Upload
dangdan
View
259
Download
6
Embed Size (px)
Citation preview
1
CHAPTER 2:
CHEMICAL KINETICS
LESSON OUTCOME
• State, write and explain the term rate of reaction,
rate constant,order of reaction
• Apply the theories/concepts in chemical kinetics
• Solves the problem based on the concepts
• Determine rate of reaction
2
• Chemical kinetics - speed or rate at which a
reaction occurs
Chemical Kinetics
Chemical kinetics: the study of reaction rate, a
quantity conditions affecting it,the molecular events
during a chemical reaction (mechanism), and
presence of other components (catalysis).
Factors affecting reaction rate:
Concentrations of reactants
Catalyst
Temperature
Surface area of solid reactants or catalyst
4
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 = -D[A]
Dt
rate = D[B]
Dt
D[A] = change in concentration of A over
time period Dt
D[B] = change in concentration of B over
time period Dt
Because [A] decreases with time, D[A] is negative.
5
A B
rate = -D[A]
Dt
rate = D[B]
Dt
6
Br2 (aq) + HCOOH (aq) 2Br- (aq) + 2H+ (aq) + CO2 (g)
time
393 nm
light
Detector
D[Br2] a D Absorption
red-brown
t1< t2 < t3
7
Br2 (aq) + HCOOH (aq) 2Br- (aq) + 2H+ (aq) + CO2 (g)
average rate = -D[Br2]
Dt= -
[Br2]final – [Br2]initial
tfinal - tinitial
slope of
tangentslope of
tangentslope of
tangent
instantaneous rate = rate for specific instance in time
8
rate a [Br2]
rate = k [Br2]
k = rate
[Br2]= rate constant
= 3.50 x 10-3 s-1
9
2H2O2 (aq) 2H2O (l) + O2 (g)
PV = nRT
P = RT = [O2]RTnV
[O2] = PRT
1
rate = D[O2]
Dt RT
1 DP
Dt=
measure DP over time
10
11
Reaction Rates and Stoichiometry
2A B
Two moles of A disappear for each mole of B that is formed.
rate = D[B]
Dtrate = -
D[A]
Dt
1
2
aA + bB cC + dD
rate = -D[A]
Dt
1
a= -
D[B]
Dt
1
b=
D[C]
Dt
1
c=
D[D]
Dt
1
d
12
Write the rate expression for the following reaction:
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (g)
rate = -D[CH4]
Dt= -
D[O2]
Dt
1
2=
D[H2O]
Dt
1
2=
D[CO2]
Dt
13
Expressing reaction rates
For a chemical reaction, there are many ways to express
the reaction rate. The relationships among expressions
depend on the equation.
Note the expression and reasons for their relations for the
reaction
2 NO + O2 (g) = 2 NO2 (g)
D[O2] 1 D[NO] 1 D[NO2]
Reaction rate = – ——— = – — ———— = — ———
D t 2 D t 2 D t
Make sure you can write expressions for any reaction and
figure out the relationships. For example, give the
reaction rate expressions for
2 N2O5 = 4 NO2 + O2
14
Calculating reaction rate
The concentrations of N2O5 are 1.24e-2 and 0.93e-2 M at 600
and 1200 s after the reactants are mixed at the appropriate
temperature. Evaluate the reaction rates for
2 N2O5 = 4 NO2 + O2
Solution:
(0.93 – 1.24)e-2 – 0.31e - 2M
Decomposition rate of N2O5 = – ————— = – ——————
1200 – 600 600 s
= 5.2e-6 M s-1.
Note however,
rate of formation of NO2 = 1.02e-5 M s-1.
rate of formation of O2 = 2.6e-6 M s-1.
The reaction rates
are expressed in 3
forms
Reaction Rates
Rates of reactions can be determined by
monitoring the change in concentration of either
reactants or products as a function of time. D[A]
vs Dt
Rxn Movie
In this reaction, the
concentration of butyl
chloride, C4H9Cl, was
measured at various
times, t.
