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Techniques to Study Kinetics (come up with mechanisms)
ultimately - conc as f(time)
1. pressure changes
- at least one component must be gas
2. spectroscopy- use Beer's law A ac
- only one species must absorb at particular l
3. Electrochemical methods- can monitor change in pH or conductivity of solution
4. Other more sophisticated methods
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dt
didt
idn
dni /V
dt i
V
ddt
dcidt
i
V
d
dt
dci
dt
Assuming closed systems:
aA + bB cC + dD
moles of i : ni= ni(t) = nio+ i
i= stoichiometric coef. - unitless - rcts, + pdts
dni=id noi= constant = extent of rxn = mol
const. V
For const. V: rate ui= = rate of species i
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dtNH
dc
dtH
dc
dtN
dc
dtd
V3
212
3121
32
1
23
1
2
2NHHNdt
Ndc
uuu
32
3
23
2
2NHNHdt
Hdc
uuu
dcNH3
dtu
NH3 2u
N2
2
3uH2
Example: N2 + 3H2= 2NH3
rate of disappearance N2=
rate of disappearance H2=
rate of formation NH3=
u(t) = rate of rxn is related to concentrations of various chemical
species present at time t. This relationship called rate law.
u= rate of reaction =
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]][[][
22 IBrkdt
BrId
]/[]['1
][2/1][][
2
22
BrHBrk
HBrk
dt
HBrd
which components of rxn are kinetically active, i.e. a
change in its concentration will alter the rate of rxn. or
included in rate lawRate law often looks like following:
u= k[A]a[B]b k = rate constant = f(T) c
Br2+ I2 = 2BrI
Br2+ H2 = 2HBr
must be determined experimentally, cannot be deduced
from stoichiometry of rxn.Example: rate law
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In general: [A]a[B]b[D]d
a = order with respect to species A
b = order with respect to species B
d = order with respect to species D
(a+b+d) = overall order of rxn
a can be fraction, negative
For 1st rxn: order w.r.t. Br2and I2is 1, overall order =2
rate law happens to reflect rxn stoichiometry
For 2nd rxn: concept of order does not apply
rate law is complicated and rxn occurs by multistep process
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If concentration of any kinetically active species is
nearly constant it can be incorporated into k and we get
pseudo nth order rxn. C = catalyst for instance:
CBAkdt
Ad
][
(pseudo 2nd order)
can use to determine the order (isolation method)
if [C]~ constant
BAkdt
Ad'
][
(3rd order)
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to find rate of reaction at any time, measure change of
concentration with time at constant T
rate at any time = slope of curve at that time
easier to compare integrated rate expression
time
Co
equilibriumConc.
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dt
Ad ][
kdtA
Ad
][][
tA
oA dtkA
Ad
0
][
][ ][][
ln[A]
[A]o kt
First Order Reactions
A products or A + B products
= k[A]
Plot ln[A] vs t slope = -k
u(t) =
[A] = [A]oe
-kt
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21][
2/][ln ktAA
o
o 2
121ln kt
kt
2ln
21
Half-life: time required for half of the specified
reactant to disappear
t = t1/2when [A] = [A]o/2
For first order reaction:
.)(21 concft
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Second Order Reactions
A products or A + B products
kdtA
Ad
2][][
tA
oA dtkA
Ad
0
][
][2][
][
1
[A
]
1
A o
kt
Plot 1/[A] vs t slope = k
dt
Ad ][ = k[A]2u(t) =
kt
o
AA 1
][
1
kt
oAA 1
][
1
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oAkt
1
21
Half-life for Second Order Reaction
t = t1/2when [A] = [A]o/2
.)(2
1 concinitialft
2
1
1
][
2kt
oo AA
oA
kt
][
1
21
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need to integrate but we have a problem
need to know how [B] varies with [A]
[A] = [A]o- /V and [B] = [B]o- /V
dt
Bd
dt
Ad ][][
= k[A][B]u(t) =
Second Order Reactions
kdtBAAd
][
/V = [A]o-[A] so [B] = [B]o-[A]o+[A]
kdt
AABA
Ad
oo
][
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method of partial fractions: for [B]o [A]o
AAB
q
A
