Cinètica Química i Dinàmica Moleculariqc.udg.edu/~vybo/DOCENCIA/CINETICA/Apunts.pdf · 2020. 2....
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Vyboishchikov (1) Cinètica Química i Dinàmica Molecular Sergei Vyboishchikov Tutories: • DESPATX: C3-106 dimarts i dijous 12:30–13:30 • No dono cites prèvies – veniu i parlem durant les hores indicades.
Transcript of Cinètica Química i Dinàmica Moleculariqc.udg.edu/~vybo/DOCENCIA/CINETICA/Apunts.pdf · 2020. 2....
Vyboishchikov (1)
Cinètica Química i Dinàmica Molecular
Sergei Vyboishchikov
Tutories:
• DESPATX: C3-106 dimarts i dijous 12:30–13:30
• No dono cites prèvies – veniu i parlem durant les hores indicades.
Vyboishchikov (2)
EVALUATION
Activity Percentage of
the total grade Date
1st partial exam (formal kinetics) 40% October 28
2nd (reaction dynamics) 40% December 20
Problem solution in the class 20% December 2
Recuperation exam 100% January 24
Classes: 2 lectures every first week (Monday and Thursday at 8:40 a.m.); 1 lecture and 1 recitation class every second week (Monday and Thursday/Friday) at 8:40 a.m. Attendance is not obligatory. For the recitation class you have to bring a printed problems booklet. Changes: November 5 instead of November 7 December 3 instead of December 9
Exams: Theory 75% + 25% problems. A problem only counts if solved completely!
Problem solution in the class: a kind of quiz (exam) where you have to solve some problems with the help of literature and/or of your teacher.
Recuperation: the partial exams along with the “Problem solution in the class” are recoverable in one single exam in which the entire content of the course enters. Language of lectures: English
recitation classes: English
exam tasks: English
exam answers: whatever
Vyboishchikov (3)
LITERATURE ATKINS P., DE PAULA J., 2006, Atkins’ Physical Chemistry (8th Ed.), Oxford University press, Oxford.
Not available on the internet; ask you professor how to get it if interested.
ARNAUT L., FORMOSINHO S., BURROWS H., Chemical Kinetics. From Molecular Structure to
Chemical Reactivity, 2006, Elsevier.
Available at http://www.sciencedirect.com/science/book/9780444521866
SCHMITZ K.S., Physical Chemistry, 2017, Chapter 15: Reaction Rates and Mechanisms Elsevier.
No longer available on the internet; ask you professor how to get it if interested.
Vyboishchikov (4)
CHEMICAL KINETICS
• Why chemical kinetics?
► Thermodynamics does not tell us how fast a reaction is.
► Phenomenological (“formal”) Chemical Kinetics studies reaction
rates (= velocitats de reacció); Rate laws are either postulated or taken from experiment; Rate prediction using molecular data only is not possible; Determination of reaction mechanism is possible in principle (not always in practice).
► Reaction dynamics (= dinàmica molecular) studies what exactly happens in an elementary process: changes in geometric configura-tion, energy redistribution etc.
Intents to predict reaction rates for elementary reactions.
Vyboishchikov (5)
Reaction mechanism
• Elementary reactions:
► Those in which the reactants transform into products directly: A + B → Products • Most observed chemical reactions are complex (not elementary) ► A reaction is usually a succession of several elementary reactions
(steps = etapes); ► A product of one of the steps that serves as reactant for a subse-
quent step is referred to as intermediate. ► The mechanism is the succession of all the elementary steps
leading from the reactants to the products. • Example: 2H2 + O2 → 2H2O
H2 + O2 → 2OH·
OH· + H2 → H2O + H·
H· + O2 → HO2·
HO2· → O· + OH·
O· + H2 → H· + OH·
HO2· + H· → 2OH·
OH· + H· → H2O
2 HO2· → H2O2 + O2
H2 + HO2· → H2O2 + H
H· + HO2· → 2OH·
H· + HO2· → H2 + O2
H· + HO2· → H2O + O·
and much more steps.
Vyboishchikov (6)
Elementary reactions
► C2H5Br → C2H4 + HBr (gas phase)
► 2 ClO → Cl2 + O2
► 2 NOI → 2NO + I2
► Diels–Alder and many other cycloaddition (and retro-) and electro-cyclic reactions
are considered elementary (= reaccions senzilles); the contrary was not proven.
• Example of a non-elementary process: H2 + I2 → 2HI
► The kinetics is first order with respect to each component:
v = k[H2][I2] ► Nevertheless, the mechanism accepted now is as follows:
I2 →← 2 I· (fast)
I· + H2 →← H2I· (fast)
H2I· + I· →← 2 HI (slow)
Vyboishchikov (7)
Chemical kinetics
• Formal (phenomenological) kinetics studies reaction rates in a descriptive way.
► Direct problem: knowing the mechanism and the rate constants, de-termine the concentrations of species (reactants, products, and inter-mediates) at any moment of time t.
– deduce a rate law for a non-elementary reaction from a mechanism for instance, deduce the rate law v = k[H2][I2] for the H2 + I2 → 2HI reaction. In general a rate law is an analytic expression for the reac-
tion rate v as a function of reactant concentrations. Sometimes product concetrations can also enter a rate law.
► Inverse problem: knowing the concentrations [Ai]t of some (or even all) species at any moment of time t, determine the mechanism and the rate constants.
The inverse problem does not have a general solution. Typically, we have to conjecture a mechanism, solve the direct problem for that me-chanism, and compare the resulting [Ai]t with experimental observa-tions.
Vyboishchikov (8)
Reaction rates (for elementary reactions)
• Reaction rate (= velocitat de reacció) is a number of elementary events (elementary acts) per unit time and unit volume (usually divided by NA).
► The reaction rate is always positive by definition.
► Average reaction rate (= velocitat de reacció mitjana) in the interval of time [t, t +∆t] (if one reactant molecule disappears in one elementary event):
el-act reactant reactant( ) ( )[ , ] · ·
· ·
[reactant] [reactant]
1 1v
t t
A A
t
N N t t N tt t t
V t VN t
t
N
+∆
+ ∆ −+ ∆ = = =
∆ ∆
−=
−
∆−
since [x]A
xn N
V N V= =
► Instantaneous reaction rate at time t:
+
→
− = − = −
0
[reactant] [reactant] [reactant]( ) lim
t t t
t
dv t
t dt
∆
∆ ∆
► If two reactant molecules disappear in one elementary event:
1
2[reactant] [reactant]
( ) ( )2
d dv t v t
dt dt= − ⇒ = −
► Units: [concentration]/[time] ⇒ M·s–1
Vyboishchikov (9)
Reaction rates (elementary and complex reactions)
► Watch the stoichiometric coefficients: νAA + νBB → νCC + νDD
► Reaction rate with respect to a certain component:
[A] [B] [C] [D]; ; ;
;
A B C D
CA B DA B C D
A B C D
d d d dv v v v
dt dt dt dt
vv v vv v v v v
ν ν ν ν
= − = − = + = +
≠ ≠ ≠ = = = =
1 [A] 1 [B] 1 [C] 1 [D]
( )A B C D
d d d dv t
dt dt dt dtν ν ν ν= − = − = + = +
Vyboishchikov (10)
0 1 2 3
0.0
0.2
0.4
0.6
0.8
1.0
vlarge
t
vsmall
[A]/[A]0
Vyboishchikov (11)
Molecularity of a reaction (elementary reactions)
• Molecularity is a number of molecules participating in an elementary
event. This notion only applies to elementary reactions. νAA + νBB + νCC → Products
(νA + νB + νC ≤ 3) is the molecularity
► Unimolecular reactions: A → Products
► Bimolecular reactions: 2A → Products; A+B → Products
► Trimolecular (termolecular) reactions (rare (though not too rare) for molecules, more common for ionic gas-phase processes):
2NO + Cl2 → 2NOCl
2NO + O2 → 2NO2 (disputed) He
+ + He + He → He2
+ + He
— Understanding trimolecular reactions: · trimolecular reactions usually do not occur through simultane-
ous collision of 3 molecules. Rather, 2 molecules collide to form a short-living species, which a very short time later collides with the third molecule.
· Very often, the third molecule serves to absorb an extra energy that the reacting molecules have:
F· + F·
+ M → F2 + M
Vyboishchikov (12)
Kinetic law of mass action (elementary reactions)
• For an elementary reaction: νAA + νBB + νCC → Products
the rate v is given by the equation:
[A] [B] [C] CA Bk νν ν=v • Recalling that
1 [A]
A
d
dtν= −v
A rate law (rate equation) is obtained:
1 [A][A] [B] [C] CA B
A
dk
dtνν ν
ν− =
Attention: a differential equation! ► k is referred to as rate constant
► Units of k: [concentration]1–(νA+νB
+νC)/[time] ⇒
M1–(νA+νB
+νC)·s–1 depend on the reaction order: s–1 for first-order reactions, M–1·s–1 for second-order reactions.
► To solve a direct kinetic problem, we need k and have to solve a differential equation.
Attention: valid for elementary reactions only!
