Jeffrey Mack California State University, Sacramento Chapter 15
Chemical Kinetics: The Rates of Chemical Reactions Slide 2 Chemical
Kinetics will now provide information about the arrow! HOW This
gives us information on HOW a reaction occurs! Reactants Products
Chemical Kinetics: The Rates of Chemical Reactions Thermodynamics
does a reaction take place? Kinetics how fast does a reaction
proceed? Slide 3 Kinetics is the study of how fast (Rates) chemical
reactions occur. Important factors that affect the rates of
chemical reactions: reactant concentration or surface area in
solids temperature action of catalysts Our goal: Our goal: Use
kinetics to understand chemical reactions at the particle or
molecular level. Chemical Kinetics Slide 4 Reactants go away with
time. Products appear with time. The rate of a reaction can be
measured by either. In this example by the loss of color with time.
Rate of Reactions Slide 5 Blue dye is oxidized with bleach. Its
concentration decreases with time. The rate the change in dye conc.
with time can be determined from the plot. Dye Conc Time
Determining a Reaction Rate A B Slide 6 aA + bB cC + dD In general
for the reaction: reactants go away with time therefore the
negative sign Reaction Rate & Stoichiometry Reaction rate is
the change in the concentration of a reactant or a product with
time (M/s). Slide 7 Determining a Reaction Rate Slide 8
concentrationtime The rate of appearance or disappearance is
measured in units of concentration vs. time. Rate = time There are
three types of rates 1.initial rate 2.average rate 3.instantaneous
rate = M s 1 or M min 1 etc... Determining a Reaction Rate Slide 9
Reactant concentration (M) The concentration of a reactant
decreases with time. Time Reaction Rates Slide 10 Reactant
concentration (M) Time Initial rate Reaction Rates Slide 11
Reactant concentration (M) Time Reaction Rates Slide 12 Reactant
concentration (M) Time Instantaneous rate (tangent line) Reaction
Rates Slide 13 Reactant concentration (M) Time Instantaneous rate
(tangent line) Initial rate Reaction Rates During the beginning
stages of the reaction, the initial rate is very close to the
instantaneous rate of reaction. Slide 14 Problem: Consider the
reaction: Over a period of 50.0 to 100.0 s, the concentration of
NO(g) drops from 0.0250M to 0.0100M. a) What is the average rate of
disappearance of NO(g) during this time? Slide 15 a) Problem:
Consider the reaction: Over a period of 50.0 to 100.0 s, the
concentration of NO(g) drops from 0.0250M to 0.0100M. a)What is the
rate of rxn? b)What is the average rate of disappearance of NO(g)
during this time? = 1.50 10 4 Ms 1 (0.0100M 0.0250M) 100.0 s 50.0 s
RATE= - [NO] tt 1 2 b) [NO]/ tt = -0.0150/50.0 M/s = -3.00 X 10 -4
M/s Slide 16 Practice example Write the rate expression for the
following reaction: CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O (g) rate
= - [CH 4 ] tt = - [O 2 ] tt 1 2 = [H 2 O] tt 1 2 = [CO 2 ] tt
Slide 17 There are several important factors that will directly
affect the rate of a reaction: Temperature The physical state of
the reactants Addition of a catalyst All of the above can have a
dramatic impact on the rate of a chemical process. Reaction
Conditions & Rate Slide 18 Bleach at 54 CBleach at 22 C
Reaction Conditions & Rate: Temperature Slide 19 Reaction
Conditions & Rate: Physical State of Reactants Slide 20
Catalyzed decomposition of H 2 O 2 2 H 2 O 2 2 H 2 O + O 2 Reaction
Conditions & Rate: Catalysts Slide 21 0.3 M HCl6 M HCl Mg(s) +
2 HCl(aq) MgCl 2 (aq) + H 2 (g) Effect of Concentration on Reaction
Rate: Concentration Slide 22 concentration The rate of reaction
must be a function of concentration: collisionsAs concentration
increases, so do the number of collisions probabilityAs the number
of collisions increase, so does the probability of a reaction.