C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq)
[C4H9Cl] M
The average rate of the
reaction over each
interval is the change in
concentration divided
by the change in time:
C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq)
Average Rate, M/s
• Note that the average rate
decreases as the reaction
proceeds.
• This is because as the
reaction goes forward,
there are fewer collisions
between reactant
molecules.
C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq)
• A plot of concentration vs.
time for this reaction
yields a curve like this.
• The slope of a line tangent
to the curve at any point is
the instantaneous rate at
that time.
C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq)
• The reaction slows down
with time because the
concentration of the
reactants decreases.
C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq)
• In this reaction, the ratio
of C4H9Cl to C4H9OH is
1:1.
• Thus, the rate of
disappearance of C4H9Cl
is the same as the rate of
appearance of C4H9OH.
C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq)
Rate =-D[C4H9Cl]
Dt=
D[C4H9OH]
Dt
Concentration and Rate
Each reaction has its own equation that gives
its rate as a function of reactant
concentrations.
this is called its Rate Law
To determine the rate law we measure the rate at
different starting concentrations.
Compare Experiments 1 and 2:
when [NH4+] doubles, the initial rate doubles.
Likewise, compare Experiments 5 and 6:
when [NO2-] doubles, the initial rate doubles.
This equation is called the rate law,
and k is the rate constant.
LESSON OUTCOME
• Write the rate law, rate equation & half life
for zero, first & second order of reaction
• Solves the problem involving rate equation
• Explain & describe factors that affect the
rate of reaction
26
27
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 cC + dD
Rate = k [A]x[B]y
Reaction is xth order in A
Reaction is yth order in B
Reaction is (x +y)th order overall
28
F2 (g) + 2ClO2 (g) 2FClO2 (g)
rate = k [F2][ClO2]
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.
1
29
F2 (g) + 2ClO2 (g) 2FClO2 (g)
rate = k [F2]x[ClO2]
y
Double [F2] with [ClO2] constant
Rate doubles
x = 1
Quadruple [ClO2] with [F2] constant
Rate quadruples
y = 1
rate = k [F2][ClO2]
30
Determine the rate law and calculate the rate constant for the
following reaction from the following data:
S2O82- (aq) + 3I- (aq) 2SO4
2- (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
rate = k [S2O82-]x[I-]y
Double [I-], rate doubles (experiment 1 & 2)
y = 1
Double [S2O82-], rate doubles (experiment 2 & 3)
x = 1
k = rate
[S2O82-][I-]
=2.2 x 10-4 M/s
(0.08 M)(0.034 M)= 0.08/M•s
rate = k [S2O82-][I-]
31
First-Order Reactions
A product rate = -D[A]
Dtrate = k [A]
k = rate
[A]= 1/s or s-1M/s
M=
D[A]
Dt= k [A]-
[A] 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
32
2N2O5 4NO2 (g) + O2 (g)
Graphical Determination of k
33
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 ?
ln[A] = ln[A]0 - kt
kt = ln[A]0 – ln[A]
t =ln[A]0 – ln[A]
k= 66 s
[A]0 = 0.88 M
[A] = 0.14 M
ln[A]0
[A]
k=
ln0.88 M
0.14 M
2.8 x 10-2 s-1=
34
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
ln[A]0
[A]0/2
k=t½
ln 2
k=
0.693
k=
What is the half-life of N2O5 if it decomposes with a rate constant
of 5.7 x 10-4 s-1?
t½ln 2
k=
0.693
5.7 x 10-4 s-1= = 1200 s = 20 minutes
How do you know decomposition is first order?
