chemical kinetics - georgia southern university...
Post on 05-Apr-2018
226 Views
Preview:
TRANSCRIPT
1
CHAPTER 15
Chemical Kinetics
2
�� We can use We can use thermodynamicsthermodynamics to tell if a reaction is to tell if a reaction is productproduct-- or reactantor reactant--favored.favored.
�� But this gives us no info on But this gives us no info on HOW FASTHOW FAST reaction reaction goes from reactants to products.goes from reactants to products.
�� KINETICSKINETICS —— the study of the study of REACTION RATESREACTION RATES and and their relation to the way the reaction proceeds, their relation to the way the reaction proceeds, i.e., its i.e., its MECHANISMMECHANISM..
�� The reaction mechanism is our goal!The reaction mechanism is our goal!
Chemical KineticsChemical Kinetics
The Rate of a Reaction
� Kinetics is the study of rates of chemical
reactions and the mechanisms by which they
occur.
� The reaction rate is the increase in concentration
of a product per unit time or decrease in
concentration of a reactant per unit time.
� A reaction mechanism is the series of molecular
steps by which a reaction occurs.
4
Reaction MechanismsReaction Mechanisms
The sequence of events at the molecular level that control the The sequence of events at the molecular level that control the
speed and outcome of a reaction.speed and outcome of a reaction.
Br from biomass burning destroys stratospheric ozone. Br from biomass burning destroys stratospheric ozone.
(See R.J. Cicerone, (See R.J. Cicerone, ScienceScience, volume 263, page 1243, 1994.), volume 263, page 1243, 1994.)
Step 1:Step 1: Br + OBr + O33 ------> > BrOBrO + O+ O22
Step 2:Step 2: ClCl + O+ O33 ------> > ClOClO + O+ O22
Step 3:Step 3: BrOBrO + + ClOClO + light + light ------> Br + > Br + ClCl + O+ O22
NET: NET: 2 O2 O33 ------> 3 O> 3 O22
The Rate of a Reaction
� Thermodynamics determines if a reaction can occur.
� Kinetics determines how quickly a reaction occurs.
� Some reactions that are thermodynamically feasible are kinetically so slow as to be imperceptible.
( ) ( )
( ) ( ) ( )
OUSINSTANTANE
kJ -79=G OHOH+H
SLOW VERY
kJ 396G COO C
o
2982
-
aq
+
aq
o
298g2g2diamond
∆→
−=∆→+
l
The Rate of Reaction� Consider the
hypothetical reaction,
A(g) + B(g) → C(g) + D(g)
� equimolar amounts of
reactants, A and B, will
be consumed while
products, C and D, will
be formed as indicated
in this graph:
0
0.2
0.4
0.6
0.8
1
1.2
0 50 100
150
200
250
300
350
Time
Concentrations of
Reactants & Products
[A] & [B]
[C] & [D]
[ ] [ ] [ ] [ ]t d
D+
t c
C+or
t b
B-
t a
A-= Rate
∆∆
=∆∆
∆∆
=∆∆
2
7
�� Reaction rate = change in concentration of a reactant or productReaction rate = change in concentration of a reactant or product
with time.with time.
�� Three Three ““typestypes”” of rates of rates
�� initial rateinitial rate
�� average rateaverage rate
�� instantaneous rateinstantaneous rate
Reaction Rates Reaction Rates
8
Determining a Reaction RateDetermining a Reaction Rate
Blue dye is oxidized with Blue dye is oxidized with
bleach. bleach.
Its concentration decreases Its concentration decreases
with time.with time.
The rate The rate —— the change in dye the change in dye
concconc with time with time —— can be can be
determined from the plot.determined from the plot.
Slope of tangent line to initial Slope of tangent line to initial
part of curve.part of curve.
Dy
e
Dy
e C
on
cC
on
c
Time
9
Determining a Reaction RateDetermining a Reaction Rate Rate Law� Example: The rate of a simple one-step, elementary reaction is
directly proportional to the concentration of the reacting substance.
� [A] is the concentration of A in molarity, k is the rate constant.
� Example: proportional to the molar concentration of the reactant raised to the power of the number of molecules involved in the reaction.