rateThis in turn increases the rate of conversion of reactants to
products. Rate Law ExpressionThe relationship between reaction rate
and concentration is given by the reaction Rate Law Expression. The
reaction rate law expression relates the rate of a reaction to the
concentrations of the reactants. The Rate Law Expression Slide 23
aA + bB cC + dD x and y are the reactant orders determined from
experiment. x and y are NOT the stoichiometric coefficients. Each
concentration is expressed with an order (exponent). The rate
constant converts the concentration expression into the correct
units of rate (Ms 1 ). For the general reaction: Reaction Order
Slide 24 A reaction order can be zero, or positive integer and
fractional number. OrderNameRate Law 0zerothrate = k[A] 0 = k
1firstrate = k[A] 2secondrate = k[A] 2 0.5one-halfrate = k[A] 1/2
1.5three-halfrate = k[A] 3/2 0.667two-thirds rate = k[A] 2/3
Reaction Orders Slide 25 Overall order: or seven halves order note:
when the order of a reaction is 1 (first order) no exponent is
written. 1 + + 2 = 3.5 =7/2 Reaction Order Slide 26 To find the
units of the rate constant, divide the rate units by the Molarity
raised to the power of the overall reaction order. k = if (x+y) =
1k has units of s -1 if (x+y) = 2k has units of M -1 s -1 Reaction
Constant Slide 27 The rate constant, k, is a proportionality
constant that relates rate and concentration. It is found through
experiment that the rate constant is a function temperature. Rate
constants must therefore be reported at the temperature with which
they are measured. The rate constant also contains information
about the energetics and collision efficiency of the reaction.
Reaction Constant Slide 28 EXAMPLE: The reaction, 2 NO (g) + 2 H 2
(g) N 2 (g) + 2 H 2 O (g) first orderthird order is experimentally
found to be first order in H 2 and third order in NO a) Write the
rate law. Slide 29 EXAMPLE: The reaction, 2 NO (g) + 2 H 2 (g) N 2
(g) + 2 H 2 O (g) first orderthird order is experimentally found to
be first order in H 2 and third order in NO a) Write the rate law.
Rate(Ms -1 ) =k[H 2 ][NO] 3 b) What is the overall order of the
reaction? Slide 30 EXAMPLE: The reaction, 2 NO (g) + 2 H 2 (g) N 2
(g) + 2 H 2 O (g) first orderthird order is experimentally found to
be first order in H 2 and third order in NO a) Write the rate law.
Rate(Ms -1 ) =k[H 2 ][NO] 3 b) What is the overall order of the
reaction? = 4Overall order =1+ 3 4 th order c) What are the units
of the rate constant? Slide 31 EXAMPLE: The reaction, 2 NO (g) + 2
H 2 (g) N 2 (g) + 2 H 2 O (g) first orderthird order is
experimentally found to be first order in H 2 and third order in NO
a) Write the rate law. Rate(Ms -1 ) =k[H 2 ][NO] 3 b) What is the
overall order of the reaction? = 4Overall order =1+ 3 4 th order c)
What are the units of the rate constant? Slide 32 doublesdoubles If
the rate doubles when [A] doubles and [B] stays constant, the order
for [A] is? Rate Law Expression Slide 33 doublesdoubles If the rate
doubles when [A] doubles and [B] stays constant, the order for [A]
is? one 1 Rate Law Expression Slide 34 doublesdoubles If the rate
doubles when [A] doubles and [B] stays constant, the order for [A]
is? one 1 Rate Law Expression Slide 35 Instantaneous rate (tangent
line) Reactant concentration (M) During the beginning stages of the
reaction, the initial rate is very close to the instantaneous rate
of reaction. Initial rate Determining a Rate Equation Slide 36 The
reaction of nitric oxide with hydrogen at 1280 C is as follows: 2NO
(g) + 2H 2 (g) N 2 (g) + 2H 2 O (g) From the following experimental
data, determine the rate law and rate constant. Determining
Reaction Order: The Method of Initial Rates Slide 37 The reaction
of nitric oxide with hydrogen at 1280 C is as follows: Notice that
in Trial 1 and 3, the initial concentration of NO is held constant
while H 2 is changed. Determining Reaction Order: The Method of
Initial Rates 2NO (g) + 2H 2 (g) N 2 (g) + 2H 2 O (g) Slide 38
Determining Reaction Order: The Method of Initial Rates The
reaction of nitric oxide with hydrogen at 1280 C is as follows: 2NO
(g) + 2H 2 (g) N 2 (g) + 2H 2 O (g) This means that any changes to
the rate must be due to the changes in H 2 which is related to the
concentration of H 2 & its order! Slide 39 Rate(M/min) = k [NO]
x [H 2 ] y The rate law for the reaction is given by: 2NO(g) + 2H 2
(g) N 2 (g) + 2H 2 O(g) Taking the ratio of the rates of Trials 3
and 1 one finds: Plugging in the values from the data: Rate (Trial
3) Rate (Trial 1) = Determining Reaction Order: The Method of
Initial Rates Slide 40 2.00 log Take the log of both sides of the
equation: log(2.00) y = 1 Rate(M/min) = k [NO] x [H 2 ] Determining
Reaction Order: The Method of Initial Rates Slide 41 x = 3