Guldberg and Waage
Vyboishchikov (13)
Rate “law” for complex reactions
• For a complex reaction: νAA + νBB + νCC → Products
no fundamental rate law exists! ► The law of mass action is still applicable to each elementary step
individually ► In many cases (but not always) the rate v of a complex reaction is ex-
pressed by an empirical equation (still called a rate law for a particu-lar reaction):
[A] [B] [C] CA B nn nk=v
where the powers nA, nB, nC are, in general, not equal to the stoichio-metric coefficients νA, νB, νC.
nA, nB, nC are orders with respect to reactants A, B, C.
(nA+nB+nC) is the total order of the reaction. nA, nB, nC may or may not be integer. Example: *IO3
– + I2 → IO3– + *I2 (isotope exchange)
v = k[IO3–]3/5[I2]
9/5[H+]9/5 ► In some cases other rate laws were found. Example: H2 + Br2 → 2 HBr
1/22 2
2
[H ][Br ]1 [HBr] / [Br ]
k
k
′=
+ ′′v
Vyboishchikov (14)
Half-life
• Halife (= temps de la semivida; not exactly the same as the mean life-time) t1/2 (τ1/2) is the period of time when the concentration of the reactant is a half of its initial value:
1/2 0
1[A] [A]
2t =
0 1 2 30.00
0.25
0.50
0.75
1.00
klarge
t1/2
[A]/
[A] 0
t
t1/2
ksmall
••
Vyboishchikov (15)
First- and second-order reactions – examples
Kinetic data for first-order reactions* Phase T, °C k, s–1 t1/2
2N2O5 → 4NO2 + O2
gas
HNO3 (liq.)
Br2(liq.)
25
25
25
3.38·10–5
1.47·10–6
4.27·10–5
5.7 h
131 h
4.5 h
C2H6 → 2 CH3· gas 700 5.36·10–4 21.6 min
gas 500 6.71·10–4 17.2 min
CH3N2CH3 → C2H6 + N2 gas 327 3.4·10–4 34 min
saccarose → glucose + fructose aqueous (H+) 25 6.0·10–5 3.2 h
Kinetic data for second-order reactions* Phase T, °C k, M–1·s–1
2NOBr → 2NO + Br2 gas 10 0.80
2NO2 → 2NO + O2 gas 300 0.54
H2 + I2 → 2HI gas 400 2.42·10–2
2I → I2 gas 23 7·109
CH3· + H2 → CH4 + H· gas 2000 8.1·108
CH3Cl + CH3O– → CH3OCH3 + Cl– methanol 25 2.3·10–6
H+ + OH– → H2O water
ice
25
–10
1.25·1011
8.6·1012
• The rate constant for a given elementary reaction under given conditions is a constant! It does not depend on other reactions occurring in the same time.
*Data from Laidler K. J., Chemical Kinetics (3rd Ed.), Harper and Row, New York, 1987.
Vyboishchikov (16)
First-order reaction kinetics
• First-order ordinary differential equation (k in s–1):
[A]
[A]d
kdt
− =
• Solution
► Variable separation:
[A][A]d
kdt− =
► Version 1 – Indefinite integration
[A][A]d
dt Ck +− =∫ ∫ ln[A] t Ck− = +
C is the integration constant, the value of which is determined
from the initial condition: [A](t=0) = [A]0 ⇒ C = –ln[A]0
0
0
0
ln[A]
[A]
ln[
[A]
A]
[A]ln
[A]exp( )kt
kt
kt
e kt−
− =
= −
= ≡ −
−
0[A] [A] kte−=
Vyboishchikov (17)
First-order reaction kinetics
• Solution
► Version 2 – Definite integration
( )
0
0
[A]
[A] 0
0
0
[A[A]
0
]0
[A][A]
ln[A]
ln[A] ln[A]
ln[A] ln[A]
[A]ln
[A]
t
t
t
t
t
t
t
dkdt
kt
kt
kt
kt
− = ′
− = ′
− − =
− = −
= −
∫ ∫
[B]
product
t1/2
[A]
t
•
reactant
0[Product] ][A] [A
t t= −
Vyboishchikov (18)
Exponential decay
• Exponential decay in the first-order kinetics: 0[A] [A] kte−=
► halflife:
− −= ⇒ = ⇒ − = −
=
1/2 1/200 1 /
1/ 2
2
ln
[A] 1[A] ln2
22 2
kt kt
t
e e kt
k
1 / 2/
0[A] [A] 2 t t−=
0 1 2 30.00
0.25
0.50
0.75
1.00
t1/2
[A]/
[A] 0
t
•
•t3/4
► t3/4 = 2 t1/2 (for first-order reactions only!)
Vyboishchikov (19)
Linearization in the first-order kinetics
• 0[A] [A] kte−=
► Linear form:
0ln[A] ln[A] kt= −
0 1 2 3
ln[A]
t
reactant
On treatment of experimental data, see Appendix 1 on page 129.
Vyboishchikov (20)
Second-order reaction kinetics – one substrate
• For a one-substrate reaction: 2A → P:
• Rate law (k in M–1·s–1):
– 21 [A]= A
2[ ]
dk
dt
• Solution
► Variable separation:
2
2
[A] [A]2 [A] 2
[A]
d dk kdt
dt− = ⇒ = −
► Integration:
2
[A]2
[A]1
2[A]
dkdt
t Ck +
− =
=
∫ ∫
► Initial condition:
( ) 0
0
1[A] 0 [A]
[A]t C= = ⇒ =
0
12
A] [A[1]
kt +=
Vyboishchikov (21)
Second-order reaction kinetics – one substrate
0
12
A] [A[1]
kt +=
0
0
[A]1 A
[A]2 [ ]k t
=+
► Half-life:
1/
0 0
0 1/2
0 1/2
20
[A] [A]2 1 2 [A]
1 2
12 [A
[ ]
]
A 2
k
tk
t
k t
=
=+
+ =
•t3/4
t1/2
[A]
t
•
reactant
► t3/4 = 3 t1/2 (for second-order reactions only!)
Vyboishchikov (22)
Second-order reaction kinetics – two substrates ([A]0 = [B]0)
• For a two-substrate reaction: A + B → P:
[A] [B][A][B]
d dk
dt dt− = − =
► If the initial concentrations are equal ([A]0 = [B]0):
2[A][A]
dk
dt− =
The same equation as for one substrate, except that there is no factor “2”.
0
1 1
[A] [A]kt= +
0
0
[A][A] [B]
1 [A]k t= =
+
► Half-life:
1/2
0
1
[A]t
k=
Vyboishchikov (23)
Second-order reaction kinetics – two substrates ([A]0 ≠ [B]0)
► If the initial concentrations are not equal ([A]0 ≠ [B]0):
− = ⇒ = −
− = ⇒ = −
[A] 1 [A][A][B] [B]
[A][B] 1 [B]
[A][B] [A][B]
d dk k
dt dt
d dk k
dt dt
Let’s subtract:
( )
( ) ( ) ( )
( ) ( )
( )
− = −
− = −
− = −
= −
1 [A] 1 [B][A] [B]
[A] [B]
ln[A] ln[B] [A] [B]
ln[A] ln[B] [A] [B]
[A]ln [A] [B]
[B]
d dk
dt dt
d dk
dt dt
dk
dt
dk
dt
Note that due to stoichiometry : [A]– [A]0 = [B]– [B]0 ⇒ [A]– [B] = [A]0–[B]0.
Thus, the right-hand side is a constant k([A]0–[B]0). Upon integration we get:
( )00 0
0
[A] [A]ln ln [A] [B]
[B] [B]k t− = −
or
0
0 0 0
1 [B] [A]ln
[A] [B] [A] [B]kt
−=
Vyboishchikov (24)
Second-order reaction kinetics – two substrates ([A]0 ≠ [B]0)
► An alternative (tedious, but more straightforward) derivation for [A]0 >[B]0
[A] [B][A][B]
d dk
dt dt− = − =
( ) ( )
( )( ) ( )( )
0 0 0 0
0 00 0
[A] [A]
[A][A
[B] [B] [A] [B] [A]
1[B] [A]
[B] [A] [A] [A]
] ][A] [Ad
k d kdtdt
= − − = + −
− = + − ⇒ = −+ −
( )( )( )
0 0 0
0 0 0 0
[A] [A] [B][A] [B] exp [
[A]B] [A]k t
−=
− −
Vyboishchikov (25)
Second-order reaction kinetics
• Linearization – second-order one-substrate reaction:
0
12
A] [ ]1
A[tk= +
t is used as abscissa and 1/[A] as ordinate. The slope will be 2k and the intercept 1/[A]0.
• Linearization – second-order two-substrate reaction, equal initial con-
centrations ([A]0 = [B]0):
0[A] [ ]
1 1A
tk= +
t is used as abscissa and 1/[A] as ordinate. The slope will be k (not 2k) and the intercept 1/[A]0.
• Linearization – second-order two-substrate reaction, different initial
concentrations ([A]0 ≠ [B]0):
( ) 00 0
0
[A][A] [B] ln
[A]ln
[ B] [ ]Btk= − +
t is used as abscissa and ln([A]/[B]) as ordinate. The slope will be k([A]0–[B]0) and the intercept ln([A]0 /[B]0).
