chapter 25 the rates of chemical reactions. contents empirical chemical kinetics 25.1 experimental...
TRANSCRIPT
Chapter 25 The Rates of chemical reactions
ContentsEmpirical chemical kinetics• 25.1 Experimental techniques• 25.2 The rates of reactions• 25.3 Integrated rate laws• 25.4 Reaction approaching equilibrium• 25.5 The temperature dependence of reaction ratesAccounting for the rate laws• 25.6 Elementary reactions• 25.7 Consecutive elementary reactions• 25.8 Unimolecular reactions
Assignment for chapter 25
• 25.4(b),25.6(a),25.10(b),25.15(a)
• 25.2,25.5,25.11,25.23
St. John’s wort is an herb that is thought to create a sense of tranquility. Herbs and other medicines have been used through the ages to cure disease and to relieve pain. In many cases, the medicine is effective because it controls the rates of reactions within the body. In this chapter, we examine the rates of chemical reactions and the mechanisms by which they take place.
Empirical chemical kinetics
Monitoring the progress of a reaction
2N2O5(g) 4NO2(g)+O2(g)Initial pressure of n moles of N2O5 is p0.
Progress of the reaction: N2O5 NO2 O2 Total
Initial n 0 0 n
At time t n(1-a) 2an 0.5an n(1+1.5a)
Progress of the reaction: N2O5 NO2 O2 Total
Initial p0 0 0 p0
At time t n(1-a) p0 2an p0 0.5an p0 n(1+1.5a) p0
Classroom exercise
2NOBr(g) 2NO(g)+Br2(g)Initial pressure of n moles of N2O5 is p0.
Progress of the reaction: NOBr NO Br2 Total
Initial n 0 0 n
At time t n(1-a) an 0.5an n(1+0.5a)
Progress of the reaction: N2O5 NO2 O2 Total
Initial p0 0 0 p0
At time t n(1-a) p0 an p0 0.5an p0 n(1+0.5a) p0
Other properties to be monitored in the progress of chemical reactions
• Absorption of radiation (spectrophotometry)• Ions in solution (Electrical conductivity)• Emission of radiation (emission spectroscopy)• Mass of ions (mass spectrometry)• Adsorption of molecules (gas chromatography)• Absorption of radiofrequency radiation (NMR spectroscopy)• Absorption of microwave radiation (ESR spectroscopy)• Etc.
Experimental techniques:flow technique
Experimental techniques:stopped-flow technique
Experimental techniques:flash photolysis
Cl
Cl
Cl
Cl
Experimental techniques:flash photolysis
ClCl
ClCl
Experimental techniques:quenching methods
• Chemical quench flow method: quench by chemical reactants such as acids
• Freeze quench method: quench by rapid cooling
Definition
kAmC+nD
Rate=change in concentration of reactant / time interval
2HI(g) H2(g)+I2(g)
tI
tHIv
][][ 2
mmol/L/s500
10017][ 2
tIv
The activity of penicillin declines over several weeks when it is stored at room temperature in the absence of stabilizers. The shape of this graph of concentration of penicillin as a function of time is typical of the behavior of chemical reactions, although the time span may vary from fractions of a second to years.
This graph shows two examples of how the rate of consumption of penicillin can be monitored while it is being stored. The red line shows the average rate calculated from measurements at 0 and 10 weeks, and the blue line shows the average rate calculated from measurements at 2.5 and 7.5 weeks. The instantaneous rate at 5 weeks is the tangent to the curve at that time (not shown).
To calculate the instantaneous reaction rate, we draw the tangent to the curve at the time of interest and then calculate the slope of this tangent. To calculate the slope, we identify any two points, A and B, on the straight line and identify the molar concentrations and times to which they correspond. The slope is then worked out by dividing the difference in concentrations by the difference in times. Notice that this graph shows the concentration of a product.