Similarly for x: Rate(M/min) = k [NO] x [H 2 ] y k k Determining
Reaction Order: The Method of Initial Rates Slide 42 The Rate Law
expression is: The order for NO is 3 The order for H 2 is 1 The
over all order is 3 + 1 =4 2NO(g) + 2H 2 (g) N 2 (g) + 2H 2 O(g)
Determining Reaction Order: The Method of Initial Rates Slide 43
The Rate constant Rate(M/min) = k [NO] 3 [H 2 ] To find the rate
constant, choose one set of data and solve: Determining Reaction
Order: The Method of Initial Rates Slide 44 It is important know
how long a reaction must proceed to reach a predetermined
concentration of some reactant or product. We need a mathematical
equation that relates time and concentration: This equation would
yield concentration of reactants or products at a given time. It
will all yield the time required for a given amount of reactant to
react. ConcentrationTime Relationships: Integrated Rate Laws Slide
45 zero For a zero order process where A goes onto products, the
rate law can be written: A products = k k has units of Ms 1 For a
zero order process, the rate is the rate constant! Integrated Rate
Laws Slide 46 This is the average rate If one considers the
infinitesimal changes in concentration and time the rate law
equation becomes: This is the instantaneous rate Zero order
kinetics A products Integrated Rate Laws = k Slide 47 [A] t [A] o =
k (t 0)= k t and [A] = [A] at time t = t where [A] = [A] o at time
t = 0 [A] t [A] o = kt Zero order kinetics Integrated Rate Laws
Slide 48 [A] o [A] t = kt whats this look like? y = mx + b [A] t =
kt + [A] o rearranging a plot of [A] t vs t looks like [A] t
(mols/L) t (time) y-intercept the y-intercept is [A] o Conclusion:
If a plot of reactant concentration vs. time yields a straight
line, then the reactant order is ZERO! slope = k k has units of
M(time) 1 Zero order kinetics Integrated Rate Laws Slide 49 first
order For a first order process, the rate law can be written: A
products This is the average rate If one considers the
infinitesimal changes in concentration and time the rate law
equation becomes: This is the instantaneous rate Integrated Rate
Laws Slide 50 Taking the exponent to each side of the equation: or
Conclusion: Conclusion: The concentration of a reactant governed by
first order kinetics falls off from an initial concentration
exponentially with time. First order kinetics Integrated Rate Laws
Slide 51 Taking the natural log of both sides ln[A] t = rearranging
= ln[A] o kt kt + ln[A] o First order kinetics Integrated Rate Laws
Slide 52 y = mx + bso a plot of ln[A] t vs t looks like ln[A] t t
(time) y-intercept the y-intercept is ln[A] o Conclusion:
Conclusion: If a plot of natural log of reactant concentration vs.
time yields a straight line, then the reactant order is FIRST!
slope = k ln[A] t = kt + ln[A] o k has units of (time) 1 First
order kinetics Integrated Rate Laws Slide 53 Problem: The
decomposition of N 2 O 5 (g) following 1 st order kinetics. If 2.56
mg of N 2 O 5 is initially present in a container and 2.50 mg
remains after 4.26 min, what is the rate constant in s 1 ? Slide 54
Problem: The decomposition of N 2 O 5 (g) following 1 st order
kinetics. If 2.56 mg of N 2 O 5 is initially present in a container
and 2.50 mg remains after 4.26 min, what is the rate constant in s
1 ? Begin with the integrated rate law for a 1 st order process:
Wait what is the volume of the container??? Do we need to convert
to moles? Slide 55 Problem: The decomposition of N 2 O 5 (g)
following 1 st order kinetics. If 2.56 mg of N 2 O 5 is initially
present in a container and 2.50 mg remains after 4.26 min, what is
the rate constant in s 1 ? Check it out! You dont need the volume
of the container! Slide 56 Problem: The decomposition of N 2 O 5
(g) following 1 st order kinetics. If 2.56 mg of N 2 O 5 is
initially present in a container and 2.50 mg remains after 4.26
min, what is the rate constant in s 1 ? taking the natural log and
substituting time in seconds: k = 9.3 10 5 s 1 Slide 57 Rate = k[A]
2 Integrating as before we find: A Products k has units of M 1 s 1
Second order kinetics Integrated Rate Laws Slide 58 y = mx + bso a
plot of 1/[A] t vs t looks like 1/[A] t t (time) y-intercept the
y-intercept is 1/[A] o Conclusion: Conclusion: If a plot of one
over reactant concentration vs. time yields a straight line, then
the reactant order is second! slope = k Second order kinetics k has
units of M 1 s 1 Integrated Rate Laws Slide 59 Summary of
Integrated Rate Laws Slide 60 Half-life of a reaction is the time
taken for the concentration of a reactant to drop to one-half of
the original value. Half-life & First-Order Reactions Slide 61
For a first order process the half life (t ) is found
mathematically from: Start with the integrated rate law expression
for a 1 st order process Bring the concentration terms to one side.