Vyboishchikov (26)
Zeroth-order reaction kinetics
• − = ⇒ = −0
[A][A] [A]
dk kt
dt
0 1 2 3
[A]
t
•
reactant
?
► Zeroth-order kinetics can never take place for elementary reactions; ► Zeroth-order kinetics is quite typical of catalytic reactions, when a
reactant is in excess; ► Zeroth-order kinetics can never hold for low reactant concentrations; it be-
comes first or second order when a small amount of the reactants remains.
Vyboishchikov (27)
Summary on rate laws (kinetic equations) – (orders 0–1–2)
Order Explicit equation Linearized form
0 = −0[A] [A] kt
1 0[A] [A] exp( )kt= − 0ln[A] ln[A] kt= −
2 (one sub-strate)
0
0
[A][A]
1 2 [A]k t=
+ 0
1 12
[A] [A]kt= +
2 (two sub-strates,
[A]0 =[B]0)
0
0
[A][A]
1 [A]k t=
+ 0
1 1[A] [A]
kt= +
2 (two sub-strates,
[A]0 ≠[B]0)
( )( )( )
0 0 0
0 0 0 0
[A] [A] [B][A] [B] exp [
[A]B] [A]k t
−=
− − ( ) 0
0 00
[A][A] [B] ln
[A]ln
[ B] [ ]Btk= − +
Vyboishchikov (28)
•Methods of determination of reaction order – one substrate
• From a single kinetic experiment [A](t) through numerical differentiation:
t
reactant concentration[A]
t
v reaction rate
v = –d[A]/dt
ln[A]
ln v
► The order n is obtained from the slope of the ln v(ln[A]) line, since
v = k[A]n ⇒ ln v = ln k + n ln[A]
► an entire high-quality smooth curve is necessary for a reliable nume-rical differentiation (a table is not enough) ⇒ permanent concentration monitoring
• From a single kinetic experiment [A](t) from characteristic times t1/2 and t3/4: ► Quite approximate; for details see page 31 below. • From a series of kinetic experiments – Method of initial rates (mètode de
velocitats inicials); for details see page 29 below. • From a series of kinetic experiments – From dependence of t1/2 upon [A]0:
for details see page 30 below. ► All the methods are applicable both to one- and two-substrate reactions.
Vyboishchikov (29)
Methods of determination of reaction order – two substrates
• Method of initial rates (mètode de velocitats inicials)
0 0 0
0 0 0l
[A] [B]
ln ln[A] ln[ ]n B
n m
k
k
v n m
v == + +
► The initial rates are obtained from finite differences:
0
00
[A] [A][A] t
t
dv
dt t∆
∆≈
=
−= −
► Multilinear fit (least-squares method) in coordinates lnv0(ln[A]0, ln[B]0) (adjust multilineal, mètode dels mínims quadrats)
► Alternatively, if multilinear fit is not available (exam!):
– Fixing a certain [B]0, then doing linear fit (least-squares method) in coordinates lnv0(ln[A]0)
– Fixing a certain [A]0, then doing linear fit (least-squares method) in coordinates lnv0(ln[B]0)
Vyboishchikov (30)
•Methods of determination of reaction order – one substrate
• For a one-substrate reaction:
[A]
[A] [A] [A]n ndk d kdt
dt−− = ⇒ = −
( )
0
[1
A
0] 0
1]
[A
[A1
[A] [] A A[ ] ]1
t
nn nd kdt ktn
− −− −−
= − ⇒ =∫ ∫
Non-linear equation with respect to n. ► Consider the half-life t1/2:
1
100 1/2
1
100 1/2
1
0 1/21
1
1/2 1
0
[A]1[A]
1 2
[A][A] ( 1)
2
1[A] 1 ( 1)
2
2 1 1
( 1) [A]
n
n
n
n
n
n
n
n
ktn
kt n
kt n
tk n
−
−
−
−
−
−
−
−
− = −
− = − − = −
−=
−
A similar derivation is valid also for any characteristic time (t3/4, t1/10 ...)
Note that t1/2 is always inversely proportional to k.
1
1/2 0
2 1ln ln ( 1)ln[A]
( 1)
n
t nk n
− −= − −
−linear equation in coordinates ln[A]0/lnt1/2!
Vyboishchikov (31)
Methods of determination of reaction order – one substrate
−
−
−=
−1 / 2
0
1
1
2 1
( 1)[A]
n
nt
k n
► a similar expression for t3/4:
1
3/4 10
4 1
( 1)[A]
n
nt
k n
−
−
−=
−
► Dividing one expression by the:
( )2113/ 4
1 11/2
1
2 14 1
2 1 2 1
2 1
nn
n n
n
t
t
−−
− −
−
−−= =
− −
= + 0 10 20 30 40 50 60 70 80
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
C (
mol·l–
1)
t (s)
t1/2
t3/4
3/ 4
1/2
2
log 1 1t
nt
= − +
This formula serves to quickly estimate n from t1/2 and t3/4; the accuracy of this method is not very high
Vyboishchikov (32)
•Methods of determination of reaction order – two substrates
• How to reduce a problem of a two-substrate reaction to one substrate: Ostwald’s isolation method:
v0 = k[A]n[B]m Three key steps:
1) Maintaining a large excess of reactant A: [A]0 >> [B]0. In this case, rela-
tive change in [A] during the reaction is small ([A]≈ const) and we can follow the kinetics with respect to [B]. Eventually, we can find the or-der m with respect to B and an effective rate constant k', which will be
k[A]n0.
2) Maintaining a large excess of reactant B: [A]0 << [B]0. Now, the relative change in [B] during the reaction is small ([B]≈ const); we now follow the kinetics with respect to [A]. Finally, we find the order n with res-pect to A and the effective rate constant k'', which will be k[B]m
0.
3) From the the effective rate constants k' and k'' and orders n and m ob-tained in two previous experiments:
k' = k[A]n0
k'' = k[B]m0.
we obtain the real rate constant k. Note: steps 1 and 2 can be repeated several times at various [B]0 and [A]0, respectively.
Vyboishchikov (33)
Complex reactions
• Main types:
Reversible:
1
1
A Bk
k�
−
Parallel:
1
2
A B
A C
k
k
→
→
Consecutive:
1 2
A B Ck k→ →
• The kinetics is described by a system of differential equations.
Vyboishchikov (34)
Reversible reactions
•
1
1
A B
B A
k
k−
→
→
( )
( )
1 1
0
1 1 0
1 1 1 0
[A][A] [B]
[B] [A] [A]
[A][A] [A] [A]
[A][A A]] [
dk k
dt
dk k
dt
dk k
dtk
= − +
= −
+
= − + −
= − +
−
−
−−
−= −1 1[B]
[A] [B]d
k kdt
► Separation of variables is possible also in this case:
1 1 1 0
1[A]
[A) []( A]d t
k k k− −=
− + +
( )
0
0
[A]
1 1 1 0[A] 0
1 1 1 01 1
[A]
[A] 0
[A][A]
[A]
1( ) [A]
1ln ( ) [A]
t
t
d dtk k k
k k k tk k
− −
− −−
=− + +
− − + + =+
∫ ∫
After a tedious algebra (see Appendix 2 on Page 133) we arrive to the final kinetic law:
1 1( )1 0 1 0
1 1 1 1
[A] [A][A] k k tk k
ek k k k
+= ++ +
−−−
− −
Vyboishchikov (35)
Reversible reactions
1 11 1 0
11
)
1
(0
1
[[A] A[A]
] k k tk
k k
k
k ke +=
+++ −
−−
− −
• Behavior at infinite time:
1 00
1 00
1 1
1 0
1
eq1
1 1
1
1
[
[A]
A][B] [A]
[B
[A]
[A][A] [A
]
]
][A
kk
k k
kK
k
k
k
k
k
k
∞
∞
∞
∞
∞
=
= = −
=
++
=
+
− =−
−
−
− −
−
( )1 1 1 1( ) ( )
0[A] [A] [B] [A] [A] [A]k k t k k te e− −− + − +∞ ∞ ∞ ∞= + = + −
► Linearization:
( ) ( )0 1 1ln [A] [A] ln [A] [A] ( )k k t∞ ∞− = − − +
−
allows us to determine k1+ k–1. To find the individual constants k1, k–1, Keq = k1/k–1= [B]∞/[A]∞ is needed.