Exercise
dt
Cd
dt
Bd
dt
Ad
dt
Rd
dt
Qd
dt
Pd
RnQnPnCnBnAn
CBARQP nnnnnn
RQPCBA
][][][][][][
......
111111
dt
Cd
dt
Bd
dt
Ad
dt
Rd
dt
Qd
dt
Pd
RnQnPnCnBnAn
Qn
RQPCBA
][?
][?
][?
][?
][][?
......
1
Instantaneous Rate of Reaction
dtdn
dtd J
Jv
1A+2B3C+D
dt
Bd
dt
Ad
dt
Cd
dt
Dd ][][][][21
31
...... RQPCBA RQPCBA
0,JJJ nn
BACD RRRdt
DdR
21
31][
v is the same for all species in a reaction.
The initial rate of reaction for the decomposition of N2O5 in five experiments. The initial rate of disappearance of a reactant is determined by drawing a tangent to the curve at the start of the reaction.
2N2O5(g)4NO2(g)+O2(g)
00 || tdtd
tv
Instantaneous Rate of Reaction
For a homogeneous reaction,
dt
Jdv
J1
VnJ J /][
dt
dnv J
J1
For a heterogeneous reaction,
dt
dv J
J
1
The surface density of species J
The plot of the initial rate of decomposition of N2O5 as a function of initial concentration for the five samples in previous figure is a straight line. The linear plot shows that the rate is proportional to the concentration. The graph also illustrates how we calculate the rate constant, k , from the slope of the straight line.
00 || tdtd
tv
][| 520 ONconstv t
Initial rate=k x initial concentration
(a) The instantaneous reaction rates for the decomposition of N2O5 at five different times during a single experiment are obtained from the slopes of the tangents to the line at each of the five points. (b) When these rates (the slopes) are plotted as a function of the concentration of N2O5 remaining, the result is a straight line with a slope equal to the rate constant. In (b), we have indicated the rates by redrawing the tangents.
rate=k x concentration
tt ONconstv ][| 52
Illustration
• In the reaction 2NOBr(g)2NO(g)+Br2(g), the rate of formation of NO is 0.16 mmol/L/s. Calculate the rate of consumption of NOBr.
0,JJJ nn
2NO 2NOBr
dt
NOBrd
dt
NOd
NOBrNO 11
1-1- smmolL 16.0dt
NOd
dt
NOBrd
NO
NOBr
Classroom exercise• In the reaction 2CH3(g)CH3CH3(g), the rate of
consumption of CH3 is -1.2 mol/L/s. Calculate the rate of the reaction and the rate of formation of CH3CH3.
23
CH 133
CHCH
dt
CHCHd
dt
CHdv
CHCHCH
333
333
11
1-1-333 smolL 6.03
33 dt
CHd
dt
CHCHd
CH
CHCH
• a: order of reaction, determined by experiment.
• k: rate constant, determined by experiment.
aionconcentratkv )(
Order of a Reaction
(a) The concentration of the reactant in a zero-order reaction falls at a constant rate until the reactant is exhausted. (b) The rate of a zero-order reaction is independent of the concentration of the reactant and remains constant until all the reactant has been consumed, when it falls abruptly to 0.
v=constant
There are no images in this section of the chapter.
More General Cases:
...
...][][
baorderOverall
BAk bav
23
][
22
3
]][][[
)(3)(3
)(6)(5)(
3
HBrBrOk
lOHaqBr
aqHaqBraqBrO
tBrO
More Complicated Rate Laws
Order=a+b+…-p-q…
...][][
...][][qp
ba
QP
BAkv nAA+nBB+nCC+…nPP+nQQ+nRR+…
More Complicated Rate Laws
Order=a for A b for B,-p for P, -q for Q…
...][][
...][][q
Qp
ba
QkP
BAkv
nAA+nBB+nCC+…nPP+nQQ+nRR+…
(a) When sulfur dioxide and oxygen are passed over hot platinum foil, they combine to form sulfur trioxide. (b) The sulfur trioxide forms dense white fumes of sulfuric acid when it comes into contact with moisture in the atmosphere. The rate law for the formation of sulfur trioxide shows that its rate of formation decreases as its concentration increases, so the sulfur trioxide must be removed as it is formed if the reaction is to proceed rapidly.2SO2(g)+O2(g)2SO3(g) SO3(g)+H2O(g)H2SO4(l)
pSOv ][ 3
The determination of rate law
• Method of initial rates:
Only one reactant is left limited, all the rest are in large excess.