Express the concentration terms as a fraction using the rules of
ln. at time = t Reaction Half-Life: 1 st Order Kinetics Slide 62
exchange [A] with [A] 0 to reverse the sign of the ln term and
cancel the negative sign in front of k Substitute the value of [A]
at the half-life Reaction Half-Life: 1 st Order Kinetics Slide 63
So, knowing the rate constant for a first order process, one can
find the half-life! The half-life is independent of the initial
concentration! Reaction Half-Life: 1 st Order Kinetics Slide 64
Problem: A certain reaction proceeds through first order kinetics.
The half-life of the reaction is 180. s. What percent of the
initial concentration remains after 900.s? Slide 65 Problem: A
certain reaction proceeds through first order kinetics. The
half-life of the reaction is 180. s. What percent of the initial
concentration remains after 900.s? Using the integrated rate law,
substituting in the value of k and 900.s we find: Slide 66 Problem:
A certain reaction proceeds through first order kinetics. The
half-life of the reaction is 180. s. What percent of the initial
concentration remains after 900.s? Using the integrated rate law,
substituting in the value of k and 900.s we find: k = 0.00385 s -1
Slide 67 Problem: A certain reaction proceeds through first order
kinetics. The half-life of the reaction is 180. s. What percent of
the initial concentration remains after 900.s? Using the integrated
rate law, substituting in the value of k and 900.s we find: k =
0.00385 s -1 = 0.0312 Slide 68 Problem: A certain reaction proceeds
through first order kinetics. The half-life of the reaction is 180.
s. What percent of the initial concentration remains after 900.s?
Using the integrated rate law, substituting in the value of k and
900.s we find: k = 0.00385 s -1 = 0.0312 Since the ratio of [A] t
to [A] 0 represents the fraction of [A] that remains, the % is
given by: 100 0.0312 = 3.12% Slide 69 Elements that decay via
radioactive processes do so according to 1 st order kinetics:
Element:Half-life: 238 U 234 Th + 14 C 14 N + 131 I 131 Xe + 4.5 10
9 years 5730 years 8.05 days Reaction Half-Life: 1 st Order
Kinetics Slide 70 Tritium decays to helium by beta ( ) decay: The
half-life of this process is 12.3 years Starting with 1.50 mg of 3
H, what quantity remains after 49.2 years. Solution: Begin with the
integrated rate law expression for 1 st order kinetics. Reaction
Half-Life: 1 st Order Kinetics Slide 71 Recall that that the rate
constant for a 1 st order process is given by: [ 3 H] t = 1.50 mg =
0.094 mg Reaction Half-Life: 1 st Order Kinetics Slide 72 Notice
that 49.2 years is 4 half-lives After 1 half life:= 0.75 mg remains
After 2 half life's:= 0.38 mg remains After 3 half life's:= 0.19 mg
remains After 4 half life's:= 0.094 mg remains Reaction Half-Life:
1 st Order Kinetics Slide 73 Arrhenius: Arrhenius: Molecules must
posses a minimum amount of energy to react. Why? (1) In order to
form products, bonds must be broken in the reactants. (2) Bond
breakage requires energy. (3) Molecules moving too slowly, with too
little kinetic energy, dont react when they collide. The Activation
energy, E a, is the minimum energy required to initiate a chemical
reaction. E a is specific to a particular reaction. A Microscopic
View of Reaction Rates Slide 74 When one writes a reaction all that
is seen are the reactants and products. This details the overall
reaction stoichiometry. mechanism How a reaction proceeds is given
by the reaction mechanism. A Microscopic View of Reaction Rates
Slide 75 The reaction of NO 2 and CO (to give NO and CO 2 ) has an
activation energy barrier of 132 kJ/mol-rxn. The reverse reaction
(NO + CO 2 NO2 + CO) requires 358 kJ/mol-rxn. The net energy change
for the reaction of NO 2 and CO is 226 kJ/mol-rxn. Activation
Energy Slide 76 reactants products Reaction Coordinate The progress
of a chemical reaction as the reactants transform to products can