Vyboishchikov (36)
Reversible reactions
► k1 > k–1:
0 1 2 30.00
0.25
0.50
0.75
1.00
product
t
[A], [B]
reactant
Vyboishchikov (37)
Reversible reactions
► k1 < k–1:
0 1 2 3
0.00
0.25
0.50
0.75
1.00
product
t
[A], [B]
reactant
Vyboishchikov (38)
Parallel (competitive) reactions
1 2A B, A Ck k
→ →
( )1 2 1 2
1
2
[A][A] [A] [A]
[B][A]
[C][A]
dk k k k
dt
dk
dt
dk
dt
= − − = − +
=
=
( )
( ) ( ) ( )( )
1 2
1 2 1 2 1 2
0
1 01 1 0 1 0
1 20
[A] [A]
[A][B][A] [A] [B] [A] 1
k k t
t
k k t k k t k k t
e
kdk k e k e dt e
dt k k
− +
− + − + − +
=
= = ⇒ = = −+∫
Analogously, ( ) ( )( )1 2 1 22 02 0
1 20
[A][C] [A] 1
t
k k t k k tkk e dt e
k k− + − += = −
+∫
Thus, 1
2
[B][C]
k
k=
at any moment of time – for same-order reactions only
► Linearization:
( )0 1 2ln[A] ln[A] k k t= − +
Vyboishchikov (39)
Parallel (competitive) reactions
0 1 2 3
0.00
0.25
0.50
0.75
1.00
product
product
t
[A], [B], [C]
reactant
Vyboishchikov (40)
Consecutive reactions
1 2A B, B Ck k
→ →
1
1 2
2
[A][A]
[B][A] [B]
[C][B]
dk
dt
dk k
dt
dk
dt
= −
= −
=
1
1
0
1 0 2
[A] [A]
[B][A] [B]
k t
k t
e
dk e k
dt
−
−
=
= −
► The equation is not separable. There are at least two ways of solving this equations: multiplying by ek
2t (Appendix 3a on page 134) or
substitution [B] = K(t)e–k2t (see Appendix 3b on page 136):
( ) 11 21 0
0 1 22 1
1
[A][B] ; [A] ( )k t k kt tk
e e k kk k
k te−− −= − =−
► The solution for [C] is found by direct integration with the initial condi-tion [C](t = 0) = 0
( ) ( )
( ) ( )
1 2 1 2
1 2
1 0 02 2 0 2 1
2 1 2 1
2 10
2 1
[C] [A] [A][B] [C] [A]
1 1[A]
k t k t k t k t
k t k t
d kk k e e k e k e
dt k k k k
k e k e
k k
− − − −
− −
= = − ⇒ = − − =− −
− − −=
−
Vyboishchikov (41)
Consecutive reactions
► Comparable k1 and k2:
0 1 2 30.00
0.25
0.50
0.75
1.00
product
intermediate
t
[A], [B], [C]
reactant
Vyboishchikov (42)
Consecutive reactions
► k1 << k2:
0.0 0.5 1.0 1.5 2.0 2.5 3.0
0.00
0.25
0.50
0.75
1.00
product
intermediate
t
[A], [B], [C]
reactant
( ) ( ) ( ) ( )1 2 1
12 2
0 02
10
2 1
1 1 1[C] [A] [A] 1 [A]
k t k t k tk t
k e e k ee
k
k
k k
− − −−− − − −
= ≈ = −−
► The product formation rate is determined by k1 ⇒ A→B is the rate-
determining step (= rate-limiting step = etapa limitant de velocitat)
Vyboishchikov (43)
Consecutive reactions – induction time
► Product formation rate
0 1 2 3
d[C]/dt
t
product
► Induction time (= induction period = període d’inducció); maximum product formation rate, maximum concentration of the intermediate.
Vyboishchikov (44)
Consecutive reactions – rate-limiting step
► k1 >> k2:
0.0 0.5 1.0 1.5 2.0 2.5 3.0
0.00
0.25
0.50
0.75
1.00
product
intermediate
t
[A], [B], [C]
reactant
( ) ( ) ( ) ( )1 2 2
21 1
0 02
20
1 1
1 1 1[C] [A] [A] 1 [A]
k t k t k tk t
e k e k ee
kk k
k − − −−− − − − −
= ≈ = −− −
► The product formation rate is determined by k2 ⇒ B→C is the rate-
determining step.
Vyboishchikov (45)
Steady-state approximation
► k1 << k2:
0.0 0.5 1.0 1.5 2.0 2.5 3.0
0.00
0.25
0.50
0.75
1.00
product
intermediate
t
[A], [B], [C]
reactant
( )21 1
1
1 0 1 10
2 2 2
[A][B] [A] [A]k tk t k tk k k
e ek k
ek k
− − −= ≈ =−
−
This is equivalent to k1[A] – k2[B]≈0. k1[A] – k2[B] = d[B]/dt this is in turn equivalent to d[B]/dt≈0: (quasi-)steady-state approximation (aproximació de
l’estat (quasi)estacionari).
Vyboishchikov (46)
0.0 0.5 1.0 1.5 2.0 2.5 3.0
0.00
0.25
0.50
0.75
1.00
product
intermediate
t
[A], [B], [C]
reactant
Steady-state approximation
• Using the steady-state approximation
► [B] << [A] ; d[B]/dt≈0.
1
2
1
1 2
2 12
21
0
[A][B]
[A][A]
[A]
[B][A] [B]
[C][B [] A]
dk
dt
dk k
dt
dk k
d kt
k
k
kk
= −
≈
≈ =
⇒
=
≈= −
1[C]
[A]d
kdt
≈
In this example: k2/k1 = 12
( )1 1
1
1 0 0
1 1 0
2 2
[C][A] [C] [A] 1
[A] [A][B]
k t k t
k t
dk e e
dt
k ke
k k
− −
−
≈ ⇒ = −
≈ =
Vyboishchikov (47)
Steady-state approximation
• General procedure (Bodenstein 1913):
► Write down the mechanism.
► Write down the system of differential kinetic equations (rate laws) for all components (not always all of them are actually needed).
► Identify short-living intermediates and equate their rate of formation to zero.
► Express the concentrations of the short-living intermediates.
► Express the formation rate for the product.
Max Bodenstein (1871–1942)
Vyboishchikov (48)
Steady-state approximation – example
• Bodenstein’s original example (H2 + Br2 → 2 HBr): see Appendix 4 page 138.
• Another example: (2 N2O5 → 4 NO2 + O2, Ogg’s mechanism)
Total reaction: 2 A → 4 B + C
N2O5 → NO2 + NO3 k1 A → B + D
NO2 + NO3 → N2O5 k–1 B + D → A
NO2 + NO3 → NO2 + O2 + NO k2 B + D → B + C + E
NO + N2O5 → 3NO2 k3 E + A → 3 B
► Product B formation rate:
1 1 3
[ ][ ] [ ] [ ][ ]3[ ]D
d Bk A k B k A
dtE−= − +
► Formation rates for intermediates D and E:
1 1 2 1 1 2
2 3
[ ] [ ] [ ] [[ ]
[ ] [ ] ] ( )[ ] 0
[ ] [ ] 0[
[ ]
] [ ][]E
E
dk A k B k B k A k k B
dt
dk B k A
dt
DD D D
D
− −= − − = − + =
= − =
Vyboishchikov (49)
Steady-state approximation – example ► Solving the system of equations with respect to [D] and [E]:
1 1 2
2 3
1
1 2
2 2 1 2 1 1 2
3 3 1 2 3 1 2 3 1 2
[ ]
[ ]
[ ]
[
[ ] ( )[ ] 0
[ ] [ ] 0
[ ]( )[ ]
[ ] [ ] [ ][ ] [ ]
[ ]
[ ]( ] ( ( )
])[ )
k A k k B
k B k A
k A
k k B
k B k B
D
k A k k k k
k A k A k k B k
D
D
E
DE
k k k k k
−
−
− − −
− + =
− =
=+
= = = =+ + +
► Product B formation rate
1 1 3
1 21 1 3
1 2
3 1 2
1
2 1 21
[ ]
(
[ ][ ] [ ] 3 [ ]
4[ ] [ ] 3
[ ]
[ ]( )
[ ] [ ])[ ]
D
k A
d Bk A k B k E A
dt
k
k
kk A k B k A A
k k
k
k kk B
k
k
−
−−−−
= − +
+ ++= − + =
► Product C formation rate:
1 1 22 2
1 2 1 2
[ ] [ ][ ] [ ] [ ]
( )[
[]
]d C k A k k
k B k B Adt k k B
Dk k− −
= = =+ +
► Reactant A consumption rate:
− −− − −
= − + − = − + − = −+ + +
1 1 2 1 21 1 3 1 1 3
1 2 3 1 2 1 2
[ ] [ ] [[ ]
] 2[ ] [ ] [ ] [ ] [ ][ ]
( )d A k A k k A k k
k A k B k A k A k k Adt k k k k k k k
D E
Vyboishchikov (50)
Steady-state approximation – useful property ► In the previous example, we derived d[A]/dt, d[B]/dt, and d[C]/dt
independently and observed that:
= − =
[C] 1 [A] 1 [B]2 4
d d d
dt dt dt
This is a general property of the quasi-steady-state approximation: • The rate laws for any reactant or product obtained within the quasi-steady-
state approximation are all the same (within a constant factor equal to the ratio of stoichiometric coefficients).
► An important consequence of this is that we can first derive a rate law for the most convenient component and subsequently obtain it for the one we need, using stoichiometric coefficients of the total reaction.
► This is not true for the exact solution. ► Consider, for example, the following reaction: A + B → C occuring according
to the following mechanism (exam 30/10/2017): 2A → X k1 2B + X → Y + C k2 Y → C k3
The derivation of d[A]/dt is very simple, because A occurs only in the first
equation: d[A]/dt = –2k1[A]2. Then, from stoichiometry we know that
d[C]/dt = – d[A]/dt. Therefore, d[C]/dt = 2k1[A]2. An independent derivation of d[C]/dt would be much more sophisticated, since expressions for [X] and [Y] would be needed first.