BAkv a
aAkv
0Bkk
True rate law:
By making B in large excess, [B]=[B]0
The rate law with respect to A can be determined.
The determination of rate law
• Isolation method:
Only one reactant is left limited, all the rest are in large excess.
BAkv a
aAkv
0Bkk
True rate law:
By making B in large excess, [B]=[B]0
The rate law with respect to A can be determined.
aAkv 00
00 log'loglog Aakv
Using the method of initial rates• The initial rates of the reaction of 2I(g)+Ar(g)I2(g)
+Ar(g) were measured as follows:[I]0/10-5
molL-1
1.0 2.0 4.0 6.0
v0/molL-1s-
1
a) [Ar]=1 mmolL-1
8.7x10-4 3.48x10-3 1.39x10-2 3.13x10-2
b) [Ar]=5 mmolL-1
4.35x10-3 1.74x10-2 6.96x10-2 1.57x10-1
c) [Ar]=10 mmolL-1
8.69x10-3 3.47x10-2 1.38x10-1 3.13x10-1
Determine the orders of reaction with respect to I and Ar and the rate constant.
00 logloglog Iakv I 00 logloglog Arakv Ar
2Ia
1Ara
00 logloglog Iakv I
-12-29 sLmol 109,87.8log kk
Log[I]0=0
Logv0=8.94,8.88,8.75
Integrated Rate Law
• Find the concentration of a reactant at time t from reaction order and initial concentration.
• First order reaction:
ktt eAA 0][][
Akdt
Ad
The characteristic shape of the graph showing the time dependence of the concentration of a reactant in a first-order reaction is an exponential decay, as shown here. The larger the rate constant, the faster the decay from the same initial concentration.
kteAA 0
How to find k (first order reaction):
ktAA t 0]ln[]ln[
Half-Lives of First Order Reactions
kk
A
A
AA
t
kt
ktt
693.02ln2/1
2/1][
][
][][
)ln(
)ln(
021
0
0
k
1Time constant:
The half-life of a reactant is short if the first-order rate constant is large, because the exponential decay of the concentration of the reactant is then faster.
Second Order Integrated Rate Laws
tAkA
tA0
0
][1][][
2Akdt
Ad
Prove above equation (classroom exercise).
The characteristic shapes (orange and green lines) of the time dependence of the concentration of a reactant during two second-order reactions. The gray lines are the curves for first-order reactions with the same initial rates. Note how the concentrations for second-order reactions fall away much less rapidly than those for first-order reactions do.
2Akdt
Ad
Second Order Integrated Rate Laws
BAkdt
Ad
ktAB
AA
BB00
0
00
/
/ln
For A+BP
01
10 AktA
A
First order reaction close to equilibrium
BA Akv AB
Bkv
BkAkdt
Ad
00 )( AkAkkAAkAkdt
Ad
0)(
Akk
kekA
tkk
kk
AkA eq
0
kk
AkAAB eq
0
0
k
k
A
BK
eq
eq
eqeq BkAk
0][][][ ABA
First order reaction close to equilibrium:general cases
...
b
b
a
a
k
k
k
kK
BAAb
b
a
a
k
k
kk ...'
''
Relaxation methods
/0
texx
ba kk 1
BA BkAkdt
Adba''
''''eqbeqa BkAk
eqbeqa BkAk
xkk
BxkAxkdt
Ad
ba
eqbeqa
)(
)()(
At T=T1
At T=T2
dtdx
dt
Ad
Exercise• The H2O(l)H+(aq)+OH-(aq) reaction relaxes to equilibrium
with a relaxation time of 37 μs at T=298 K and pH=7, pKw=14.01. Given that the forward reaction is first-order and the reverse is second-order overall, calculate the rate constants for the forward and reverse reactions.