[ ] [ ]Ak = Rateor ARate
C + BA (g)(g)(g)
∝
→
( ) ( ) ( )
[ ] [ ]22
ggg
Xk = Rateor XRate
Z+ YX 2
∝
→
11
Concentrations, Rates, & Concentrations, Rates, & Rate LawsRate Laws
In general, for In general, for
a A + b B a A + b B ----> x X> x X
Rate = k [Rate = k [A]A]mm[B][B]nn
K is the rate constantK is the rate constant
The exponents m, n, and p The exponents m, n, and p
•• are the are the reaction orderreaction order
•• can be 0, 1, 2 or fractionscan be 0, 1, 2 or fractions
•• must be determined by experimentmust be determined by experiment
THEY ARE NOT BALANCED EQUATION THEY ARE NOT BALANCED EQUATION COEFFICIENTSCOEFFICIENTS 12
Interpreting Rate Interpreting Rate Laws Laws
Rate = k [Rate = k [A]A]mm[B][B]nn
�� If m = 1, If m = 1, rxnrxn. is 1st order in A. is 1st order in A
Rate = k [A]Rate = k [A]11
If [A] doubles, then rate goes up by factor of __ If [A] doubles, then rate goes up by factor of __
�� If m = 2, If m = 2, rxnrxn. is 2nd order in A.. is 2nd order in A.
Rate = k [A]Rate = k [A]22
Doubling [A] increases rate by ________Doubling [A] increases rate by ________
�� If m = 0, If m = 0, rxnrxn. is zero order.. is zero order.
Rate = k [A]Rate = k [A]00
If [A] doubles, rate ________If [A] doubles, rate ________
3
The Rate of Reaction – Rate Law
� Rate Law Expressions must be determined experimentally.
� The rate law cannot be determined from the final
balanced chemical equation, only from simple, one step
mechanistic equations
�The order of a reaction can be expressed in terms of
either: each reactant in the reaction or the overall reaction.
�Order for the overall reaction is the sum of the orders for each
reactant in the reaction.
�For example: ( ) ( ) ( )
[ ]
overall.order first and
ONin order first isreaction This
ONk= Rate
O + NO4ON 2
52
52
g2g2g52 →
The Rate of Reaction� Example: ( ) ( ) ( )
[ ]
overallorder third and ,Oin order first
NO,in order second isreaction This
isreaction This Ok[NO]=Rate
NO 2O+NO 2
2
2
2
g2g2g →
( ) ( ) ( )
[ ]2ggg
Ak = Rate
CBA 2 +→�Example:
�Because it is a second order rate-law
expression:
�If the [A] is doubled the rate of the reaction
will increase by a factor of ?????
�If the [A] is halved the rate of the reaction
will decrease by a factor of ??????
Factors That Affect Reaction
Rates
� There are several factors that can influence the
rate of a reaction:
1. The nature of the reactants.
2. The concentration of the reactants.
3. The temperature of the reaction.
4. The presence of a catalyst.
� We will look at each factor individually.
Nature of Reactants� This is a very broad category that encompasses the different
reacting properties of substances.
� For example sodium reacts with water explosively at room
temperature to liberate hydrogen and form sodium hydroxide.
( ) ( ) ( ) ( )
burns. and ignites H The
reaction. rapid and violent a is This
HNaOH 2OH 2Na 2
2
g2aq2s +→+l
( ) ( ) reaction No OH Mg 2s →+l
•The reaction of magnesium with water at room temperature is so
slow that that the evolution of hydrogen is not perceptible to the
human eye.
Concentrations of Reactants: The
Rate-Law Expression
� This is a simplified representation of the effect of
different numbers of molecules in the same
volume.
� The increase in the molecule numbers is indicative of
an increase in concentration.
A(g) + B (g) →→→→ Products
A B
A B
A B
B
A B
A B
A B
A B
4 different possible
A-B collisions
6 different possible
A-B collisions
9 different possible
A-B collisions18
Deriving Rate LawsDeriving Rate Laws
ExptExpt. . [CH[CH33CHO]CHO] Disappear of CHDisappear of CH33CHOCHO
(mol/L)(mol/L) (mol/L(mol/L••sec)sec)
11 0.100.10 0.0200.020
22 0.200.20 0.0810.081
33 0.300.30 0.1820.182
44 0.400.40 0.3180.318
Derive rate law and k for Derive rate law and k for CHCH33CHO(g) CHO(g) ----> CH> CH44(g) + CO(g)(g) + CO(g)
from experimental data for rate of disappearance from experimental data for rate of disappearance of CHof CH33CHOCHO
4
19
Deriving Rate LawsDeriving Rate Laws
Rate of Rate of rxnrxn = ??= ??