p
AABA oooo
1
1 = p([B]o-[A]o+[A] ) + q[A] = p([B]o-[A]o) +(p+q)[A]
1= p([B]o-[A]o) 0 = p+q
oo ABp
1
oo ABq
1
A
oA AoAoBA
1
A
oA
oooo
A
oA
o AABAB
Ad
A
o
BA
Ad
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oo
ooo
ooo
oo BABA
ABkt
BAAABA
ABln1ln1
ktB
AAB
ABA
A
AB ooo
ooooo
ln1
ln1
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Second Order Autocatalytic Reaction
AB
[A] = [A]o- /V and [B] = [B]o+ /V
dt
Bd
dt
Ad ][][ = k[A][B]u(t) =
kdtBA
Ad ][
/V = [A]o-[A] so [B] = [B]o+ [A]o- [A]
kdt
AABA
Ad
oo
][
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method of partial fractions: for [B]o [A]o
AAB
q
A
p
AABA oooo
1
1 = p([B]o+[A]o-[A] ) +q[A] = p([B]o+[A]o) - (p-q)[A]
1= p([B]o+ [A]o) 0 = p-q p = q
oo ABp
1
A
oA AoAoBA
1
A
o
A
A
o
A oAoB
o
AoB
Ad
oA
o
BA
Ad
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Determining the Rate Law
1. Take c vs. t , guess n, plot as discussed
1st order ln c vs. t straight line n=12nd order 1/c vs. t straight line n = 2
- perform experiment at const. T where [B] and [D] are in excess
ba DBAkdt
Ad
DBAkdt
Adlogloglogloglog ba
2. Isolation method:
aAkdt
Ad ' where k' includes [B] and [D]
- can determine k' and a
- do for other components as well
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2 . Method of Initial Rates
- measure rates at initial concentrations
- run experiment at series of initial concentration
rates plot log ovs. log [A]o slope = a
3. Half-life Method
nth order: =k[A]n for n > 2
Determining rate law: need 2 different [A]o at const. T (k is same)
solve for n knowing [A]o(1), t (1), [A]o(2), t(2) from experiments
rate law must be of form = k[A]n
dt
Ad ][
1
o
o
1-n
o
1-n
o
1-n
o
1)-(n
1-n
o
1)-(n
21
21
1[A]
2[A]
1[A]
2[A]
21)[A]-k(n1)-(2
11)[A]-k(n
1)-(2
2
1
n
o
o
At
At
1-no
1)-(n
21
1)[A]-k(n
1)-(2t
au oo Ak'
dt
Ad ][
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A = B k1, k-1AB k1B A k-1
Reversible Reactions
- usually study reactions far from equilibrium where
reverse reaction is not important
For forward rxn:
[A] = [A]o- 1/V + 2/V and
[B] = [B]o+ 1/V - 2/V = [B]o+[A]o-[A] assume [B]o= 0
For reverse rxn:
For both rxns:
][][][
1
Akdt
Bd
dt
Ad
][][][
1 Bkdt
Bd
dt
Ad
][][][][
11 BkAkdt
Bd
dt
Ad
= -k1[A] + k-1([B]o+[A]o-[A]) = -(k1+k-1)[A] + k-1[A]odt
Ad ][
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tA
oA
dtAd
01-1o1-
)[A]k(k-[A]k
tAk A
oAo
|A)k(kln)k(k
11-11
1-1
tAk
Ak
o
o)k(k 1-1
1
1-11 A)k(kln
Solve for [A]
)k(k
)k(k
)k(k
)k(k
1
1-1
1
1-1
1
1
1-1
1
1-1
o
oo
Ak
t
ekAkteAkA
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Ea= activation energy f(T) - energy needed for rct to proceed to products
A = preexponential factor or frequency factor
Effect of Temperature on Reaction Rate
- most common relationship is exponential one
- reaction rate increases by a factor of 2 or 3 for each 10orise in T
expt'ly we see: plot of ln k vs. 1/T is linear
Arrhenius eqn. - empirical relationship2
ln
RT
E
dT
kd a
integrate ARTEk a lnln RT
Ea
Aek
or
will talk more about significance of A and Ealater
A has units of k
1st order k = s-1
2nd order k = M-1s-1
The higher Ea, the stronger the T dependence of k
Ea= 0 means rate independent of T
Ea< 0 means rate as T
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Elementary Reactions
- rxns that occur in a single step - only involve 1 or 2
molecules- most rxns do not take place in a single collision but have
mechanisms that involve several elementary reactions
molecularity of rxn - # of particles involved in simple collisionalprocess for elementary rxn
- theoretical concept, whereas order is empirical
unimolecular - one that takes place w/o collision
- spontaneous disruption or transformation
- radioactive disintegration
- absorption of radiation to give energy
bimolecular - 2 molecules collide
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tktko ekek
kkAC 21 12
21
11
Can calculate time at which [B] = max by setting 0dt
Bd
time
Conc.
