AP Chemistry
Mr. Markic
Page 1 of 16
Chapter 13 - Chemical Kinetics
Chemical Kinetics
Thermodynamics – does a reaction take place?
Kinetics – how fast does a reaction proceed?
Reaction rate is the change in the concentration of a reactant or a product with time (M/s).
A  B
Rate = Rate =
𝛥 [𝐴]
 [A] = change in concentration of A over time period t
𝛥𝑡
𝛥 [𝐵]
 [B] = change in concentration of B over time period t
𝛥𝑡
Because [A] decreases with time,  [A] is negative.
A  B
Rate = Rate =
𝛥 [𝐴]
𝛥𝑡
𝛥 [𝐵]
𝛥𝑡
Br2 (aq) + HCOOH (aq)

2Br- (aq) + 2H+ (aq) + CO2 (g)
393 nm
Detector
light
Δ[Br2] α Δ Absorption
instantaneous rate = rate for
specific instance in time
Average rate = = -
𝛥 [𝐵𝑟2 ]
𝛥𝑡
=-
[𝐵𝑟2 ]𝑓𝑖𝑛𝑎𝑙 − [𝐵𝑟2 ]𝑖𝑛𝑖𝑡𝑎𝑙
𝑡 𝑓𝑖𝑛𝑎𝑙 − 𝑡𝑖𝑛𝑖𝑡𝑖𝑎𝑙
AP Chemistry
Mr. Markic
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rate α [Br2]
rate = k [Br2]
k=
𝑟𝑎𝑡𝑒
𝐵𝑟2
= 3.50 x 10-3 s-1
= rate constant
Reaction Rates and Stoichiometry
2A 
B
Two moles of A disappear for each mole of B that is formed.
1 𝛥 [𝐴]
Rate = - 2
𝛥𝑡
aA + bB
1 𝛥 [𝐴]
Rate = - 𝑎
𝛥𝑡
𝛥 [𝐵]
Rate =

cC + dD
1 𝛥 [𝐵]
=-𝑏
𝛥𝑡
𝛥𝑡
=
1 𝛥 [𝐶]
𝑐
𝛥𝑡
=
1 𝛥 [𝐷]
𝑑
𝛥𝑡
Sample Exericse
Write the rate expressions for the following reactions in terms of the disappearance of the reactants and the
appearance of the products:
(a) I-(aq) + OCl-(aq)  Cl-(aq) + OI-(aq)
(b) 4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g)
Write the rate expression for the following reaction: CH4 (g) + 2O2 (g)  CO2 (g) + 2H2O (g)
AP Chemistry
Mr. Markic
Page 3 of 16
Consider the reaction
4NO2(g) + O2(g)  2N2O5(g)
Suppose that, at a particular moment during the reaction, molecular oxygen is reacting at the rate of 0.024 M/s.
(a) At what rate is N2O5 being formed?
(b) At what rate is NO2 reacting?
Consider the reaction
4PH3(g)  P4(g) + 6H2(g)
Suppose that, at a particular moment during the reaction, molecular hydrogen is being formed at the rate of
0.078 M/s.
(a) At what rate is P4 being formed?
(b) At what rate is PH3 reacting?
Review of Concepts
Write a balanced equation for a gas-phase reaction whose rate is given by
Rate = -
1 ∆ [𝑁𝑂𝐶𝑙]
2
∆𝑡
=
1 ∆ [𝑁𝑂]
2
∆𝑡
=
∆ [𝐶𝑙2 ]
∆𝑡
The Rate Law
The rate law expresses the relationship of the rate of a reaction to the rate constant and the concentrations of the
reactants raised to some powers.
aA + bB

Rate = k [A]x[B]y
cC + dD
reaction is xth order in A
reaction is yth order in B
reaction is (x +y)th order overall
F2 (g) + 2ClO2 (g) 
2FClO2 (g)
Double [F2] with [ClO2] constant
Rate doubles
x=1
rate = k [F2][ClO2]
Quadruple [ClO2] with [F2] constant
rate = k [F2]x[ClO2]y
Rate quadruples
y=1
AP Chemistry
Mr. Markic
Page 4 of 16
Rate Laws
Rate laws are always determined experimentally.