Vyboishchikov (51)
Pre-equilibrium approximation
• Consider a complex reaction
1 2
1
A B Ckk
k�
−
→
in which k–1 >> k2. Let us apply the steady-state approximation to this system:
11 1 2
1 2
[B] [B][A] [B] [B] 0
[A]d k
k k kdt k k−
−= − − ≈ ⇒ =
+
Neglecting k2 in the denominator, we get
1eq
1
[B][A]
kK
k−≈ =
This means that [B] can be obtained simply from the equilibrium constant
of the A →← B process without writing down a full equation for the steady state: [B] = Keq[A].
► Product formation rate:
1
2 2 2 eq1
[C][B] [A] [A]
d kk k k K
dt k−= = =
• The pre-equilibrium (prior equilibrium, quasi-equilibrium) approximation can be used for more complex reactions:
1 2 1
1 21 2A B B ... B C
n n
n
k k k k
nk k k
+
− − −
→� � �
Vyboishchikov (52)
If k–n >> kn+1 , we can simply write down the [Bn] concentration and d[C]/dt
as follows:
1 2
1 1 1 eq1 2
...[C][B] [A] [A]
...n
n n nn
k k kdk k k K
dt k k k+ + +− − −
= = =
► Note that we can apply the pre-equilibrium approximation not in all cases where we encounter a reversible reaction step. We have to be sure that the condition k–n >> kn+1 is fulfilled in order to apply it.
► It is not required that kn+1 >> kn. This is different from the steady-state approximation.
Vyboishchikov (53)
Dependence of rate constants on temperature
• A simple rule of thumb: Increase in temperature by 10 °C (=10 K) typical-
ly results in an increase in reaction rate by a fac-
tor of 2 to 4: k = k(T) = abT /10K (2<b<4).
• Arrhenius equation (1889) – originally due to
Hood and van’t Hoff (1884)): ( )aE
RTk k T Ae−
= =
► Linearized form:
1ln ln ·a
EA
TRk = −
Jacobus Henricus van’t
Hoff (1852–1911)
Svante Arrhenius
(1859–1927)
Vyboishchikov (54)
Arrhenius equation – interpretation
• Not all the molecules A are able to react; only those “active” molecules A* with energy higher than a certain threshold (=llindar) Ea will do. This Ea is called activation energy.
► Arrhenius’ idea: “active” and “non-active” molecules are in equilibrium:
A →← A*:
*[A ][A]
aE
RTe=−
where exp(–Ea/RT) gives a share of molecules that have an energy greater than Ea.
Vyboishchikov (55)
Activation energies – examples
Vyboishchikov (56)
Modified Arrhenius equation
• A more accurate representation of the dependence of rate constants on temperature (cf. Llibret de problemes, problem 25, page 6):
( )E
m RTk k T aT e′
−= =
(modified Arrhenius equation – due to Kooij and van’t Hoff (1893))
► Relation to the Arrhenius equation is as follows:
ln ln ln
Em RT
Ek aT e k a m T
RT
−⇒ −
′ ′= = +
ln ln
aEaRT
Ek Ae k A
RT
−= ⇒ = −
Differentiating both equations yields the following:
2 2
lnd k m E mRT E
dT T RT RT
+′ ′= + =
2
ln aEd k
dT RT=
Comparing these two equations we conclude that Ea = E' + mRT .
Vyboishchikov (57)
Catalysis – general concepts
• Catalysis is the phenomenon of increasing reaction rate due to a presence of a substance (catalyst) that is not consumed in the reaction (Ostwald's definition) Concept coined by Berzelius in 1836:
conversion of starch to sugars by acids, combustion of hydrogen over platinum, decomposition of hydrogen peroxide by metals are early examples.
► A catalyst cannot alter the equlibrium constant (or the reaction free energy ∆rGº) – this would be against the 1st law of thermodynamics (think why!).
► A catalyst decreases the activation energy Ea (and
the activation energy free energy ∆‡Gº – to be dis-cussed later).
► A catalyst alters the reaction mechanism. Thus, the catalyst does participate in the reaction by form-ing intermediates. Heterogeneous catalysis always includes adsorption (chemisorption or physisorption) of a reactant on the catalyst surface.
Jöns Jacob Berzelius
(1779–1848)
Wilhelm Ostwald
(1853–1932)
Vyboishchikov (69)
Reaction dynamics
• The ultimate purpose of all the theories of the elementary act of reaction
(= reaction dynamics) is to calculate the rate constant from some
molecular data.
• In other words, Reaction Dynamics is the study of the mechanism of an
elementary chemical reaction at molecular level.
Vyboishchikov (70)
Potential energy surface – where does it come from?
• In Quantum Mechanics, a molecule is a collection of interacting nuclei and electrons that behaves as a wave and is described by a wavefunction Ψ(R1,..RN, r1,...rn)
where Ri = Xi, Yi, Zi are coordinates of ith nucleus ri = xi, yi, zi are coordinates of ith electron.
Time-independent Schrödinger equation: ĤΨ = EΨ where Ĥ(R1,..RN, r1,...rn) = TN(R1,..RN) + T e(r1,...rn) + VNN(R1,..RN) +
+VeN(R1,..RN, r1,...rn) + Vee(r1,...rn) is the molecular Hamiltonian Such a description is exact and universal, but hardly useful.
• It is very common to use the Born–Oppenheimer approximation, in which the motion of nuclei and electrons is separated:
Ĥe(R1,..RN|r1,...rn)Ψe(R1,..RN|r1,...rn)= Ee(R1,..RN)Ψe(R1,..RN|r1,...rn)
Now the electronic energy Ee depends on nuclear coordinates as parame-ters, i.e., every nuclear configuration R1,..RN corresponds to its own electronic energy Ee(R1,..RN). This Ee serves as potential in the Schrödinger equation for the nuclear motion:
ĤN(R1,..RN)ΨN(R1,..RN) = EN ΨN(R1,..RN)
where ĤN(R1,..RN) = TN(R1,..RN) + Ee(R1,..RN)
Vyboishchikov (71)
Potential energy surface – where does it come from?
• The function Ee(R1,..RN) is referred to as the potential energy surface (PES, superfície d’energia potencial). Its physical meaning is that every nuclear configuration R1,..RN corresponds to a certain energy Ee.
• In general, the motion of nuclei is described by the nuclear Schrödinger equation (see page 70 above), but sometimes we can consider the nuclear motion as classical (not quantum-mechanical) motion in the same potential.
• Partial derivatives of the PES with respect to each coordinate (with oppo-site sign) gives the corresponding component of force acting on the atoms:
1 1 11 1 1
; ;x y z
E E EF F F
x y z
∂ ∂ ∂= − = − = −
∂ ∂ ∂ in Cartesian coordinates
Note: the vector of all the partial derivatives of a function ƒ(x1,y1,z1,x2,y2,z2, ..,xN
,yN,z
N) is referred to as gradient of ƒ:
grad ƒ(x1,y1,z1,x2,y2,z2, ..,xN,yN,zN) ≡ ∇ƒ(x1,y1,z1,x2,y2,z2, ..,xN,y
N,z
N) =
= (∂ƒ /∂x1, ∂ƒ /∂y1, ∂ƒ /∂z1, ∂ƒ /∂x2, ∂ƒ /∂y2, ∂ƒ /∂z2,..,∂ƒ /∂xN,∂ƒ /∂y
N,∂ƒ /∂z
N)
Vyboishchikov (72)
Potential energy surface – basic idea
• Typically, in the course of reaction there is some bond formation or bond
breaking or both.
► Some characteristic distances in the molecule change strongly in the
course of reaction since the distance between bonded atoms is usually
much smaller than between non-bonded ones.
► We need some geometric parameters to characterize these changes.
Vyboishchikov (73)
Potential energy surface – internal coordinates and degrees of freedom • Consider an entire reacting molecular system, consisting of two molecules (for a
bimolecular reaction) or of a single molecule (for a unimolecular reaction).
► If this system has N atoms, we have 3N degrees of freedom and need 3N Cartesian coordinates {x1, y1, z1, x2,y2,z2, ..,xN
,yN,z
N} to describe its structure.
► Instead, we can choose: – some 3 coordinates (e.g., X,Y,Z of the center of mass) to characterize the
translational motion of the entire system; and – some 3 coordinates (e.g., angles φ, θ, ψ) to characterize the rotational
motion of the entire system; – For a linear system (e.g., a diatomic molecule) there are only 2 (not 3)
rotational coordinates;
– some 3N–6 (or 3N–5 in the linear case) coordinates are needed to describe vibrations and chemical changes;
These are (non-linear) combinations of the Cartesian coordinates are called internal coordinates; commonly used are:
– interatomic distances dAB; angles αABC; dihedral angles βABCD.
• Instead of talking of the motion of N atoms in our usual 3D physical space, we can unite all the 3N coordinates (or 3N–6) and consider they as a coordinate of a single figurative point in a 3N-dimensional space (or 3N–6 dimensional)
► The motion of the entire molecule or a reacting system is then represented by the motion of that figurative point.
Vyboishchikov (74)
Potential energy surface – diatomic molecule
• For a diatomic molecule AB, upon separation of 3 center-of-mass coordinates and 2 rotational angles, only one internal coordinate dAB remains.