Here the rate goes up by ______ when initial Here the rate goes up by ______ when initial conc. doubles. Therefore, we say this reaction is conc. doubles. Therefore, we say this reaction is _________________ order._________________ order.
Concentrations of Reactants: The
Rate-Law Expression� Example: The following rate data were obtained at 25oC
for the following reaction. What are the rate-law expression and the specific rate-constant for this reaction?
2 A(g) + B(g) → 3 C(g)
2.0 x 10-40.200.103
4.0 x 10-40.300.202
2.0 x 10-40.100.101
Initial rate of
formation of C
(M/s)
Initial [B]
(M)
Initial [A]
(M)
Experiment
Number
[ ] [ ]yxBAk=Rate
Concentrations of Reactants: The
Rate-Law Expression2 A(g) + B(g) + 2 C(g) → 3 D(g) + 2 E(g)
2.0 x 10-40.100.100.201
1.8 x 10-30.400.300.604
2.0 x 10-40.300.100.203
6.0 x 10-40.200.300.202
Initial rate
of
formation
of D (M/s)
Initial
[C]
(M)
Initial [B]
(M)
Initial [A]
(M)
Experimen
t
Concentrations of Reactants: The
Rate-Law Expression
� Example : consider a chemical reaction between
compounds A and B that is first order with respect to A,
first order with respect to B, and second order overall.
From the information given below, fill in the blanks.
?0.403.2 x 10-23
0.050?1.6 x 10-22
0.0500.204.0 x 10-31
Initial [B]
(M)
Initial [A]
(M)
Initial Rate
(M/s)Experiment
Concentration vs. Time: The Integrated Rate Equation
� The integrated rate equation relates time and concentration for
chemical and nuclear reactions.
� From the integrated rate equation we can predict the amount
of product that is produced in a given amount of time.
� First order reactions: 1st order in the reactant and 1st order
overall.
A → products
This is a common reaction type for many chemical reactions and all simple radioactive decays.
�Two examples of this type are:
2 N2O5(g) → 2 N2O4(g) + O2(g)
238U → 234Th + 4He 24
Concentration/Time Concentration/Time RelationsRelations
What is concentration of reactant as function of What is concentration of reactant as function of
time? time?
Consider Consider FIRST ORDER REACTIONSFIRST ORDER REACTIONS
The rate law isThe rate law is
Rate ==== -∆∆∆∆[A]
∆∆∆∆time = k [A]
5
25
Concentration/Time RelationsConcentration/Time Relations
Integrating Integrating -- ((∆∆ [A] / [A] / ∆∆
time) = k [A], we gettime) = k [A], we get[A] = - k tln[A]o
naturallogarithm [A] at time = 0
[A] = - k tln[A]o
naturallogarithm [A] at time = 0
[A] / [A][A] / [A]00 =fraction remaining after time t =fraction remaining after time t
has elapsed.has elapsed.
Called the Called the integrated firstintegrated first--order rate laworder rate law..
oAktA ]ln[]ln[ +−=26
Concentration/Time RelationsConcentration/Time Relations
Sucrose decomposes to simpler sugarsSucrose decomposes to simpler sugars
Rate of disappearance of sucrose = k Rate of disappearance of sucrose = k
[sucrose][sucrose]
GlucoseGlucose
If k = 0.21 hrIf k = 0.21 hr--11
and [sucrose] = 0.010 Mand [sucrose] = 0.010 M
How long to drop 90% How long to drop 90%
(to 0.0010 M)?(to 0.0010 M)?
27
Using the Integrated Rate LawUsing the Integrated Rate Law
The integrated rate law suggests a way to tell the order based oThe integrated rate law suggests a way to tell the order based on n
experiment. experiment.