[A]o
[A]
[B]
[C]
eqm.
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Parallel 1st Order Reactions
AU
A
VAW
AkAkkkAkAkAk
dt
Ad 321321
ktAA
o
][][ln [A] = [A]oe-kt
ktoeAkAk
dt
Ud 11
t
kt
o
U
U
dteAkUd
o 0
1 o1kt-
o1 A
k
keA
k
k-U-[U] o
k k
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products are in a constant ratio to each other
dependent on time, [A]oand k's.
okt-1 Ae-1k
kU[U] o okt-2 Ae-1
k
kV[V] o
okt-3 Ae-1k
kW[W] o If [U]o= [V]o= [W]o= 0
13
k
k
U
[W]
12
k
k
U
[V]and
time
Conc.
[A]o
[A]
[V]
[U]
eqm.
[W]
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steady state approximation - assumes that the conc. of
all rxn intermediates (B) remain constant thruout exp't.
Steady State Approximation
A = B k1, k-1B C k2
- usually intermediates are very reactive & thus are
present in small conc.
BkAk
dt
Ad11
BkBkAk
dt
Bd211
Bk
dt
Cd2
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0
dt
Bd
BkBkAk
dt
Bd211
21
1
kk
AkB
2121
kk
Akk
dt
Cd
dt
Cd
kk
Akk
kk
kAk
kk
AkkAk
dt
Ad
21
21
21
11
21
111 1
t
kk
kk
oeAA 2121
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when k2
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3 main rxn types:
1. initiation - unimolecular or bimolecular (can be
photoinitiated)
2. chain propagation - often bimolecular - one
intermediate and one substrate - intermediate is
product
3. termination - converts chain propagating
intermediates to stable molecules4. inhibition - rxn where product is destroyed, i.e. is rct
(intermediate is destroyed & no new one produced)
Chain Reactions
-rxn where intermediates are produced which generate
more intermediates thus propagating the rxn.
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C2H6 2CH3. k1 initiation
CH3.+ C2H6 CH4+ C2H5
.k2 initiation
C2H5.
C2H4+ H
.
k3 propagationH.+ C2H6 H2+ C2H5. k4 propagation
H.+ C2H5 C2H6 k5 termination
methane is a byproduct
Example: Ethane decomposition
C2H6C2H4+ H2
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dt
HCd 62
dt
Hd
dt
CHd 3
dt
HCd 52
k3[C2H5] -k4[H][C2H6] -k5[C2H5][H] =0
[CH3] = 2k1/k22k1[C2H6] - k2[CH3][C2H6] = 0
2k1[C2H6]=2k5[C2H5][H]
-k1[C2H6] - k2[CH3][C2H6] -k4[H][C2H6] +k5[C2H5][H]
k2[CH3][C2H6]-k3[C2H5]+k4[H][C2H6]-k5[C2H5][H]=0
1.
2.
3.
