular
Rate = [A]
Rate = [A]2
Rate = [A][B]
a reaction mechanism must satisfy two 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.
UNIT 3 Chapter 6: Rates of Reaction
Section 6.3
The Rate-determining Step
the slowest of the elementary steps in a complex
reaction is called the rate-determining
step.
This reaction occurs in three elementary steps:
Step 2 is the rate-determining step:
• it is the slowest elementary step
• the overall rate of the reaction is dependent on this step
• the E for this step is higher than Ea for each of the other steps
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Potential Energy graph of Reaction
Mechanism
B
A
C
D
E
UNIT 3 Chapter 6: Rates of Reaction
Section 6.3
A Proposed Reaction Mechanism
• Experiments show that this reaction is zero order with
respect to OH– (i.e. its rate does not depend on [OH–])
• This can be explained by a two-step mechanism
Step 2 is very fast and
depends on completion of
Step 1, not on the
concentration of OH–.
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Potential Energy graph of Reaction Mechanism
Example:
The balanced equation for a reaction of the gases nitrogen dioxide and fluorine is
2NO2(g) + F2(g) → 2NO2F(g)
Experimentally determined rate law is
Rate = k[NO2][F2]
The suggested mechanism for this reaction is:
NO2 + F2 → NO2F + F slow
F + NO2 → NO2F fast
Is this an acceptable mechanism?
Does it satisfy the requirements for reaction mechanism?
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.
3. The reactant (s) of the slowest reaction gives the rate law equation.
Solution:
First:The sum of the steps should give the balanced equation
NO2 + F2 → NO2F + F
F + NO2 →
NO2F
___________________
NO2 + F2 + F + NO2 →NO2F + F + NO2F
Overall reaction : 2NO2 + F2 → 2NO2F
Second: Mechanism must agree with the experimentally
determine rate. Since the first step is the slowest (the rate
determining step), it must determine the reaction rate
Rate = k[NO2][F2]