1. UNIT 3: Energy Changes and Rates of Reaction
Chapter 6: Rates of Reaction

2. 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.
Image source: MHR, Chemistry 12 © ISBN ; page 276
The light produced by a firefly depends on the speed of a particular chemical reaction that occurs in its abdomen.
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3. 6.1 Chemical Reaction Rates
UNIT 3
Chapter 6: Rates of Reaction
Section 6.1
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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4. Determining Reaction Rates
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.
Image source: MHR, Chemistry 12 © ISBN ; page 354, Table 6.1
The change in concentration of reactant A was monitored over time.
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5. Determining Reaction Rates
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.
Image source: MHR, Chemistry 12 © ISBN ; page 355
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6. Average and Instantaneous Reaction Rates
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)
Image source: MHR, Chemistry 12 © ISBN ; page 356
Average rate of reaction and instantaneous rate of reaction can be determined from a graph of concentration vs. time.
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7. Expressing Reaction Rates in Terms of Reactants or Products
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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8. Answer on the next slide
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.
Answer on the next slide
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9. The mole ratio of ammonia to hydrazine is 4:3
UNIT 3
Chapter 6: Rates of Reaction
Section 6.1
The mole ratio of ammonia to hydrazine is 4:3
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10. Methods for Measuring Rates of Reaction
UNIT 3
Chapter 6: Rates of Reaction
Section 6.1
Methods for Measuring Rates of Reaction
Image source: MHR, Chemistry 12 © ISBN ; page 362, Table 6.2
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11. Calculating Reaction Rates from Experimental Data
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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12. Calculating Reaction Rates from Experimental Data
UNIT 3
Chapter 6: Rates of Reaction
Section 6.1
Calculating Reaction Rates from Experimental Data
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13. Calculating Reaction Rates from Experimental Data
UNIT 3
Chapter 6: Rates of Reaction
Section 6.1
Calculating Reaction Rates from Experimental Data
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14. Section 6.1 Review UNIT 3 Chapter 6: Rates of Reaction Section 6.1
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15. 6.2 & 6.3: Collision Theory and Factors Affecting Rates of Reaction
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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16. Effective Collision Criteria 1: The Correct Orientation of Reactants
UNIT 3
Chapter 6: Rates of Reaction
Section 6.2
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).
Image source: MHR, Chemistry 12 © ISBN ; page 365
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.
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17. Effective Collision Criteria 2: Sufficient Activation Energy
UNIT 3
Chapter 6: Rates of Reaction
Section 6.2
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.
Image source: MHR, Chemistry 12 © ISBN ; page 366
The shaded part of the Maxwell-Boltzmann distribution curve represents the fraction of particles that have enough collision energy for a reaction (ie the energy is ≥ Ea).
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18. Representing the Progress of a Chemical Reaction
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)
Image source: MHR, Chemistry 12 © ISBN ; page 366
Exothermic
Endothermic
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19. Activation Energy and Enthalpy
UNIT 3
Chapter 6: Rates of Reaction
Section 6.2
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.
Image source: MHR, Chemistry 12 © ISBN ; page 367
Potential energy diagram for the combustion of octane.
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20. Activation Energy for Reversible Reactions
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
Image source: MHR, Chemistry 12 © ISBN ; page 368
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21. Analyzing Reactions Using Potential Energy Diagrams
UNIT 3
Chapter 6: Rates of Reaction
Section 6.2
Analyzing Reactions Using Potential Energy Diagrams
Image source: MHR, Chemistry 12 © ISBN ; page 369
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.
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22. Answer on the next slide
UNIT 3
Chapter 6: Rates of Reaction
Learning Check
Describe the relative values of Ea(fwd) and Ea(rev) for an exothermic reaction
Answer on the next slide
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23. Learning Check UNIT 3 Ea(rev) is greater than Ea(fwd)
Chapter 6: Rates of Reaction
Section 6.2
Learning Check
Ea(rev) is greater than Ea(fwd)
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24. Factors Affecting Reaction Rate
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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25. Factors Affecting Reaction Rate
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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26. A Catalyst Influences the Reaction Rate
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).
Image source: MHR, Chemistry 12 © ISBN ; page 374
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27. Catalysts in Industry 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.
Image source: MHR, Chemistry 12 © ISBN ; page 375
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28. Catalysts in Industry 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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29. Section 6.2 Review UNIT 3 Chapter 6: Rates of Reaction Section 6.2
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30. 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
Reaction rates will decrease as time progresses. This is because of the presence of more and more product. Once product is present, the reverse reaction can occur. Therefore, the rate of a reaction at any point after time zero is going to be affected by the rate of the reverse reaction.
Image source: MHR, Chemistry 12 © ISBN ; page 380
Initial rate is found by determining the slope of a line tangent to the curve at time zero.
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31. The Rate Law rate = k[A]m[B]n UNIT 3
Chapter 6: Rates of Reaction
Section 6.3
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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32. Graphing Reaction Rate in Terms of Concentration
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
Image source: MHR, Chemistry 12 © ISBN ; page 381
Initial rates are determined (A) and these are plotted against concentration (B).
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33. First-order Reactions
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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34. 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)