be described graphically by a Reaction Coordinate. Potential Energy
Reaction Progress reactants products H RXN E act In order for the
reaction to proceed, the reactants must posses enough energy to
surmount a reaction barrier. Transition State Activation Energy
Slide 77 The temperature for a system of particles is described by
a distribution of energies. At higher temps, more particles have
enough energy to go over the barrier. E > E a E < E a Since
the probability of a molecule reacting increases, the rate
increases. Activation Energy Slide 78 Molecules need a minimum
amount of energy to react. activation energy, E a. Visualized as an
energy barrier - activation energy, E a. Reaction coordinate
diagram Activation Energy Slide 79 Orientation factors into the
equation The orientation of a molecule during collision can have a
profound effect on whether or not a reaction occurs. reactive or
effective collision When the green atom collides with the green
atom on the molecule, a reactive or effective collision occurs. The
reaction occurs only when the orientation of the molecules is just
right Activation Energy Slide 80 Orientation factors into the
equation non-reactive or ineffective collision When the green atom
collides with the red atom on the molecule, this leads to a
non-reactive or ineffective collision occurs. In some cases, the
reactants must have proper orientation for the collision to yield
products. This reduces the number of collisions that are reactive!
Activation Energy Slide 81 Arhenius discovered that most
reaction-rate data obeyed an equation based on three factors: (1)
The number of collisions per unit time. (2) The fraction of
collisions that occur with the correct orientation. (3) The
fraction of the colliding molecules that have an energy greater
than or equal to E a. Arrhenius equation From these observations
Arrhenius developed the aptly named Arrhenius equation. The
Arrhenius Equation Slide 82 Both A and E a are specific to a given
reaction. k is the rate constant E a is the activation energy R is
the ideal-gas constant (8.314 J/K mol) T is the temperature in K
Afrequencypreexponential factor A is known the frequency or
preexponential factor In addition to carrying the units of the rate
constant, A relates to the frequency of collisions and the
orientation of a favorable collision probability The Arrhenius
Equation Slide 83 Temperature Dependence of the Rate Constant:
Increasing the temperature of a reaction generally speeds up the
process (increases the rate) because the rate constant increases
according to the Arrhenius Equation. Rate (Ms -1 ) = k[A] x [B] y
As T increases, the value of the exponential part of the equation
becomes less negative thus increasing the value of k. The Arrhenius
Equation Slide 84 Reactions generally occur slower at lower T.
Iodine clock reaction. H 2 O 2 + 2 I - + 2 H + 2 H 2 O + I 2 Room
temperature In ice at 0 o C Effect of Temperature Slide 85
Determining the Activation Energy E a may be determined
experimentally. First take natural log of both sides of the
Arrhenius equation: ln ln k y = mx + b The Arrhenius Equation Slide
86 One can determine the activation energy of a reaction by
measuring the rate constant at two temperatures: Writing the
Arrhenius equation for each temperature: Subtracting k 1 from k 2
we find that: Determining the Activation Energy The Arrhenius
Equation Slide 87 Knowing the rate constants at two temps yields
the activation energy. or Knowing the E a and the rate constant at
one temp allows one to find k(T 2 ) Determining the Activation
Energy The Arrhenius Equation Slide 88 Problem: The activation
energy of a first order reaction is 50.2 kJ/mol at 25 C. At what
temperature will the rate constant double? (1) (2) (3) Slide 89
Problem: (4) (5) T 2 = 308 K A 10 C change of temperature doubles
the rate!! algebra! Slide 90 Catalysts speed up reactions by
altering the mechanism to lower the activation energy barrier. Dr.