1+3 k2[CH3][C2H6]-2k5[C2H5][H]=0 Plug in for [CH3]
][][ 525624
523
HCkHCk
HCkH
From (1)
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5253525624621
5
62152
][][
2
2
HCkk
HCkHCkHCk
Hk
HCkHC
rearrange k3k5[C2H5]2 - k1k5[C2H6][C2H5] - k1k4[C2H6]2 = 0
][2
][4][][62
53
2
625431
2
62516251
52 HCkkk
HCkkkkHCkkHCkkHC
53
2
5431
2
5151
2
4
kk
kkkkkkkkk
'
][][ 625624
623 kHCkkHCk
HCkkH
]Hk[Ck'k]Hk[Ck]H[Ck
]H][CHk[Ckk-]H[C
k
2kk-]H[C-k 625
625624
62623462
2
12621
62
dt
HCd
=k[C2H6] first order
dt
HCd 62
Fl E t d St d St t A i ti
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Fluorescence Expt and Steady State Approximation
tris(2,2-bipyridine) ruthenium +2= R+2
R+2 + h R+2* absorption (rate: Iabs)
R+2* R+2 fluorescence
R+2* + Fe+3 R+3+ Fe+2 quenching (rate constant:kq)
03*2*2
*2
FeRkRkIdt
Rdqsabs
3
*2
FekkI
Rqs
abs
Intensity of fluorescence a[R+2*]
I
Io intensity when no quencher present ([Fe+3]=0)= intensity when quencher present
sqs
qsabs
sabso
k
Fekk
FekkI
kI
I
I ][
][/
/ 3
3
][1 3 Fe
k
k
I
I
s
qo
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I
Io
sqs
qsabs
sabso
k
Fekk
FekkI
kI
I
I ][
][/
/ 3
3
intensity when no quencher present ([Fe+3
]=0)=intensity when quencher present
][1 3 Fekk
II
s
qo
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Lindemann-Hinshelwood Mechanism
1. A + A A* + A k1 activation
2. A* + A A + A k-1 deactivation
3. A* P______ k2 spontaneous decay- unimolecular
A P k overall rxn
Unimolecular Reaction
(Lindemann-Hinshelwood Mechanism)
- unimolecular rxns seen as complicated rxn involving
multistep mechanism- collisions are bimolecular, how can a rxn be first order
- from exptl data, it has been seen that rate laws are often
1storder at high concentrations, but 2ndorder at low
concentrationslook at Lindemann-HinshelwoodMechanism tp help explain this
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step 1: excitation due to collision, A's translational
kinetic E or with an inert molecule N2or Ar
step 2: loss of energy due to collision of A* with A
step 3: truly unimolecular rxn.assume steady state approximation
0[A*]k-A*][k-][Ak
*21-
2
1 Adt
Ad
0kAk-
][Ak*
21-
2
1
A 21-
2
212
k][k
][Akk[A*]k
Adt
Pd
1. low P: fewer collisions, A* has time to decompose k2>>k-1[A]uB = k1[A]
2 second order
2. high P - more collisions [A] large k-1[A]>>k2first order (3rd step rate determining)
neither 1st or 2nd order2 limiting cases:
1
21
k
Akk
dt
Pdu
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heterogeneous catalyst - catalyst in different phase,
i.e. solid in gaseous rxn.
Homogeneous Catalysis (same phase)
catalyst - changes rate of desired reaction- changes mechanism to one with lower Ea
enzymes - biological catalysts - read about Michaelis-
Menten rxn.
homogeneous catalyst - catalyst in same phase as rxn mixture
G Ph C l i
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Gas Phase Catalysis
NO2+ SO2 NO + SO3 bimolecular
NO2oxidizes SO2in low Eabimolecular process
we spoke of autocatalysis - product accelerates rxn.
2SO2+ O22SO3 slow
termolecular rxn needs to take place in order to occur
2NO + O22NO2 fast
termolecular rxn with NO well known
In general: uncatalyzed A + B P slow uP=k[A][B]
catalyzed A + C X fast if k2>>k1
X + B P + C fast uP=k1[A][C]
catalyzed rxns often found expt'ly to be 1st order w.r.t. [catalyst]
C = catalystk1
k2
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Michaelis-Menten mechanism
-enzyme catalyzed reaction
Substrate = S products = P enzyme = E ES = intermediate
E + S = ES P + E ESkdt
Pd2
k2
k-1
k1
0][][ 211 ESkESkSEkdtESdSteady-state approximation
21
1
kk
SEkES
[E]o= [E] + [ES]
[S]o= [S] + [ES]negligible
only a little enzyme is added so [S] >> [E]
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21
1
kk
SESEkES oo
ESkk
Sk
kk
SEkES ooo
21
1
21
1
21
1
21
11kk
SEk
kk
SkES ooo
o
oo
Skkk
kk
kk
SEkES
121
21
21
1
ooo
Skkk
SEk
ES121
1
ESk
dt
Pd2
o
o
oo EkSkkk
ESkk
121
21 Since [S] = [S]o~ const.
oM
o
SK
Skk
2
1
21
k
kkK
M
Michaelis const. mol/L
oMoo
SK
ESk
dt
Pd
2
Rate varies linearly with enzyme conc.