2 N2 N22OO55(g) (g) ------> 4 NO> 4 NO22(g) + O(g) + O22(g)(g)
Time (min)Time (min) [N[N22OO55]]00 (M) (M) lnln [N[N22OO55]]00
00 1.001.00 00
1.01.0 0.7050.705 --0.350.35
2.02.0 0.4970.497 --0.700.70
5.05.0 0.1730.173 --1.751.75
28
Using the Integrated Rate LawUsing the Integrated Rate Law
2 N2 N22OO55(g) (g) ------> 4 NO> 4 NO22(g) + O(g) + O22(g)(g)
[N2 O5 ] vs . t i me
t i me
1
0
0 5
l n [N2 O5 ] vs. t i me
t i me
0
-2
0 5
Data of conc. vs. Data of conc. vs.
time plot do not fit time plot do not fit
straight line.straight line.
Plot of Plot of lnln [N[N22OO55] vs. ] vs.
time is a straight time is a straight
line!line!
29
Using the Integrated Rate LawUsing the Integrated Rate Law
All 1st order reactions have straight line plot for All 1st order reactions have straight line plot for lnln [A] [A]
vs. time. vs. time.
(2nd order gives straight line for plot of 1/[A] vs. time)(2nd order gives straight line for plot of 1/[A] vs. time)
l n [N2 O5 ] vs. t i me
t i me
0
-2
0 5
Plot of Plot of lnln [N[N22OO55] vs. time ] vs. time
is a straight line! is a straight line!
EqnEqn. for straight line: . for straight line:
y = y = mxmx + b + b
ln [N 2O5] = - kt + ln [N 2O5]o
conc at time t
rate const = slope
conc at time = 0
Rate = k [NRate = k [N22OO55]]
30
HalfHalf--LifeLifeSection 15.4 & Screen 15.8Section 15.4 & Screen 15.8
HALFHALF--LIFELIFE is is
the time it the time it
takes for 1/2 a takes for 1/2 a
sample is sample is
disappear.disappear.
For 1st order For 1st order
reactions, the reactions, the
concept of concept of
HALFHALF--LIFE LIFE
is especially is especially
useful.useful.Active Figure 15.9
6
31
�� Reaction is 1st order Reaction is 1st order
decomposition of decomposition of
HH22OO22..
HalfHalf--LifeLife
32
HalfHalf--LifeLife
�� Reaction after 1 Reaction after 1
halfhalf--life.life.
�� 1/2 of the 1/2 of the
reactant has been reactant has been
consumed and consumed and
1/2 remains.1/2 remains.
33
HalfHalf--LifeLife
�� After 2 halfAfter 2 half--lives lives
1/4 of the reactant 1/4 of the reactant
remains.remains.
34
HalfHalf--LifeLife
�� A 3 halfA 3 half--lives 1/8 of lives 1/8 of
the reactant the reactant
remains.remains.
35
HalfHalf--LifeLife
�� After 4 halfAfter 4 half--lives lives
1/16 of the reactant 1/16 of the reactant
remains.remains.
Concentration vs. Time: The
Integrated Rate Equation
� Solve the first order integrated rate equation for t.
� Define the half-life, t1/2, of a reactant as the time required
for half of the reactant to be consumed, or the time at
which [A]=1/2[A]0.
[ ][ ]
−=
oA
Aln
k
1t
7
Concentration vs. Time: The
Integrated Rate Equation
� At time t = t1/2, the expression becomes:
[ ][ ]
( )
k
693.0t
5.0lnk
1t
A
A2/1ln
k
1t
1/2
1/2
0
01/2
=
−=
−=
38
HalfHalf--LifeLife
Sugar is fermented in a 1st order process (using Sugar is fermented in a 1st order process (using
an enzyme as a catalyst).an enzyme as a catalyst).
sugar + enzyme sugar + enzyme ----> products> products
Rate of disappear of sugar = k[sugar]Rate of disappear of sugar = k[sugar]
k = 3.3 x 10k = 3.3 x 10--44 secsec--11
What is the What is the halfhalf--lifelife of this reaction?of this reaction?
39
HalfHalf--LifeLifeSection 15.4 & Screen 15.8Section 15.4 & Screen 15.8
Rate = k[sugar] and k = 3.3 x 10Rate = k[sugar] and k = 3.3 x 10--44 secsec--11. Half. Half--life life is 35 min. Start with 5.00 g sugar. How much is is 35 min. Start with 5.00 g sugar. How much is left after 2 hr and 20 min (140 min)?left after 2 hr and 20 min (140 min)?
40
HalfHalf--LifeLifeSection 15.4 & Screen 15.8Section 15.4 & Screen 15.8
Radioactive decay is a first order process. Radioactive decay is a first order process.