James Cusumano, Catalytica Inc. What is a catalyst? Catalysts and
society Catalysts and the environment Catalysis Slide 91 In auto
exhaust systems Pt, NiO 2 CO + O 2 2 CO 2 2 NO N 2 + O 2 Catalysis
Slide 92 2.Polymers: H 2 C=CH 2 polyethylene 3.Acetic acid:CH 3 OH
+ CO CH 3 CO 2 H 4. Enzymes biological catalysts Catalysis Slide 93
Catalysis and activation energy Uncatalyzed reaction Catalyzed
reaction MnO 2 catalyzes decomposition of H 2 O 2 2 H 2 O 2 2 H 2 O
+ O 2 Catalysis Slide 94 Iodine-Catalyzed Isomerization of
cis-2-Butene Slide 95 Slide 96 The overall stoichiometry of a
chemical reaction is most often the sum of several steps: (1) 2AB A
2 B 2 (2) A 2 B 2 + C 2 A 2 B + C 2 B (3) A 2 B + C 2 A 2 + C 2 B
2AB + A 2 B 2 + A 2 B + 2C 2 A 2 B 2 + A 2 B + A 2 + 2C 2 B Net:
2AB + 2C 2 A 2 + 2C 2 B The sequence of steps (1-3) describes a
possible reaction mechanism. Reaction Mechanisms Slide 97 The
overall stoichiometry of a chemical reaction is most often the sum
of several steps: (1) 2AB A 2 B 2 (2) A 2 B 2 + C 2 A 2 B + C 2 B
(3) A 2 B + C 2 A 2 + C 2 B 2AB + A 2 B 2 + A 2 B + 2C 2 A 2 B 2 +
A 2 B + A 2 + 2C 2 B Net: 2AB + 2C 2 A 2 + 2C 2 B The species that
cancel out (not part of the overall reaction) are called reaction
intermediates. Reaction Mechanisms Slide 98 (1) 2AB A 2 B 2 (2) A 2
B 2 + C 2 A 2 B + C 2 B (3) A 2 B + C 2 A 2 + C 2 B 2AB + 2C 2 A 2
+ 2C 2 B elementary step. Each step in the mechanism is called an
elementary step. molecularity The number of reactants in an
elementary step is called the molecularity. bimolecular process In
this example each step (1-3) is a bimolecular process. (2
reactants) unimolecular process A 2 2A is a unimolecular process (1
reactant) termolecular process 2A + B is a termolecular process (3
reactants) Reaction Mechanisms Slide 99 The rate of the overall
reaction can never be faster than the slowest step in the
mechanism. If reaction (1) is the slowest of the three steps in the
mechanism Then it is known as the rate determining step slow fast
(1) 2AB A 2 B 2 (2) A 2 B 2 + C 2 A 2 B + C 2 B (3) A 2 B + C 2 A 2
+ C 2 B 2AB + 2C 2 A 2 + 2C 2 B Reaction Mechanisms Slide 100 slow
fast (1) 2AB A 2 B 2 (2) A 2 B 2 + C 2 A 2 B + C 2 B (3) A 2 B + C
2 A 2 + C 2 B 2AB + 2C 2 A 2 + 2C 2 B The slowest step controls the
rate of the reaction. It determines the rate law! Rate (Ms -1 ) =
k[AB] 2 The rate law is not based on the overall reaction: Rate (Ms
-1 ) = k[AB] 2 [C 2 ] 2 Reaction Mechanisms Slide 101 Consider the
following reaction: 2NO 2 (g) + F 2 (g) 2FNO 2 (g) If the reaction
proceeded by the overall reaction, the rate law for the reaction
would be 3 rd order overall. The actual rate law is found to be:
Rate = k[NO 2 ][F 2 ] Indicating that the slowest step in the
mechanism is a bimolecular reaction between NO 2 and F 2. Reaction
Mechanisms Slide 102 2NO 2 (g) + F 2 (g) 2FNO 2 (g) Rate = k[NO 2
][F 2 ] To explain the observed kinetics, a possible mechanism is
proposed: Reaction Mechanisms Slide 103 2NO 2 (g) + F 2 (g) 2FNO 2
(g) Rate = k[NO 2 ][F 2 ] Since the 1 st step in the reaction is
the slow step, it determines the kinetics and rate law for the
reaction: Reaction Mechanisms Slide 104 Validating a Reaction
Mechanism: A mechanism is a proposal of how the reaction proceeds
at the molecular level. 1.The individual elementary steps must sum
to yield the overall reaction with correct stoichiometry. 2.The
predicted reaction rate law must be in agreement with the
experimentally determined rate law. 3.Note that there may be more
than one mechanism that is in agreement with the reaction
stoichiometry and kinetics. Reaction Mechanisms