]][[2212 OHHkOHk
dt
OHd
eqOHOH ][][ 22
xOHOHxHH eqeq ][][,][][
eqbeqa BkAk
xOHHkk
xkOHHkOHkxOHHkkdt
dx
eqeq
eqeqeqeqeq
)}][]([{
][][)}][]([{
21
2222121
)][]([1
21 eqeq OHHkk
eqeqeq OHHkOHk 221
1-6.55][H
)L mol( L mol2
21-
22
1 W
eq
W
eq
eqeq KO
K
OH
OHH
kk
1-2
7
1-2/12/12
1-2
L mol 100.2
L mol }{}][][)L mol {(1
k
KKKkOHHKk WWeqeq
-1-111
L mol 102.0 s 107.31
2 s mol L 104.11-7-5 k
-15-121 s 104.2L mol Kkk
An Arrhenius plot is a graph of ln k against 1/T. If, as here, the line is straight, then the reaction is said to show Arrhenius behavior in the temperature range studied. The activation energy for the reaction is obtained by setting the slope of the line equal to Ea/R.
RTEaAk lnln
Example
T/K 700 730 760 790 810 840 910 1000
k/L mol-1s -1 0.011 0.035 0.105 0.343 0.789 2.17 20.0 145.0
The rate of the second-order decomposition reaction of CH3CHO:
Find its activation energy and pre-exponential factor.
103 K /T 1.43 1.37 1.32 1.27 1.23 1.19 1.10 1.00
ln(k/L mol-1s -1) -4.51 -3.35 -2.25 -1.07 -0.24 0.77 3.00 4.98
RT
EAk alnln
1114 mol kJ 189molK J 3145.8K1027.2 aE
112117.27 s mol L 101.1smol L eA
More values are given in the Data Section
The variation of the rate constant with temperature for different values of the activation energy. Note that the higher the activation energy, the more strongly the rate constant varies with temperature.
dT
kdRTEa
ln2
)(
ln1T
ad
kdRE
RTEaAek /
This sequence of images illustrates the motion of two reactant molecules (red and green) in solution. The blue spheres represent solvent molecules. We see the reactants drifting together, lingering near each other for some time, and then drifting apart again. Reaction may occur during the relatively long period of encounter. To highlight the positions of the reactant molecules, the insets show the solvent molecules as a solid blue background.
We join this sequence of images at the moment when the reactant molecules are in the middle of their encounter. They may acquire enough energy by impacts from the solvent molecules to form an activated complex, which may go on to form products. Once again, the insets highlight the reactants by showing the solvent molecules as a solid blue background.
The reaction profile for the activated complex theory of reactions in solution, in which the reactants form an activated complex, provided they encounter each other with at least the activation energy.
A catalyst provides a new reaction pathway with a lower activation energy, thereby allowing more reactant molecules to cross the barrier and form products. Notice that Ea for the reverse reaction is also lowered on the catalyzed path.
A small amount of catalyst—in this case, potassium iodide in aqueous solution—can accelerate the decomposition of hydrogen peroxide to water and oxygen. This effect is shown (a) by the slow inflation of the balloon when no catalyst is present and (b) by its rapid inflation when a catalyst is present. )()()( 2232 gHgOHaqOH KI
The reaction between ethene, CH2CH2, and hydrogen on a catalytic metal surface. In this sequence of images, we see the ethene molecule approaching the metal surface to which hydrogen molecules have already adsorbed: when they adsorb, they dissociate and stick to the surface as hydrogen atoms. After sticking to the surface, the ethene molecule meets a hydrogen atom and forms a bond. At this stage (center), a ·CH2CH3 radical is attached to the surface by one of its carbon atoms. Finally, the radical and another hydrogen atom meet, ethane is formed, and the product escapes from the surface.