0.5 1.0 1.5 2.0 2.5 3.0
-1000
-800
-600
-400
-200
0
e
e– D
rA+B
E(dAB
)
minimum AB
dAB
•
• Minimum position corresponds to the equilibrium distance re and disso-ciation energy De.
► When dAB >r0, –∂E /∂dAB <0 and there is a attraction force returning the
molecule to the equilibrium distance. ► When dAB <r0, –∂E /∂dAB >0 and there is a repulsion force also returning
the molecule to the equilibrium distance.
Thus, vibrations take place.
Vyboishchikov (75)
Potential energy surface – diatomic molecule
• Analytic approximations to the PES for a diatomic molecule AB: ► Harmonic approximation (suitable to describe small vibrations, but not
dissociation): r = dAB
( )2( )2 ek
E r r r= − k =d2E(r) /dr is the force constant (ν is the vibrational frequency)
12
kν
π µ=
0.5 1.0 1.5 2.0 2.5 3.0
-1000
-800
-600
-400
-200
0
e
e– D
rA+B
E(dAB
)
dAB
•
► Morse potential (also an approximation, but properly describes dissociation as well as anharmonicity):
( ) ( )2 ( ) ( ) 2( )( ) 12 e eer r r r r
er
e eE r D e D ee Dββ β− − − −− −= −= − −
r E(r
Vyboishchikov (76)
Potential energy surface – triatomic system
• The PES for a triatomic system is a function of 3×3–6=3 variables ⇒ no gra-phical representation is possible. In this case we consider a reduced poten-
tial energy surface, choosing a subset of variables (usually interatomic dis-tances, but possibly angles or dihedral angles) whose changes represent the process under study.
►For a chemical process A + BC → AB + C, it is common to consider the re-duced PES in coordinates E(dAB, dBC) (bond B–C breaking, bond A–B forming):
dAB
dBC
dAB
dBC
dAB
dBC dAB
dBC
atoms
+
Vyboishchikov (77)
Potential energy surface – various trajectories
• Various trajectories:
A–B vibrations A + BC → AB + C reaction
• The reaction coordinate (reaction path(way)) (= coordenada de reac-
ció, camí de reacció) is a path (curve) on the PES along which the en-ergy is a minimum with respect to any lateral displacement.
• The transition state (= estat de transició) is the highest-energy point on the reaction coordinate.
► Usually, two minima are linked by a single reaction path conatining a transition state. However, sometimes, alternative pathways are possible.
Vyboishchikov (78)
Potential energy surface – reaction profile
exothermic reaction:
–∆rE
∆‡
E1 ∆‡
E–1
reactants
products
transition state
reaction coordinate
E
endothermic reaction:
∆rE
∆‡
E1 ∆‡
E–1
reactants products
transition state
reaction coordinate
E
• ∆‡E is referred to as the reaction energy barrier. Sometimes it is also called activation energy – be careful, this is not correct!
• ∆rE = ∆‡E1 – ∆‡E–1: reaction energy is related to the energy barriers for the forward and reverse reaction.
• The same philosophy applies to other types of energies: ∆‡G, ∆‡H... For instance, ∆rG = ∆‡G1 – ∆‡G–1.
Vyboishchikov (79)
Potential energy surface – stationary points
• Equilibrium configurations of molecules correspond to local energy minima: – Forces (gradient) vanish in a minimum: ∇E = 0; – At small deviation from a minimum, there are forces returning the system
to the minimum. • Transition states correspond to saddle points: – Forces (gradient) also vanish in a transition state; – A transition state is an energy maximum with respect to the reaction
coordinate, but a minimum with respect to all other coordinates:
• All the points on the PES where gradient vanishes (∇E = 0) are called station-
ary points. These are minima, transition states, higher-order saddle points, and maxima. The latter two types play no role in chemistry (see page 84).
Vyboishchikov (80)
Potential energy surface – search for and characterization of stationary points • The criterion of a stationary point is the zero gradient: ∇E(q1,q2,q3,..,qN) = 0,
where q1,q2,q3,.. are internal coordinates.
1
2
0
0
....
0N
E
q
E
q
E
q
∂ = ∂
∂= ∂
∂= ∂
The system of N equations with N variables is used to locate the stationary points. ►In one-dimensional case (e.g., for a diatomic molecule), a single equation appears:
0
dE
dq=
In this case, the characterization of the stationary point is based on the second derivative (force constant):
2
20
d E
dq>
2
20
d E
dq<
2
20
d E
dq=
minimum maximum inflexion point
Vyboishchikov (81)
Potential energy surface – search for and characterization of stationary points
• Characterization of a stationary point for a multi-dimensional function E(q1,q2,q3,..,qN) is more sophisticated. It is not enough to find the second-
order derivatives (force constants) ∂2E /∂q12, ∂2E /∂q2
2, ∂2E /∂q22,.., ∂2E /∂q
N
2.
► We need the entire second-derivative matrix, called Hessian matrix or force-constant matrix:
2 2 2
21 2 11
2 2 2
22 1 22
2 2 2
21 2
···
···
··· ··· ··· ···
···
N
N
N N N
E E E
q q q qq
E E E
q q q qq
E E E
q q q q q
∂ ∂ ∂ ∂ ∂ ∂ ∂∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂= ∂
∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂
H
This a symmetric N×N matrix. ► Eigenvalues λ i (i=1,...,N) of the Hessian matrix must be calculated. All
eigenvalues of a symmetrix matrix are always real (not complex), but may be positive or negative. Then, the following criteria apply:
Number of negative
eigenvalues 0 1 2 ··· N
Type minimum transition
state second-order saddle point
higher-order saddle point
maximum
Vyboishchikov (82)
Potential energy surface – search for and characterization of stationary points
• Example of calculation of eigenvalues for a 2×2 matrix:
2 2
2 211 121 1
21 122 221 22
22 1 2
;
E E
h hq qh h
h hE E
q q q
∂ ∂ ∂ ∂ = = = ∂ ∂ ∂ ∂ ∂
H
► Constructing the characteristic determinant:
( )( )11 12
11 22 12 2121 22
h hh h h h
h h
λλ λ
λ= −
−− −
−
► Equating the characteristic determinant to zero and solving the resulting quadratic equation (characteristic equation) with respect to λ:
( )( )( )
( )
( )
11 22 12 21
11 22 11 22 12 21
2
11 22 11 22 11 22 12 21
2
11 22 11 22 11 22 12 2
2
1
1
2
0
( ) 0
( ) ( ) 4
2
( ) ( ) 4
2
h h h h
h h h h h h
h h h h h h h h
h h h h h h h h
λ λ
λ λ
λ
λ
− =
− + + − =
+
− −
− + − −=
+ + −+ −=
Vyboishchikov (83)
Potential energy surface – search for and characterization of stationary points
• For larger N×N matrices, the characteristic equation will be of grade N, and we would need to find all the N roots (solutions) in order to characterize the stationary points.
► This is never done in practice. Usually, efficient methods of linear algebra (such as Jacobi or Householder diagonalization methods) are used to find the eigenvalues mith the help of computer programs.
• Note that we actually need the signs of the eigenvalues sign(λi) rather than λi’s themselves.
To determine the signs, the following simple rule (Sylvester criterion) can be employed:
11 12 1
21 22 2
1 2
···
···
··· ··· ··· ···
···
N
N
N N NN
h h h
h h h
h h h
=
H
where M1, M2,.., MN are principal
minors of H.
1 1 11
11 12
1 2 212 22
11 12 13
1 2 3 3 21 22 23
31 23 33
1 2 3
sign( ) sign( ) sign( )
sign( ) sign( ) sign
sign( ) sign( ) sign
sign( ... ) sign( ) sign(det( ))N N
M h
h hM
h h
h h h
M h h h
h h h
M
λ
λ λ
λ λ λ
λ λ λ λ
= =
= = = =
= =
···
H
Using the Sylvester criterion, we have to calculate N determinants of various size (from 1 to N), but we do not need to find the eigenvalues.
Note that the Sylvester criterion is not a method for finding eigenvalues!
Vyboishchikov (84)
Potential energy surface – Murrell–Laidler theorem
• A second-order (or higher-order) saddle point is a stationary point that
is a maximum in two (or more) coordinates, but minimum in all other
coordinates.
• The Murrell–Laidler theorem: If two local minima are connected by
a path involving a second-order (or higher-order) saddle point, then
another, energetically lower energy path exists that passes through a
transition state.
►This does not mean that a trajectory through a higher-order saddle
point is impossible. It just means that there is a better (=lower-
lying) trajectory leading through a transition state (and possible
other minima).
►Limitations of the Murrell–Laidler theorem are known, mainly in
highly symmetric systems.
Vyboishchikov (85)
Potential energy surface – various trajectories
• Reaction coordinate profile
A + BC → AB + C reaction
∆‡E is the reaction barrier.