Reaction order is always defined in terms of reactant (not product) concentrations.
The order of a reactant is not related to the stoichiometric coefficient of the reactant in the
balanced chemical equation.
F2 (g) + 2ClO2 (g) 
2FClO2 (g)
rate = k [F2][ClO2]1
Sample Exercise
• The reaction of nitric oxide with hydrogen at 1280°C is
2NO(g) + 2H2(g)  N2(g) + 2H2O(g)
Experiment
[NO] (M)
[H2] (M)
Initial Rate (M/s)
1
5.0 x 10-3
2.0 x 10-3
1.3 x 10-5
2
10.0 x 10-3
2.0 x 10-3
5.0 x 10-5
3
10.0 x 10-3
4.0 x 10-3
10.0 x 10-5
From the following data collected at this temperature, determine
(a) The rate law
(b) The rate constant
(c) The rate reaction when [NO] = 12.0 x 10-3 M and [H2] = 6.0 x 10-3 M
AP Chemistry
Mr. Markic
Page 5 of 16
Determine the rate law and calculate the rate constant for the following reaction from the following data:
S2O82- (aq) + 3I- (aq) 
2SO42- (aq) + I3- (aq)
Experiment [S2O82-]
[I-]
Initial Rate
(M/s)
1
0.08
0.034
2.2 x 10-4
2
0.08
0.017
1.1 x 10-4
3
0.16
0.017
2.2 x 10-4
Review of Concepts
The relative rates of the reaction 2 A + B  products shown in the diagrams (a) – (c) are 1:2:4. The red
spheres represent A molecules and the green spheres represent B molecules. Write a rate law for this reaction.
First-Order Reactions
A 
k=
𝑟𝑎𝑡𝑒
[𝐴]
product
=
𝑀/𝑠
𝑀
= 1/s or s-1
rate = -
𝛥 [𝐴]
𝛥𝑡
𝛥 [𝐴]
𝛥𝑡
= k [A]
rate = k [A]
[A]t is the concentration of A at any time t
[A]0 is the concentration of A at time t=0
[A] = [A]0e (-kt)
ln[A] = ln[A]0 - kt
AP Chemistry
Mr. Markic
Page 6 of 16
Sample Exercise
The conversion of cyclopropane to propene in the gas phase is a first-order reaction with a rate
constant of 6.7 x 10-4 s-1 at 500°C.
(a) If the initial concentration of cyclopropane was 0.25M, what is the concentration after 8.8 min?
(b) How long (in minutes) will it take for the concentration of cyclopropene to decrease from 0.25M to 0.15M?
(c) How long (in minutes) will it take to convert 74 percent of the starting material?
Practice Exercise
The reaction 2A  B is first order in A with a rate constant of 2.8 x 10-2 s-1 at 800C. How long will it take for
A to decrease from 0.88 M to 0.14 M ?
Graphical Determination of k
2N2O5  4NO2 (g) + O2 (g)
AP Chemistry
Mr. Markic
Page 7 of 16
First-Order Reactions
The half-life, t½, is the time required for the concentration of a reactant to decrease to half of its initial
concentration.
t½ = t when [A] = [A]0/2
[𝐴]0
[𝐴]0 /2
ln
t½ =
𝑘
=
ln 2
𝑘
=
0.693
𝑘
First-order reaction
A 
product
# of
half-lives
[A] = [A]0/n
1
2
2
4
3
8
4
16
Sample Exercise
• The decomposition of ethane (C2H6) to methyl radicals is a first-order reaction with a rate constant of 5.36 x
10-4 s-1 at 700C:
C2H6(g)  2CH3(g)
Calculate the half-life of the reaction in minutes.
What is the half-life of N2O5 if it decomposes with a rate constant of 5.7 x 10-4 s-1?
How do you know decomposition is first order?
AP Chemistry
Mr. Markic
Page 8 of 16
Review of Concepts
Consider the first-order reaction A  B in which A, molecules (blue-spheres are converted to B molecules
(orange spheres).
(a) What are the half-life and rate constant for the reaction?
(b) How many molecules of A and B are present at t = 20 s and t = 30 s?