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o
Ekdt
Pd][
rate of formation of product 1storder w.r.t. enzyme not so with substrate
oM
o
SK
Skk
2
max2
][ u oEk
dt
Pd0 order in S: so much substrate, rate is constant
- rate of formation of products is maximum
- maximum velocity of enzymolysis
- S forces eqm. to right so only dependent on P formation
when KM>> [S]o [S]olow reaction rate a[S]o and [E]o
M
oo
K
ESk
dt
Pd2
when [S]o>> KMk= k2
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Th t i l d l f d ibi h h
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2. Kinetic molecular theory
- calculate # of molecular collisions per unit volume per time and fraction of
those collisions whose energy exceeds some threshold energy
- atoms treated as rigid spheres - if collision betw. 2 spheres has enough
energy will react to form products
- does not deal with how large threshold energy must be or effectiveness of a
collision as a function of relative molecular orientation
Theoretcial models for describing how chem rxns occur
1. Activated Complex Theory or Transition State Theory
- whether or not rxn occurs determined by movement of atomicnuclei on a multidimensional potential energy surface
- valleys or holes = energy minima = stable molecular species
- heights or saddle points or cols = energy barrier to rxn
- rate = concentration of activated species x frequency with which
they pass over addle point
- need statistical mechanics to determine [activated species]
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Kinetic Model of Gases
3 assumptions
1. gas consists of molecules of mass m in ceaseless
random motion
2. size of molecules negligible dmol
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Look at volume swept out by sphere of radius d
where d = (dA+dB)/2
d
dA
dB
dBdB
dA= diam of molecule A
dB= diam of molecule B
assume center of one A molecule is at center of cylinder
when the center of a B molecule is within this volume, A & B collide
a collision occurs when the distance betw. centers becomes assmall as d
B molecule at rest, A molecule moves thru volume
d 2 3/
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Volume swept out per unit time = relcd2 = m3/s
218
Tkc Brel
BA
BA
mm
mm
relcrelative mean speedmean speed with which one molecule
approaches another
= reduced mass
kB=R/NA= 1.38 x 10-23J/K
s = d2= collision cross-section = m2
If there are (nBNA/V) B molecules per unit volume then the # of
collisions of one A molecule with the B molecules in this V is given
by the collision frequency, z
nB= moles BNA= Avogadros #
s
molecules
Tk
Pc
V
Nncz
B
relAB
rel
ss
If th i t b t ( N /V) A l l i thi l th
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If there is not one, but (nANA/V) A molecules in this volume then
the # of collisions betw. A and B molecules is given by the
collision frequency, ZAB
sm
moleculesBANTkVNn
VNncZ A
BAAABrelAB 3
222
1
8
ss
22
21
2
8
2
1AN
m
TkZ A
A
BAA
s
For a single gas molecule A = B
22
2
A
A
A m
m
m
# of collisions of A with other A molecules
in this volume. The is included so as
not to count each collision twice
3410AAZ collisoins/m3s for typical gas at std. conditions
H l t t t l ?
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How can we relate to rate laws?
A + B P u= k2[A][B]
Only a fraction of collisions successful, f
- energy must exceed a minimum value E > Ea
rate of rxn. = fZAB=
dt
Ad
NBAN
Tk
f AAB
22
1
8
s
BAN
Tkf
dt
AdA
B2
1
8
s
Experimentally we found: BAkdt
Ad2
AB NTk
fk2
1
2
8
s
Ea
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According to ArrheniusRT
a
Aek
2
Ea= activation energyA = preexponential factor
There exists a threshold energy below which rexn will
not proceed. For most rxns E >>kBT
The fraction of molecules with E > Ethresholdis given by
Boltzmann distribution
0
1
1
dEeTk
dEeTk
TkE
B
E
TkE
B
B
th
B
Eanyinmoleculesthatprob.
EEinmoleculesthatprob. th
E
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Tk
ETkE
TkE
E
TkE
B
thB
th
B
th
B
ee
e
e
10
0
f = fraction of molecules with E > Eth=RT
NETk
EAth
B
th
ee
RTNE
AB
AB AtheN
TkkN
Tkfk
/2
1
2
21
2
88
s
s
= Ae-Ea/RT
?