–∆rE
∆‡
E1 ∆‡
E–1
reactants
products
transition state
reaction coordinate
E
• Not all the trajectories (even those with sufficient energy) lead to the reaction:
d BC
dAB
d BC
dAB non-reactive collision reactive collision; a vibrationally
excited product is formed
Vyboishchikov (86)
Potential energy surface – Hammond’s postulate
• Hammond’s postulate (=Hammond’s principle = Leffler’s assumption):
The transition state bears the greater resemblance (is more similar geo-metrically) to the less stable species (reactant or product):
exothermic reaction – early transition state
–∆rE
∆‡
E1 ∆‡
E–1
reactants
products
transition state
reaction coordinate
E
endothermic reaction – late transition state
∆rE
∆‡
E1 ∆‡
E–1
reactants products
transition state
reaction coordinate
E
Thus, for an exothermic reaction (see picture above), the transition state more resembles the reactant, and for an endothermic reaction, the product.
► Hammond’s postulate is not a fundamental law, but is usually fulfilled.
Vyboishchikov (87)
Potential energy surface – various trajectories
A + BC → AB + C reaction
exothermic reaction – early transition state
endothermic reaction – late transition state
dBC
dAB
dBC
dAB Large vibrational energy – non-
reactive trajectory Large translational energy – reactive
trajectory
Large translational energy – non-reactive trajectory
Large vibrational energy – reactive trajectory
Vyboishchikov (88)
reactants
products
transition state
reaction cordinate
E •
complex
activated
Activated complex vs. transition state
• A transition state is a certain stationary point (first-order saddle point) on the potential energy surface (see Pages 77–81).
► It is a single point, and the reacting system has a zero probability of being exactly in this point.
• An activated complex is a small area surrounding a transition state.
► It has some extension in all directions and some volume (in the 3N–6-dimen-sional configuration space), though its exact size is not well defined and is arbitrary.
► The molecular system has a finite pro-
bability of being within the activated
complex ⇒ We can talk about the con-
centration of activated complex at a given moment of time.
Vyboishchikov (89)
Transition-state theory
• The Transition-state theory (activated complex theory, theory of abso-lute reaction rate, teoria de l’estat de transició = TET) uses the potential energy surface to calculate the rate constant of an elementary reaction.
• Assumptions of the TST: ► A molecular system moving on the potential energy surface toward the
product(s) and having reached the transition state, cannot go back to the reactant(s) – it will proceed to the product(s).
► The activated complex is in equilibrium with the reactant(s). • Therefore:
unimolecular bimolecular
A →← A‡ K‡ A + B →← (A···B)‡ K‡
A‡ → C k1 (A···B)‡ → C k1 ► The concentration of activated complex can be calculated from the equilibrium constant K‡:
‡‡ ‡ ‡[A ]
[A ] [A][A]
K K= =⇒ ( ) ( )
‡ ‡‡ ‡(A···B) / º
(A···B) [A][B][A] / º [B] / º º
[ ][ ]
c KK
c c c= =⇒
where cº is the standard concentration (1 M for solutions, 1 atm for gases). [A]/cº etc. are activities.
Vyboishchikov (90)
Transition-state theory – the fundamental equation
• The product formation rate:
unimolecular bimolecular
‡ ‡
1 1[C] / [A ] [A]d dt k k K= = = =‡ ‡1 1[C] / (A···B) / º [A][B][ ] )(d dt k k K c
A‡ → C k1 (A···B)‡ → C k1 ► The rate constant: k = k1K
‡ k = k1K‡/cº
► Equations of thermodynamics (or statistical thermodynamics) are applied
to obtain K‡: K‡ = e–∆‡Gº/RT.
· k1 is shown to be kBT/h for all reactions (h is the Planck constant). For
ambient temperatures, kBT/h is about 1013 s–1.
► For a full derivation, see next Page 91.
• We obtain the Eyring–(Evans)–Polanyi equation for the rate constant:
unimolecular bimolecular
‡B º /G RTk Tk e
h∆−=
‡B º /1º
G RTk Tk e
c h∆−=
► ∆‡Gº= Gº(TS) – Gº(reactants) is the standard activation free energy,
i.e., calculated at standard concentration cº (normally 1 M for solutions, 1 atm = 101325 Pa for gases).
Vyboishchikov (91)
transition state
reaction cordinate
E •
activated complex
δ
Transition-state theory – fundamental equation – derivation
• The Eyring–(Evans)–Polanyi equation derivation:
Consider again the expression for the activated-complex concentration for a bimolecuar reaction: [A···B‡] = K‡[A][B]/cº and express K‡ through partition functions:
‡
‡‡ /
A B
E RTQK e
Q Q∆−=
The total Q is a product of translational, vibrational and rotational partition functions: Q = QtransQrotQvib. Let us now separate from the Q‡ a part corres-ponding to a motion along the reaction coordinate. This motion is considered as motion of a particle in a box, i.e., as translational motion.
B
‡‡ ‡ 2 ˆ·m k T
Q Qh
πδ=
where δ is the size (length) of the activated complex along the reaction coordinate; Q‡ in-corporates all other contributions to Q‡.
Take into account that the mean relative velo-city v̄ of a one-dimensional motion is given by
B
‡2
k Tv
mπ=
Vyboishchikov (92)
The time necessary to traverse the activated complex of length δ with velo-city v̄ is:
B
tr
‡2t
v k
m
T
δ πδ= =
Then the rate v of this process (the number of molecules traversing the acti-vated complex per unit time) is:
-
-
-
= = = =
=
‡
‡
‡
B B B
B
B
B
tr
‡
‡‡ ‡
/‡ ‡
‡ ‡ ‡
‡
‡
/
/
[A][B][A···B ] [A···B ] [A][B] / º 1
º2 2 2
[A][B]1 1 = [A][B]
º º
2 ˆ· ˆ
2
E RT
A B
E RT
E RT
A B
A B
cv
t cm m m
k T k T k T
c cm
Q
k Tm k T
Q Qhh
eQ QK
ee
Q QQ Q
k T
π π πδ δ δ
πδ
πδ
∆
∆
∆
=
The rate found above is our reaction rate (!), because all the molecules cross-ing the activated complex are supposed to convert to the products. Thus, we obtain the expression for the rate constant:
‡B
‡ /
ˆ1º
E RT
A B
k T Qk e
c h Q Q∆−=
Vyboishchikov (93)
This is the Eyring–Polanyi equation for a bimolecular reaction expressed through partition functions.
Note that Q‡ in the above equation is the full partition function of the ac-tivated complex (transition state) except that it does not contain a degree of freedom corresponding to the motion along the reaction coordinate. This degree of freedom is actually an imaginary vibration with an imagi-nary frequency and a negative Hessian eigenvalue. Later on, we will skip the “hat” over Q‡, but we will always keep in mind that Q‡ contains one degree of freedom less (i.e., 3N–1 degrees of freedom: 3 translational + 3 rotational (or 2 for a linear transition state) + 3N–7 vibrational (or 3N–6 for a linear transition state)).
‡ ‡‡ ‡ ‡/ º /
ˆSince and E RT G RT
A B
QK e K e
Q Q∆ ∆− −= = , we obtain another
form of the Eyring–Polanyi equation, which was already mentioned above:
‡B º /1
ºG RTk T
k ec h
∆−=
The derivation for a unimolecular reaction is analogous.
Vyboishchikov (94)
Transition-state theory – fundamental equation
• The Eyring–(Evans)–Polanyi equation: unimolecular bimolecular
−=‡B º /G RTk T
k eh
∆
‡B º /1
ºG RTk T
k ec h
∆−=
• At a given temperature, the Eyring–(Evans)–Polanyi equation establishes a 1:1 correspondence between ∆‡Gº and the rate constant. Hence, instead of talking about rate constant, we can talk about the free energy surface (which is different from the potential energy surface).
► Units: – For unimolecular reactions, kBT/h gives s–1 automatically.
– For bimolecular reactions, the correct units (M–1·s–1) are obtained due to the 1/cº factor.
· If cº = 1 M (standard for solutions), ∆‡Gº has to be calculated at 1 M · If ∆‡Gº has been calculated by the Gaussian™ program, the standard
pressure Pº = 1 atm = 101325 Pa ⇒ cº = Pº/RT = 0.0409 M (at 298 K).
‡B º /1000º
G RTk TRTk e
P h∆−=
The factor 1000 originates from the conversion of m3 to liters. ¡ Pº is not the pressure under which the reaction occurs!
Vyboishchikov (95)
Alternatively, we can include the factor 1000 RT/Pº under the exponent:
( )
( )
‡ ‡
‡
B B
B
1000 /º / l º
ln 1000 /
º /
/º
n
( º )
10 0º
0 RT PG RT G RT
RG R RTT T P
k T k Tk e e e
h hP
k Te
RT
h
∆ ∆
∆
− −
−−
= =
=
Thus, passing from 1 atm to 1 M for a bimolecular reaction is equivalent to decreasing ∆‡Gº by RT ln(1000 RT/P º). At 298 K this corresponds to a ∆‡Gº decrease of 1.89 kcal·mol
–1.
This idea can be represented differently: changing a standard state from cº to cº' corresponds to a ∆G change of RT ln(cº'/cº). Setting cº' = 0.0409 M and cº = 1 M, we obtain the same ∆‡Gº correction of –1.89 kcal·mol
–1 for a bimolecular reaction.