35. 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

36. 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)

37. Second-order Reactions
UNIT 3
Chapter 6: Rates of Reaction
Section 6.3
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
Image source: MHR, Chemistry 12 © ISBN ; page 283
rate = k[A]2
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38. 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

39. Solution:
The rate equation is 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

40. 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

41. 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 Initial rate of formation
of products
[A] [B]
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 mol/L.s k(0.1mol/L)m
2.0 = (2.0)m

42. Meaning, if [A] is doubled the rate doubles (#1 & 2)
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 mol/L.s K(0.10 mol/L)n n
= 2
When [B] is doubled the rate increases by factor of 4 which equals 22

43. [A] [B] Rates 1) 2) 3) 4) 5)
rate 5 = 5.4 mol/L.s = k(0.30 mol/L)n
rate mol/L.s k(0.10 mol/L)n
= 3

44. 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 K=Rate [A][B] K =0.20 mol L-1s-1 (0.10 mol L-1)(0.10 mol L-1) K=0.2 mol L-1s mol2 L-2 K=2.0 x 101 L mol-1 s-1 Let’s check other sample on P
[A] [B] Rates

45. Initial Concentrations Initial Rate of Disappearance of NO [NO] [H2]
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)
[NO] [H2]
1) X 10-3
2) X 10-3
3) X 10-3
What is the rate law for this reaction?
What is the unit of the constant?
Solution:
1. Find the order of the reaction
2. Solve for K value and unit
3. Then write the rate law
Equation

46. 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 x 10-3mol/L 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 x 10-3mol/L mol/L
2.0 = [2]n, ie [2]1 , Thus m = 1
Therefore, r = k[NO]2[H2]1

47. Using trial # 1, find k
mol/L.s = k x mol/L x (0.100 mol/L)2
k = mol/L.s = 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

48. 6.6: Reaction Mechanisms 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:
The oxygen atom formed in the first elementary step is called an intermediate. An intermediate is a chemical species that appears in an elementary step, but not in the overall chemical equation.
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49. 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

50. A unimolecular step is always first order,bimolecular, second order Termolecular steps are uncommon
Elementary step Molecularity Rate law
A → Product Unimolecular Rate = [A]
A + A → Products Bimolecular Rate = [A]2
A + B → Products Bimolecular Rate = [A][B]
a reaction mechanism must satisfy two requirements:
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.

51. The Rate-determining Step
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 Ea for this step is higher than Ea for each of the other steps
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52. Potential Energy graph of Reaction MechanismB A C D E

53. A Proposed Reaction Mechanism
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
Image source: MHR, Chemistry 12 © ISBN ; page 386
Step 2 is very fast and depends on completion of Step 1, not on the concentration of OH–.
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54. Potential Energy graph of Reaction Mechanism

55. 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.

56. Solution: First:The sum of the steps should give the balanced equation
NO F2 → NO2F F
F + NO2 → NO2F ___________________
NO 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]