Tritium Tritium ------> electron + helium> electron + helium33HH 00
--11ee 33HeHe
tt1/21/2 = 12.3 years= 12.3 years
If you have 1.50 mg of tritium, how much is left If you have 1.50 mg of tritium, how much is left
after 49.2 years? after 49.2 years?
41
HalfHalf--Lives of Radioactive ElementsLives of Radioactive Elements
Rate of decay of radioactive isotopes given in Rate of decay of radioactive isotopes given in
terms of 1/2terms of 1/2--life. life.
238238U U ----> > 234234Th + HeTh + He 4.5 x 104.5 x 1099 yy
1414C C ----> > 1414N + betaN + beta 5730 y5730 y
131131I I ----> > 131131Xe + betaXe + beta 8.05 d8.05 d
Element 106 Element 106 -- seaborgiumseaborgium263263Sg Sg 0.9 s0.9 s
Concentration vs. Time: The
Integrated Rate Equation
� Example: Cyclopropane, an anesthetic,
decomposes to propene according to the
following equation.
The reaction is first order in cyclopropane with k = 9.2
s-1 at 10000C. Calculate the half life of cyclopropane
at 10000C.
CH2
CH2
CH2
CH2
CH3
CH
(g)(g)
8
Concentration vs. Time: The
Integrated Rate Equation
� Example: Cyclopropane, an anesthetic,
decomposes to propene according to the
following equation.
The reaction is first order in cyclopropane with k = 9.2
s-1 at 10000C. Calculate the half life of cyclopropane
at 10000C.
CH2
CH2
CH2
CH2
CH3
CH
(g)(g)
Concentration vs. Time: The
Integrated Rate Equation
� Example: Refer to previous. How much of a 3.0
g sample of cyclopropane remains after 0.50
seconds?
� The integrated rate laws can be used for any unit that
represents moles or concentration.
� In this example we will use grams rather than mol/L.
Concentration vs. Time: The
Integrated Rate Equation
� For reactions that are second order with respect to a
particular reactant and second order overall, the rate
equation is:
� Where:
[A]0= mol/L of A at time t=0. [A] = mol/L of A at time t.
k = specific rate constant. t = time elapsed since
beginning of reaction.
[ ] [ ]k t
A
1
A
1
0
=−
46
Properties of ReactionsProperties of Reactions
Reaction Mechanisms and the
Rate-Law Expression
� Use the experimental rate-law to postulate a
mechanism.
� The slowest step in a reaction mechanism is the
rate determining step.
� Consider the iodide ion catalyzed decomposition
of hydrogen peroxide to water and oxygen.
( ) ( ) ( )g22
I
22 O + OH 2 OH 2-
ll→
Reaction Mechanisms and the
Rate-Law Expression
� This reaction is known to be first order in H2O2 , first
order in I- , and second order overall.
� The mechanism for this reaction is thought to be:
[ ][ ]-22
2222
-
2222
-
2
--
22
IOHk=R law rate alExperiment
O+OH 2OH 2reaction Overall
I+O+OHOH+ IO stepFast
OH+IOI+OH step Slow
→
→
→
UNIMOLECULAR UNIMOLECULAR -- only one reactant is involved.only one reactant is involved.
BIMOLECULARBIMOLECULAR —— two different molecules must collide two different molecules must collide ----> products> products
9
49
Collision Theory of
Reaction Rates
� Three basic events must occur for a reaction to
occur the atoms, molecules or ions must:
1. Collide.
2. Collide with enough energy to break and form bonds.
3. Collide with the proper orientation for a reaction to
occur.
50
Collision Theory of
Reaction Rates
� One method to increase the number of collisions and the
energy necessary to break and reform bonds is to heat the
molecules.
� As an example, look at the reaction of methane and
oxygen:
� We must start the reaction with a match.
� This provides the initial energy necessary to break the first few
bonds.
� Afterwards the reaction is self-sustaining.
kJ 891 OH CO O CH (g)22(g)2(g)4(g) ++→+
51
Collision Theory of
Reaction Rates
� Illustrate the proper orientation of molecules that
is necessary for this reaction.