AB NTk
A2
1
8
s
Ea= E
thN
A
Arrhenius assumes A not f(T) here
shows slight dependence
l Ekd
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Arrhenius found 2ln
RT
E
dT
kd a If A ~ constant
A
BAth NkTRTNEk s
21
8lnln21ln
22
2
1ln
RT
E
TRT
NE
dT
kd aAth
Ea= EthNA+ RT if Ea>> RT can neglect
Then Ea= EthNA
AB NTk
A2
1
8
s
Arrhenius assumes A not f(T) here
shows slight dependence
Steric Effects
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Steric Effects
- experimental values of A often much less than theoretical
- orientation of molecules may be a factor in rxn
CH3I (g) + Rb(g)
Rb CH
H
ICH
H
I Rb
(not good) (good)
Instead of suse s* reactive cross-section s* = sp
p = steric factor p
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Activated Complex Theory
- look at reaction in terms of potential energy of rcts and
pdts & progress of rxn (reaction coordinate)
- formulate reaction rate in terms of properties of rcts &
activated complex for that reaction
- apply statistical mechanics to reactants & activated
complexes
Partition function for molecule:
- probability that molecule will exist in any one of its
possible energy states i aq (q is unitless)
i
TkE Bieq /
q gives indication of the average # of states that are
thermally accessible to a molecule at temperature T
C it titi f ti i t f titi f ti f
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q = qvqrqtqe
CBA
qom= standard molar partition function
DEo= difference in ground state of
products & reactants
= Eo(C) - Eo(A)Eo(B)
Can write partition function in terms of partition functions of
vibration, rotation, translation, electronic
BA
CC
eNqNq
Nq
K RTE
AmBAmA
AmC o
D
/
,,
,
//
/
D i ti f R t E ti ith A ti t d C l Th
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Derivation of Rate Equation with Activated Complex Theory
Rate of rxn. = # of activated complexes passing over top of
potential energy barrier per second
= x average frequency with which complex moves to
product side,
PCBA
Ck
bimolecular unimolecular decay
C
A+B
P
rxn coordinate
PotentialE
assume rcts in eqm withC
dtdPCk u
transition state
transition statecertain energetic configuration of maximum
potential energy
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BACC
K
RT
PC
barP 1where
P
RTBAKC BA
PRTKk
dtPd
BAk
dt
Pd2Experimentally
P
RTKkk2
k = frequency of vibration of= transmission coef. = success of vibration to
push complex thru transition state ~ 0.5 - 1
C
RTE
mBmA
mCARTE
AmBAmA
AmC oo eqq
qNe
NqNq
NqK
/
,,
,/
,,
,
//
/D
D
C Complex moves to products when a vibration
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C Co p e o es o p oduc s e a b a opushes it thru transition state
qv 1 eh / kBT 1Look at partition function of vibration
If hkBT
1 then
eh / kBT 1 h
kBT1
2
h
kBT
2
...1 h
kBT
uh
Tkq Bv Then CvC qqq
where
etrC qqqq
qC
kBT
hqC
Then write K
kBT
hK
where
RTE
mBmA
mCA oeqq
qNK
/
,,
, D
has one vibrational mode missing
and
RT k T RT
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k2kK
RT
P
kBT
h
RT
PK
KPRThTkk B2 Eyring eqn.
Thermodynamics of Activated Complex Theory
- look at Gibbs Energy, Enthalpy & Entropy of activation
D KRTG ln eqm const without vibrational mode of C#
RTGeK /D
RTGB eP
RT
h
Tkk /2
D
STHG DDD RTHRS eBek //2 DD
P
RT
h
TkB B Assume ~1where
RTTk
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RTH
RST
hPRkk B
DD 22 lnlnln
2
2 2ln
RT
H
TdT
kd D
2
2ln
RT
E
dT
kd a Arrhenius found
RTHEa 2D
RTERS aeeBek /2/
2
D
2/ eBeA RSD
RTEaAek /
2
experimentally
preexpoential factor
RTHRS eBek //2 DD P
RT
h
TkB B
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D 2ln/ln 22B
AReTRkhAPRS B
RTEH a 2D
STHG DDD
PRT
hTkB B A = preexponential factor
For bimolecular collision 2 molecules form complex 0DS
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