Second-Order Reactions
A 
k=
product
𝑟𝑎𝑡𝑒
=
[𝐴]2
𝑀/𝑠
𝑀2
rate = 1
= 𝑀𝑠
-
𝛥 [𝐴]
𝛥𝑡
𝛥 [𝐴]
𝛥𝑡
= k [A]2
rate = k [A]2
[A]t is the concentration of A at any time t
[A]0 is the concentration of A at time t=0
1
1
[A]t
= [𝐴} + kt
0
t½ = t when [A] = [A]0/2
1
t½ = 𝑘[𝐴]
0
Sample Exercise
• Iodine atoms combine to form molecular iodine in the gas phase
I(g) + I(g)  I2(g)
This reaction follows second-order kinetics and has the high rate constant 7.0 x 109 /M·s at 23°C.
(a) If the initial concentration of I was 0.086 M, calculate the concentration after 2.0 min.
(b) Calculate the half-life of the reaction if the initial concentration of I is 0.60 M and if it is 0.42 M.
AP Chemistry
Mr. Markic
Page 9 of 16
The reaction 2A  B is second order with a rate constant of 51 / M· min at 24°C.
(a) Starting with [A]0 = 0.0092 M, how long will it take for [A]t = 3.7 x 10-3 M?
(b) Calculate the half-life of the reaction.
Review of Concepts
Consider the reaction A
 products. The half-life of the reaction depends on the initial concentration of A.
Which of the following statements is inconsistent with the given information?
(a) The half-life of the reaction decreases as the initial concentration increases.
(b) A plot of ln [A]t versus t yields a straight line.
(c) Doubling the concentration of A quadruples the rate.
Zero-Order Reactions
A 
k=
𝑟𝑎𝑡𝑒
[𝐴]0
product
=
𝑀
𝑠
rate = -
𝛥 [𝐴]
𝛥𝑡
=k
𝛥 [𝐴]
𝛥𝑡
rate = k [A]0 = k
[A]t is the concentration of A at any time t
[A]0 is the concentration of A at time t=0
[A] = [A]0 – kt
t½ = t when [A] = [A]0/2
t½ =
[𝐴]0
2𝑘
AP Chemistry
Mr. Markic
Page 10 of 16
Summary of the Kinetics of Zero-Order, First-Order and Second-Order Reactions
Order
Rate Law
Concentration-Time Equation
Half-Life
0
rate = k
[A] = [A]0 – kt
t½ =
1
rate = k [A]
ln[A] = ln[A]0 – kt
t½ =
2
rate = k [A]2
•
1)
2)
3)
1
[A]t
1
= [𝐴} + kt
0
[𝐴]0
2𝑘
ln 2
𝑘
1
t½ = 𝑘[𝐴]
0
These equations can be used to determine:
The concentration of the reactant remaining at any time after the reaction has started
The time required for a given fraction of a sample to react
The time required for a reactant concentration to fall to a certain level
Using Initial Rates to Determine Rate Laws
The exponents in the rate law are 0, 1, or 2 (reaction orders)
If a reaction is zero order in a particular reactant, changing its concentration will have no effect in rate because
any concentration raised to the zero power is 1
When a reaction is 1st order in a reactant, changes in the concentration of that reactant will produce proportional
changes in the rate, therefore doubling the concentration will double the rate
When the rate law is 2nd order in a particular reactant, doubling its concentration increases the rate by a factor 22
= 4, tripling its concentration causes the rate to increase by a factor of 32 = 9
The Change of Concentration with Time
A rate law tells is how the rate of a reaction changes at a particular temperature as reactant concentrations are
changed
A+B

ABǂ

C+D
Exothermic Reaction
Endothermic Reaction
The activation energy (Ea ) is the minimum amount of energy required to initiate a chemical reaction.
AP Chemistry
Mr. Markic
Page 11 of 16
Activation Energy and Temperature Dependence of Rate Constants
• The rates of most chemical reactions increase as the temperature rises
• The faster the rate at higher temperature is due to an increase in the rate constant with increasing
temperature
Importance of Molecular Orientation
effective collision
ineffective collision
Reaction Mechanism
The overall progress of a chemical reaction can be represented at the molecular level by a series of simple
elementary steps or elementary reactions.