units of k (s-1)
35
A product
First-order reaction
# of
half-lives [A] = [A]0/n
1
2
3
4
2
4
8
16
36
Second-Order Reactions
A product rate = -D[A]
Dtrate = k [A]2
k = rate
[A]2= 1/M•s
M/sM2=
D[A]
Dt= k [A]2-
[A] is the concentration of A at any time t
[A]0 is the concentration of A at time t=0
1
[A]=
1
[A]0+ kt
t½ = t when [A] = [A]0/2
t½ =1
k[A]0
37
Zero-Order Reactions
A product rate = -D[A]
Dtrate = k [A]0 = k
k = rate
[A]0= M/s
D[A]
Dt= k-
[A] is the concentration of A at any time t
[A]0 is the concentration of A at time t = 0
t½ = t when [A] = [A]0/2
t½ =[A]02k
[A] = [A]0 - kt
38
Summary of the Kinetics of Zero-Order, First-Order
and Second-Order Reactions
Order Rate Law
Concentration-Time
Equation Half-Life
0
1
2
rate = k
rate = k [A]
rate = k [A]2
ln[A] = ln[A]0 - kt
1
[A]=
1
[A]0+ kt
[A] = [A]0 - kt
t½ln 2
k=
t½ =[A]02k
t½ =1
k[A]0
Factors That Affect Reaction Rates• Concentration of Reactants
– As the concentration of reactants increases, so does the likelihood that reactant molecules will collide.
• Temperature
– At higher temperatures, reactant molecules have more kinetic energy, move faster, and collide more often and with greater energy.
• Catalysts
– Speed rxn by changing mechanism.
• Concentrations
Factors Affecting Rates
Rate with 0.3 M HCl
Rate with 6.0 M HCl
• Physical state of reactants
Factors Affecting Rates
Catalysts: catalyzed decomp of H2O2
2 H2O2 --> 2 H2O + O2
Factors Affecting Rates
43
A catalyst is a substance that increases the rate of a
chemical reaction without itself being consumed.
Ea k
ratecatalyzed > rateuncatalyzed
Ea < Ea′
Uncatalyzed Catalyzed
)/( RTEaeAk
44
In heterogeneous catalysis, the reactants and the catalysts
are in different phases.
In homogeneous catalysis, the reactants and the catalysts
are dispersed in a single phase, usually liquid.
• Haber synthesis of ammonia
• Ostwald process for the production of nitric acid
• Catalytic converters
• Acid catalysis
• Base catalysis
45
Catalysis
A catalyst is a substance that changes
the rate of a reaction by lowing the
activation energy, Ea. It participates a
reaction in forming an intermediate, but
is regenerated.
Enzymes are marvelously selective
catalysts.
A catalyzed reaction,
NO (catalyst)
2 SO2 (g) + O2 — 2 SO3 (g)
via the mechanism
i 2 NO + O2 2 NO2
(3rdorder)
ii NO2 + SO2 SO3 + NO
Uncatalyzed
rxn
Catalyzed
rxn
rxn
Energy
46
Catalyzed decomposition of ozone
The CFC decomposes in the atmosphere:
CFCl3 CFCl2 + Cl
CF2Cl3 CF2Cl + Cl.
The Cl catalyzes the reaction via the mechanism:
i O3 + h v O + O2,
ii ClO + O Cl + O2
iii O + O3 O2 + O2.
The net result or reaction is
2 O3 3 O2
47
Homogenous vs. heterogeneous catalysts
A catalyst in the same phase (gases and solutions) as the
reactants is a homogeneous catalyst. It effective, but
recovery is difficult.
When the catalyst is in a different phase than reactants (and
products), the process involve heterogeneous catalysis.
Chemisorption, absorption, and adsorption cause reactions
to take place via different pathways.
Platinum is often used to catalyze hydrogenation
Catalytic converters reduce CO and NO emission.
• Temperature
Factors Affecting Rates
LESSON OUTCOME
• Explain the term of activation energy
collision theory, Arrhenius equation, reaction
mechanism, elementary step
• Calculation involving Arrhenius equation
• Write the rate law from reaction mechanism
49
The Collision Model
• In a chemical reaction, bonds are broken
and new bonds are formed.
• Molecules can only react if they collide
with each other.
The Collision Model
Furthermore, molecules must collide with the
correct orientation and with enough energy to
cause bond breakage and formation.