The internal structure of a zeolite is like a honeycomb of passages and cavities. As a result, a zeolite presents a huge surface area. It can also permit the entry and exit of molecules of a certain size into the active regions within the holes. This zeolite is ZSM-5.
The structure of a typical catalytic converter for an automobile exhaust. The gases flow through a honeycomblike porous ceramic support covered with catalyst.
The lysozyme molecule shown here is a typical enzyme molecule. Lysozyme occurs in a number of places, including tears and the mucus in the nose. One of its functions is to attack the cell walls of bacteria and destroy them. In this illustration the winding ribbon represents the long chain that makes up the molecule.
In the lock-and-key model of enzyme action, the correct substrate is recognized by its ability to fit into the active site like a key into a lock. In a refinement of this model, the enzyme changes its shape slightly as the key enters.
(a) An enzyme poison (represented by the mottled green rectangle) can act by attaching so strongly to the active site that it blocks the site, thereby taking the enzyme out of action. (b) Alternatively, the poison molecule may attach elsewhere, so distorting the enzyme molecule and its active site that the substrate no longer fits.
Case Study (a) A plot of Michaelis-Menten enzyme kinetics. At low substrate concentrations, the rate of reaction is directly proportional to substrate concentration. However, at high substrate concentrations, the rate is constant, as the enzyme molecules are “saturated” with substrate.
Case Study (b) A diagram of a synapse. The triangles represent neurotransmitters that travel from the neuron on the left to the receptors in the neuron on the right. The concentration of neurotransmitters in the synapse is controlled by enzymes.
Dopamine
• NH2
• OH• OH
Case Study (c) The neurons in the human brain affect how we think and feel and how we perceive reality, including chemistry books.
Accounting for the rate laws
• 25.6 Elementary reactions
• 25.7 Consecutive elementary reactions
• 25.8 Unimolecular reactions
Elementary reaction:unimolecular reaction
PA
Akdt
Ad
Elementary reaction:bimolecular reaction
BAkdt
Ad
PBA
OCHCHICHkv 233
CH3I(alc)+CH3CH2O-(alc)CH3OCH2CH3(alc)+I-(alc)
CH3I+CH3CH2O-CH3OCH2CH3+I-
Consecutive elementary reactions
Akdt
Ada
IkAkdt
Idba Ik
dt
Pdb
tkaeAA 0 tk
abaeAkIk
dt
Id 0
0Aeekk
kI tktk
ab
a ba
01 Akk
ekekP
ab
tkb
tka
ab
PIA ba kk PuNpU239day 2.35239min 2.35239
Rate Determining Step(RDS)
The elementary reaction that is much slower than the rest and governs the overall reaction
babtktk kkkee ab ,
01 AeP tka
The rate of a reaction is controlled by the rate-determining step (RDS). (a) If the rate-determining step is the second step, then the rate law for that step determines the rate law for the overall reaction. The orange curve shows the reaction profile for such a mechanism, with a high activation energy for the slow step. The concentrations of intermediates can usually be expressed in terms of reactants and products by taking into account the steps preceding the RDS. (b) If the rate-determining
step is the first step, then the rate law for that step must match the rate law for the overall reaction. Later steps do not affect the rate or the rate law. (c) If two parallel paths lead to products, the faster one (in this case, the lower one) determines the rate of the reaction.