The sequence of elementary steps that leads to product formation is the reaction mechanism.
2NO (g) + O2 (g)  2NO2 (g)
N2O2 is detected during the reaction!
Elementary step: NO + NO  N2O2
+ Elementary step: N2O2 + O2  2NO2
Overall reaction: 2NO + O2  2NO2
2NO (g) + O2 (g)  2NO2 (g)
Mechanism:
AP Chemistry
Mr. Markic
Page 12 of 16
Intermediates are species that appear in a reaction mechanism but not in the overall balanced equation.
An intermediate is always formed in an early elementary step and consumed in a later elementary step.
Elementary step: NO + NO  N2O2
+ Elementary step: N2O2 + O2  2NO2
Overall reaction: 2NO + O2  2NO2
The molecularity of a reaction is the number of molecules reacting in an elementary step.
•
•
•
Unimolecular reaction – elementary step with 1 molecule
Bimolecular reaction – elementary step with 2 molecules
Termolecular reaction – elementary step with 3 molecules
Rate Laws and Elementary Steps
Unimolecular reaction
Bimolecular reaction
Bimolecular reaction
A  products
A + B  products
A + A  products
rate = k [A]
rate = k [A][B]
rate = k [A]2
Writing plausible reaction mechanisms:
• The sum of the elementary steps must give the overall balanced equation for the reaction.
• The rate-determining step should predict the same rate law that is determined experimentally
The rate-determining step is the slowest step in the sequence of steps leading to product
formation.
Sequence of Steps in Studying a Reaction Mechanism
Measuring the rate of a reaction  formulating the rate law  postulating a reasonable reaction mechanism
Figure 13.22
Potential energy profile for a two-step reaction in which the first step is rate
determining. R and P represent reactants and products, respectively
Activation Energy
• If molecules are moving too slow, with little K.E.. They merely bounce off one another without changing
• Colliding molecules must have a total K.E. equal or greater than some minimal value
• The minimum energy required to initiate a chemical reaction is called the activation energy (Ea)
• The value of Ea varies from reaction to reaction
• The energy difference between the starting molecule and the highest energy along the reaction pathway is
the activation energy, Ea
AP Chemistry
Mr. Markic
•
•
•
Page 13 of 16
The particular arrangement of atoms at the top of the barrier is called the activated complex, or transition
state
The lower Ea is, the faster the reaction
The activation barrier for the reverse reaction is equal to the sum of E and Ea for the forward reaction
Sample Exercise
The gas phase decomposition of nitrous oxide (N2O) is believed to occur via two elementary steps:
𝑘1
Step 1: N2O → N2 + O
𝑘2
Step 2: N2O + O → N2 + O2
Experimentally the rate law is found to be rate = k [N2O]
(a) Write the equation for the overall reaction
(b) Identify the intermediates
(c) What can you say about the relative rates of steps 1 and 2?
Practice Exercise
The experimental rate law for the reaction between NO2 and CO to produce NO and CO2 is rate = k[NO2]2. The
reaction is believed to occur via two steps:
Step 1:
NO2 + NO2  NO + NO3
Step 2:
NO3 + CO  NO2 + CO2
What is the equation for the overall reaction?
What is the intermediate?
What can you say about the relative rates of steps 1 and 2?
AP Chemistry
Mr. Markic
Page 14 of 16
Review of Concepts
The rate law for the reaction H2 + 2IBr  I2 + 2HBr is rate = k [H2][IBr]. Given that HI is an intermediate,
write a two-step mechanism for the reaction.
Catalysts
• A substance that changes the rate of reaction, without itself being changed
• Lowers the activation energy
• The sites on the catalyst at which a reaction occurs are called the active sites
• The specific reactant molecules involved in an enzymatic reaction are called substances
• The Lock-and-Key Model for enzyme catalysts, substrate molecules bind very specifically to the active site
of the enzyme
k = A • e ( -Ea/RT )
Ea ↓
k↑
Uncatalyzed
uncatalyzed
catalyzed
catalyzed
ratecatalyzed > rateuncatalyzed
E’a < Ea
Temperature Dependence of the Rate Constant
𝐸𝑎
k = A 𝑒 (−𝑅𝑇)
(Arrhenius equation)
Ea is the activation energy (J/mol)
R is the gas constant (8.314 J/K•mol)
T is the absolute temperature
A is the frequency factor
Alternate format: ln k = -
𝐸𝑎 1
𝑅 𝑇
+ ln A
AP Chemistry
Mr. Markic
Enzyme Catalysis
Review of Concepts
Which of the following is false regarding catalysis:
(a) Ea is lower for a catalyzed reaction.