X2(g) + Y2(g) →2 XY(g)
� Some possible ineffective collisions are :
X
X
Y YY
Y
X X X X Y Y
52
Collision Theory of
Reaction Rates
� An example of an effective collision is:
X Y
X Y
X Y
X Y
X Y
+
X Y
53
Collision TheoryCollision Theory
Reactions require Reactions require
(a) activation energy and (a) activation energy and
(b) correct geometry. (b) correct geometry.
OO33(g) + NO(g) (g) + NO(g) ------> O> O22(g) + NO(g) + NO22(g)(g)
2. Activation energy 2. Activation energy
and geometryand geometry1. Activation energy 1. Activation energy
54
MechanismsMechanisms
OO33 + NO reaction occurs in a single + NO reaction occurs in a single
ELEMENTARYELEMENTARY step. Most others involve a step. Most others involve a
sequence of elementary steps.sequence of elementary steps.
Adding elementary steps gives NET reaction.Adding elementary steps gives NET reaction.
10
55
MechanismsMechanisms
Most Most rxnsrxns. involve a sequence of elementary . involve a sequence of elementary
steps.steps.
2 I2 I-- + H+ H22OO22 + 2 H+ 2 H++ ------> I> I22 + 2 H+ 2 H22OO
Rate = k [IRate = k [I--] [H] [H22OO22]]
NOTENOTE
1.1. Rate law comes from experimentRate law comes from experiment
2.2. Order and Order and stoichiometricstoichiometric coefficients not coefficients not necessarily the same!necessarily the same!
3.3. Rate law reflects all chemistry down Rate law reflects all chemistry down to and including the slowest step in to and including the slowest step in multistepmultistep reaction.reaction.
56
MechanismsMechanisms
Proposed MechanismProposed Mechanism
Step 1 Step 1 —— slowslow HOOH + IHOOH + I-- ----> HOI + OH> HOI + OH--
Step 2 Step 2 —— fastfast HOI + IHOI + I-- ----> I> I22 + OH+ OH--
Step 3 Step 3 —— fastfast 2 OH2 OH-- + 2 H+ 2 H++ ----> 2 H> 2 H22OO
Rate of the reaction controlled by slow step Rate of the reaction controlled by slow step ——
RATE DETERMINING STEPRATE DETERMINING STEP, , rdsrds..
Rate can be no faster than Rate can be no faster than rdsrds!!
Most Most rxnsrxns. involve a sequence of elementary steps.. involve a sequence of elementary steps.
2 I2 I-- + H+ H22OO22 + 2 H+ 2 H++ ------> I> I22 + 2 H+ 2 H22OO
Rate = k [IRate = k [I--] [H] [H22OO22]]
57
MechanismsMechanisms
Elementary Step 1Elementary Step 1 is is bimolecularbimolecular and involves Iand involves I--
and HOOH. Therefore, this predicts the rate law and HOOH. Therefore, this predicts the rate law
should beshould be
Rate Rate ∝∝ [I[I--] [H] [H22OO22] ] —— as observed!!as observed!!
The species HOI and OHThe species HOI and OH-- are are reaction reaction intermediates.intermediates.
2 I2 I-- + H+ H22OO22 + 2 H+ 2 H++ ------> I> I22 + 2 H+ 2 H22OO
Rate = k [IRate = k [I--] [H] [H22OO22]]
Step 1 Step 1 —— slowslow HOOH + IHOOH + I-- ----> HOI + OH> HOI + OH--
Step 2 Step 2 —— fastfast HOI + IHOI + I-- ----> I> I22 + OH+ OH--
Step 3 Step 3 —— fastfast 2 OH2 OH-- + 2 H+ 2 H++ ----> 2 H> 2 H22OO
58
Rate Laws and Rate Laws and MechanismsMechanisms
NO2 + CO reaction:
Rate = k[NO2]2
Single step
Two possible Two possible
mechanismsmechanismsTwo steps: step 1
Two steps: step 2
59
Ozone Decomposition Ozone Decomposition over Antarcticaover Antarctica
2 O2 O33 (g) (g) ------> 3 O> 3 O22 (g)(g) 60
Ozone Decomposition Ozone Decomposition MechanismMechanism
Proposed mechanismProposed mechanism
Step 1: fast, equilibriumStep 1: fast, equilibrium
O3 (g) + O2 (g) + O (g)
Step 2: slow O3 (g) + O (g) ---> 2O2 (g)
2 O2 O33 (g) (g) ------> 3 O> 3 O22 (g)(g)
Rate = k [O3]
2
[O2 ]
11
Transition State Theory
� Transition state theory postulates that reactants
form a high energy intermediate, the transition
state, which then falls apart into the products.