Activation Energy
• In other words, there is a minimum amount of energy
required for reaction: the activation energy, Ea.
• Just as a ball cannot get over a hill if it does not roll up the
hill with enough energy, a reaction cannot occur unless the
molecules possess sufficient energy to get over the
activation energy barrier.
53
Exothermic Reaction Endothermic Reaction
The activation energy (Ea ) is the minimum amount of
energy required to initiate a chemical reaction.
A + B AB C + D++
54
Temperature Dependence of the Rate Constant
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
ln k = -Ea
R
1
T+ lnA
(Arrhenius equation)
)/( RTEaeAk
Alternate format:
55
Alternate Form of the Arrhenius Equation
At two temperatures, T1 and T2
or
15 Chemical Kinetics 56
From k = A e – Ea / R T, calculate A, Ea, k at a specific
temperature and T.
The reaction:
2 NO2(g) -----> 2NO(g) + O2(g)
The rate constant k = 1.0e-10 s-1 at 300 K and the
activation energy
Ea = 111 kJ mol-1. What are A, k at 273 K and T when k =
1e-11?
Method: derive various versions of the same formula
k = A e – Ea / R T
A = k e Ea / R T
A / k = e Ea / R T
ln (A / k) = Ea / R T
57
The reaction:
2 NO2(g) -----> 2NO(g) + O2(g)
The rate constant k = 1.0e-10 s-1 at 300 K and the activation energy
Ea = 111 kJ mol-1. What are A, k at 273 K and T when k = 1e-11?
Use the formula derived earlier:
A = k eEa / R T = 1e-10 s-1 exp (111000 J mol-1 / (8.314 J mol-1K –1*300 K))
= 2.13e9 s-1
k = 2.13e9 s-1 exp (– 111000 J mol-1) / (8.314 J mol-1 K –1*273 K)
= 1.23e-12 s-1
T = Ea / [R* ln (A/k)] = 111000 J mol-1 / (8.314*46.8) J mol-1 K-1
= 285 K
58
Importance of Molecular Orientation
effective collision
ineffective collision
59
Reaction Mechanisms
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
+
60
2NO (g) + O2 (g) 2NO2 (g)
Mechanism:
61
Elementary step: NO + NO N2O2
Elementary step: N2O2 + O2 2NO2
Overall reaction: 2NO + O2 2NO2
+
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.
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
62
Unimolecular reaction A products rate = k [A]
Bimolecular reaction A + B products rate = k [A][B]
Bimolecular reaction A + A products rate = k [A]2
Rate Laws and Elementary Steps
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.
63
Sequence of Steps in Studying a Reaction Mechanism
64
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?
NO2+ CO NO + CO2
What is the intermediate?
NO3
What can you say about the relative rates of steps 1 and 2?
rate = k[NO2]2 is the rate law for step 1 so
step 1 must be slower than step 2
Reaction Mechanisms
The sequence of events that describes the
actual process by which reactants become
products is called the reaction mechanism.
Reactions may occur all at once or through
several discrete steps.
Each of these processes is known as an
elementary reaction or elementary process.
• The molecularity of a process tells how many molecules are involved in the process.
• The rate law for an elementary step is written directly from that step.
Multistep Mechanisms
• In a multistep process, one of the steps will be
slower than all others.
• The overall reaction cannot occur faster than this
slowest, rate-determining step.
Slow Initial Step
• The rate law for this reaction is found experimentally to
be
Rate = k [NO2]2
• CO is necessary for this reaction to occur, but the rate
of the reaction does not depend on its concentration.
• This suggests the reaction occurs in two steps.
NO2 (g) + CO (g) NO (g) + CO2 (g)
Slow Initial Step
• A proposed mechanism for this reaction is
Step 1: NO2 + NO2 NO3 + NO (slow)
Step 2: NO3 + CO NO2 + CO2 (fast)
• The NO3 intermediate is consumed in the second step.
• As CO is not involved in the slow, rate-determining step, it does not
appear in the rate law.
• Rate = k [NO2]2