0
(b) 𝐻𝑟𝑥𝑚
is lower for a catalyzed reaction.
(c) A catalyzed reaction has a different mechanism.
Page 15 of 16
AP Chemistry
Mr. Markic
Page 16 of 16
Big Idea 4: Rates of chemical reactions are determined by details of the molecular collisions.
Duration: Late February
Textbook Chapter: 13
Enduring Understanding
Essential Knowledge
4.A: Reaction rates that depend on
4.A.1: The rate of a reaction is influenced by the concentration or pressure of
temperature and other environmental factors reactants, the phase of the reactants and products, and environmental factors such
are determined by measuring changes in
as temperature and solvent.
concentrations of reactants or products over
4.A.2: The rate law shows how the rate depends on reactant concentrations.
time.
4.A.3: The magnitude and temperature dependence of the rate of reaction is
contained quantitatively in the rate constant.
4.B: Elementary reactions are mediated by
collisions between molecules. Only
collisions having sufficient energy and
proper relative orientation of reactants lead
to products.
4.C: Many reactions proceed via a series of
elementary reactions
4.B.1: Elementary reactions can be unimolecular or involve collisions between
two or more molecules.
4.B.2: Not all collisions are successful. To get over the activation energy barrier,
the colliding species need sufficient energy. Also, the orientations of the reactant
molecules during the collision must allow for the rearrangement of reactant bonds
to form product bonds.
4.B.3: A successful collision can be viewed as following a reaction path with an
associated energy profile.
4.C.1: The mechanism of a multistep reaction consists of a series of elementary
reactions that add up to the overall reaction.
4.D: Reaction rates may be increased by the
presence of a catalyst.
4.C.2: In many reactions, the rate is set by the slowest elementary reaction, or
rate-limiting step.
4.C.3: Reaction intermediates, which are formed during the reaction but not
present in the overall reaction, play an important role in multistep reactions.
4.D.1: Catalysts function by lowering the activation energy of an elementary step
in a reaction mechanism, and by providing a new and faster reaction mechanism.
4.D.2: Important classes in catalysis include acid-base catalysis, surface catalysis,
and enzyme catalysis.
Learning Objective
4.1- The student is able to design and/or interpret the results of an experiment regarding the factors (i.e., temperature,
concentration, surface area) that may influence the rate of a reaction.
4.2 - The student is able to analyze concentration vs. time data to determine the rate law for a zeroth-, first-, or second-order
reaction.
4.3 - The student is able to connect the half-life of a reaction to the rate constant of a first-order reaction and justify the use of this
relation in terms of the reaction being a first-order reaction.
4.4 - The student is able to connect the rate law for an elementary reaction to the frequency and success of molecular collisions,
including connecting the frequency and success to the order and rate constant, respectively.
4.5 - The student is able to explain the difference between collisions that convert reactants to products and those that do not in
terms of energy distributions and molecular orientation.
4.6 - The student is able to use representations of the energy profile for an elementary reaction (from the reactants, through the
transition state, to the products) to make qualitative predictions regarding the relative temperature dependence of the reaction
rate.
4.7 - The student is able to evaluate alternative explanations, as expressed by reaction mechanisms, to determine which are
consistent with data regarding the overall rate of a reaction, and data that can be used to infer the presence of a reaction
intermediate.
4.8 - The student can translate among reaction energy profile representations, particulate representations, and symbolic
representations (chemical equations) of a chemical reaction occurring in the presence and absence of a catalyst.
4.9 - The student is able to explain changes in reaction rates arising from the use of acid-base catalysts, surface catalysts, or
enzyme catalysts, including selecting appropriate mechanisms with or without the catalyst present.