Example
law. rate theFind
slow (g)IHCl(g)ICl(g)(g)H
fast HCl(g)HI(g)ICl(g)(g)H
:mechanism Proposed
(g)IHCl(g)2ICl(g)2(g)H
22
2
22
.[HI][ICl]Rate
is then law rate overall The
[HI]HCl][[HI]][[ICl]
:same theare ratesreaction reverse and forward thestage,fast in the that assumemay We
:Solution
[HI][ICl]Rate(slow) (g)IHCl(g)ICl(g)HI(g) :2 Step
[HI][HCl]Rate(fast) (g)HICl(g)HCl(g)HI(g) reverse. :1 Step
][[ICl]Ratefast HCl(g)HI(g)ICl(g)(g)H forward. :1 Step
[HCl]][H[ICl]
[HCl]
][H[ICl]2
[HCl]
][ICl][H1'
21
22
1'
2
212
22
1'
22
21
1'
21
overallk
kk
k
k
kk
kHk
k
k
Hk
The steady state approximation
0dt
Id
AkkI ba
AkIkdt
Pdab
00
0 1 AedteAkPt
tktka
aa
Using steady state approximation
Pre-equilibrium
PIBA
BA
IK
a
a
k
kK
BAKkIkdt
Pdbb
BAkdt
Pd
a
bab k
kkKkk
The kinetic isotope effect
DCvhcHCvhcNHCEDCE Aaa
~2
1~2
1)()(
21
1~2
1)()(
CD
CHAaa HCvhcNHCEDCE
eHCk
DCk
)(
)(
2/1
12
)(~
CD
CH
kT
HCvhc
DCvhcDCvhcHCvhcHCvhcNHCEDCE Aaa~
2
1~2
1~2
1~2
1)()(
The kinetic isotope effect
2/1
1)(~~2
1)()(
CD
CHAaa HCvHCvhcNHCEDCE
eHCk
DCk
)(
)(
2/1
12
)(~~
CD
CH
kT
HCvHCvhc
The kinetic isotope effect
The chemical equations for elementary reactions show the individual events that take place when atoms and molecules encounter one another. This illustration shows two of the steps believed to occur during the formation of hydrogen iodide from hydrogen and iodine vapor. In one, I2 I2 I2 I I, a collision between two iodine molecules results in the dissociation of one of them. In the second step, I H2 H HI, one of the I atoms produced in the first step attacks a hydrogen molecule and forms a hydrogen iodide molecule and a hydrogen atom.
• Many reactions occur by a series of elementary reactions. The molecularity of an elementary reaction indicates how many species are involved in that step.
reaction.cular termoleOO OO O
is reverse whoseOO OOO
or O32O :Overall
reaction.r bimolecula OOOO :2 Step
reaction.ar unimolecul OOO :1 Step
:ozone ofion Decomposit
33222
22233
23
223
23
In a chain reaction, the product of one reaction step is a reactant in a subsequent step, which in turn produces species that can take part in subsequent reaction steps.
Initiation
Propagation
Example (I)
2
2
2
radiationor heat 2
BrBrBr :nTerminatio
BrHBrBrH
HHBrHBr :nPropagatio
BrBrBr :Initiation
This flame front was caught during the rapid combustion that occurs inside an internal combustion engine every time a spark plug ignites gasoline vapor. This radical chain reaction occurs in automobile engines. The products, which are hot gases, push a piston out, initiating a chain of events that ultimately moves the vehicle.
Example (II): Branching Chain Reaction
HHOHO :Branching
OHOOH :Branching
HHH :Initiation
2
2
spark2
Unimolecular reactions
63HCcyclokv
AAAA
2Akdt
Ada
AAAA
AAkdt
Ada
PA
Akdt
Adb
02
AkAAkAkdt
Adbaa
Akk
AkA
ab
a
2
2363 CHCHCHHCcyclo
Akk
AkkAk
dt
Pd
ab
bab
2
Akdt
Pd
aba kkkk /
2Akdt
Pda
Akdt
Pd
Akk
Akkk
ab
ba
Akkk
k
k aba
a 11
[][or *][][* ''baba kAkAkAAk
The activation energy of a composite reaction
RTaE
a
RTbEb
RTaEa
a
ba
a
ba
eA
eAeA
k
kkk
/)(
/)(/)(
RTaEbEaEa
a
ba aaeA
AA /)()()(
)()()( aEbEaEE aaaa