� For a reaction to occur, the reactants must acquire
sufficient energy to form the transition state.
� This energy is called the activation energy or Ea.
� Look at a mechanical analog for activation energy
Transition State Theory
∆Epot = mg∆h
Cross section
of mountain
Boulder
Eactivation
∆h
h2
h1
Epot=mgh2
Epot=mgh1
Height
Transition State Theory
Potential
Energy
Reaction Coordinate
X2 + Y2
2 XY
Eactivation - a kinetic quantity
∆E ≈∆H
a thermodynamic
quantity
Representation of a chemical reaction.
Transition State Theory
� The relationship between the activation energy for
forward and reverse reactions is
� Forward reaction = Ea
� Reverse reaction = Ea + ∆E
� difference = ∆E
65
Activation Energy and TemperatureActivation Energy and Temperature
Reactions are Reactions are faster at higher Tfaster at higher T because a larger fraction of because a larger fraction of
reactant molecules have enough energy to convert to product reactant molecules have enough energy to convert to product
molecules.molecules.
In general, In general,
differences in differences in activation energyactivation energycause reactions to cause reactions to
vary from fast to vary from fast to
slow.slow.
66
More About Activation EnergyMore About Activation Energy
k = Ae-E
a/RT
ln k = - (Ea
R)(
1
T) + ln A
Arrhenius equation Arrhenius equation ——
Rate Rate
constantconstant
Temp (K)Temp (K)
8.31 x 108.31 x 10--33 kJ/KkJ/K••molmolActivation Activation
energyenergyFrequency factorFrequency factor
Frequency factor related to frequency of collisions
with correct geometry.
Plot Plot lnln k vs. 1/T k vs. 1/T
------> >
straight line. straight line.
slope = slope = --EEaa/R/R
12
Temperature:
The Arrhenius Equation
� If the Arrhenius equation is written for two temperatures,
T2 and T1 with T2 >T1.
ln k ln A -E
RT
and
ln k ln A -E
RT
1a
1
2a
2
=
=
Temperature:
The Arrhenius Equation
1. Subtract one equation from the other.
ln k k A - ln A -E
RT
E
RT
ln k kE
RT-
E
RT
2 1a
2
a
1
2 1a
1
a
2
− = − −
− =
ln ln
ln
Temperature:
The Arrhenius Equation
2. Rearrange and solve for ln k2/k1.
ln k
k
E
R T T
or
ln k
k
E
R
T - T
T T
2
1
a
1 2
2
1
a 2 1
2 1
= −
=
1 1
Temperature:
The Arrhenius Equation
� Consider the rate of a reaction for which Ea=50 kJ/mol, at 20oC (293 K) and at 30oC (303 K). � How much do the two rates differ?
Temperature:
The Arrhenius Equation
� For reactions that have an Ea≈50 kJ/mol, the rate approximatelydoubles for a 100C rise in temperature, near room temperature.
� Consider:
2 ICl(g) + H2(g) → I2(g) + 2 HCl(g)
� The rate-law expression is known to be R=k[ICl][H2].
At 230 C, k = 0.163 s
At 240 C, k = 0.348 s
k approximately doubles
0 -1 -1
0 -1 -1
M
M
Catalysts� Catalysts change reaction rates by providing an
alternative reaction pathway with a different activation
energy.
� Homogeneous catalysts exist in same phase as the
reactants.
� Heterogeneous catalysts exist in different phases than the
reactants.
13
Catalysts
� Examples of commercial catalyst systems include:
( ) ( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( )
systemconverter catalytic Automobile
ONNO 2
CO 2O+CO 2
OH 18CO16O 25+HC
g2g2
Pt and NiO
g
g2
Pt and NiO
g2g
g2g2
Pt and NiO
g2g188
+ →
→
+ →
74
CATALYSISCATALYSIS
Catalysis and activation energyCatalysis and activation energy
UncatalyzedUncatalyzed reactionreaction
Catalyzed reactionCatalyzed reaction
MnOMnO22 catalyzes catalyzes
decomposition of Hdecomposition of H22OO22
2 H2 H22OO22 ------> 2 H> 2 H22O + OO + O22
top related