Vyboishchikov (96)
Transition-state theory – various forms of the equation
• To include the possibility that not all molecules that have reached the transition state, pass to the products, the transmission coefficient κ can be included:
‡B º /G RTk T
k eh
κ ∆−= but we will not use κ in further equations. • Since ∆‡Gº = ∆‡Hº – T∆‡Sº, we obtain:
unimolecular bimolecular
‡ ‡B º / º /S R H RTk Tk e e
h∆ ∆−=
−=‡ ‡B º / º /1
ºS R H RTk T
k e ec h
∆ ∆
⇒ both activation enthalpy ∆‡Hº and activation entropy ∆‡Sº affect the
reaction rate: k increases with decreasing ∆‡Hº or increasing ∆‡Sº. • Linearized form (“Eyring plot”) for temperature dependence k(T) :
unimolecular bimolecular
B‡ ‡º º
·1
nln lk S H
ThT R R
k ∆ ∆
=
−
+
B‡ ‡º 1
n ln ·lº ºk S H
R Th R
kc
T
∆ ∆= +
−
For a bimolecular reaction, cº = 1 M and does not numerically affect the calcu-lation. This means that ∆‡Sº and ∆‡Gº obtained correspond to cº = 1 M. The use of the SI units is important; k or kcº should be in s–1! ln(kB /h)≈23.76.
Vyboishchikov (97)
Transition-state theory – role of entropy
• In the Eyring equation:
‡ ‡
Bº ºS H
R RTk T
k e eh
∆ ∆−
=
∆‡Hº and ∆‡Sº are activation enthalpy and entropy, correspondingly. ∆‡Hº is related to the energy barrier: ∆‡Hº≈ ∆‡E+∆‡nRT (see page 98 for more detail), while ∆‡Sº can be considered as a measure of change of disorder.
► For a bimolecular reaction: The entropy of the activated complex (one particle = transition state) is lower than that of the reactants (two reacting molecules, less ordered). Therefore, ∆‡Sº <0 (typically about –40 cal·mol–1·K–1 at room tempera-ture) mostly due to translational contributions (6 translational degrees of freedom in reactants, 3 ones in the activated complex). The effect is less pronounced in solution than in the gas phase.
► For a unimolecular reaction: The activation enthropy is usually relatively low: ∆‡Sº≈ 0 , since the translational contribution in the activated complex and the reactant is the same, and rotational is almost the same. The small non-zero ∆‡Sº originates mostly from the vibrational contributions.
Vyboishchikov (98)
Transition-state theory – link to Arrhenius equation
• Link (approximate) to the Arrhenius equation k(T): ► For all reactions:
∆‡E≈Ea –RT (See page 56)
In this expression, “–RT” originates from the kBT factor in the Eyring–Polanyi equation
► For unimolecular reactions:
∆‡Hº≈ ∆‡E≈ Ea – RT
► For bimolecular reactions:
∆‡Hº≈∆‡E– RT≈ Ea – 2 RT ► In general:
∆‡Hº≈Ea – (1– ∆‡n)RT , where ∆‡n is the change in the number of particles (0 for unimolecular reactions, –1 for bimolecular).
– In this expression, “1” originates from the kBT1 factor in the Eyring–
Polanyi equation, whereas ∆‡n comes from the change in volume (“activation volume”):
∆‡Hº = ∆‡E +P∆‡V = ∆‡E +∆‡nRT = (Ea – RT)+∆‡nRT = Ea – (1– ∆‡n)RT
► The Eyring–Polanyi equation then adopts the following form:
‡‡ ‡ ‡ ‡‡B B B
(1 )º º º º1
a aE n RT ES H S SnR RT R RT R RT
k T k T k Tk e e e e e e e
h h h
∆∆ ∆ ∆ ∆∆
− −− − −−= = =
Vyboishchikov (99)
∆Gº ∆Gº
Transition-state theory for complex reactions
• The Eyring–Polanyi equation applies directly to elementary reactions only. For complex reactions, it should be applied to every elementary step sepa-rately to get indivudual rate constants.
► The potential energy surface for a complex reaction contains multiple minima and transition states. Minima correspond to reactants, products, and intermediates.
► Formally, every reaction (or elementary step) is considered as reversible. Practically irreversible are reactions (or steps) for which k←<< k→. In other words, ∆‡Gº→ << ∆‡Gºº← ⇔ ∆rGº<< 0 (strongly exergonic reaction) which is consistent with thermodynamics.
reaction coordinate
∆‡G°
1
∆‡G°
2
reactant A
intermediate B
product C
Irreversible consecutive reaction A→B→C,
k1>>k2, ∆‡Gº1 << ∆‡Gº2 , B→C limiting step
∆‡G°
1
∆‡G°
2
reactant A
intermediate B
product C
Irreversible consecutive reaction A→B→C,
k1<< k2, ∆‡Gº1 >> ∆‡Gº2, A→B limiting step
reaction coordinate
Vyboishchikov (100)
∆‡Gº1
∆‡Gº2
A
B
reaction coordinate
∆‡Gºtot ∆Gº
C
TS2
TS1
Transition-state theory for complex reactions
• Reversible consecutive reactions A →← B→C, case Gº(B)>Gº(A):
► Due to pre-equilibrium, ktot = Keq(A→←B)·k2
Keq(A→←B) = exp(–(Gº(B) – Gº(A)/RT) k2 = (kBT/h) exp(–(Gº(TS2) – Gº(B)/RT)
⇒ ktot = K eq(A→←B)·k2 =
= exp(–(Gº(B) – Gº(A))/RT)·(kBT/h) exp(–(Gº(TS2) –Gº(B)/RT) = = (kBT/h) exp(–(Gº(TS2) – Gº(A))/RT)
Hence, the actual reaction free energy barrier ∆‡Gºtot = Gº(TS2) – Gº(A)
Vyboishchikov (101)
Transition-state theory for complex reactions
• Reversible consecutive reactions A →← B → C, case Gº(B)<Gº(A):
∆‡Gº2
A
B
reaction coordinate
∆‡Gºtot = ∆‡Gº2 ∆Gº
C
TS1
∆‡Gº1
TS2
► In this case, almost all of the reactant(s) A swiftly transforms to B ([B]≈[A]0), and the rate-determining step will be the B→C step.
Vyboishchikov (102)
Transition-state theory for complex reactions
A
B
reaction coordinate
C
TS2
TS1
∆‡Gº 2
∆Gº
∆‡Gº 1
• Summary: in all cases ∆‡Gºtot should be evaluated starting from the lowest
(not necessarily the nearest) left-hand minimum.
Vyboishchikov (103)
Transition-state theory for complex reactions
• Analysis of the potential energy surface for a complex reaction – an example:
-40
-35
-30
-25
-20
-15
-10
-5
0
5
10
TS4TS3
TS2
TS1
E
D
C
B
A
–37
–15
–30
–21
–25
0
–3
+10
Gº,
kca
l/m
ol
reaction coordinate
–10
► Checking all the transition states: – TS1: The only minimum to the left is A, thus ∆‡G1º = 10 kcal/mol; – TS2: The lowest minimum to the left is A, thus ∆‡G2º = 20 kcal/mol; – TS3: The lowest minimum to the left is C, thus ∆‡G3º = 4 kcal/mol; – TS4: The lowest minimum to the left is D, thus ∆‡G4º = 15 kcal/mol. ► Comparing ∆‡G1º, ∆‡G2º, ∆‡G3º, and ∆‡G4º: ∆‡G2º is the largest ⇒ A→TS2 is
the rate-deermining step. The total reaction ∆‡Gºtot = 20 kcal/mol.
Vyboishchikov (104)
Reactions at equilibrium and Microscopic Reversibility
• Consider an elementary reversible reaction of any molecularity: aA + bB →← cC + dD
The forward v1 and reverse v–1 reaction rates are:
v1 = k1[A]a[B]b ; v–1 = k–1[C]c[D]d
► At equilibrium (i.e., at t =∞) the rates must be equal: v1 = v–1. Therefore,
1
1 –1 eq–1
[C] [D][A] [B] [C] [D]
[A] [B]
c da b c d
a b
kk k K
k= ⇒ = =
Thus, we have obtained the standard expression for the equilibrium constant (also known as the thermodynamic law of mass action).
► This derivation provides a link between the chemical kinetics and thermodynamics.
• Now consider a more general case of a complex reversible reaction that has arrived to the equilibrium: aA + bB →← cC + dD →← eE + ƒF →←... →← wW + xX →← yY + zZ
► TOLMAN’s Microscopic Reversibility Principle (1928) states that at equilibrium the rate of every forward elementary reaction must be equal to the rate of the corresponding reverse reaction:
k1[A]a[B]b = k–1[C]c[D]d ; k2[C]c[D]d = k–2[E]e[F]ƒ ; ...; kn[W]w[X]x = k–n[Y]y[Z]z
Multiplying all the above equations, we obtain:
k1k2...kn[A]a[B]b[C]c[D]d...[W]w[X]x = k–1k–2...k–n[C]c[D]d[E]e[F]ƒ...[Y]y[Z]z
which gives, upon cancellation of identical terms, the following expression for the equilibrium constant:
Vyboishchikov (105)
A B C
1 2
eq–1 –2 –
... [Y] [Z]
... [A] [B]
y zn
a bn
k k kK
k k k= =
Therefore, the equilibrium constant expression is determined only by the reac-
tant concentrations [A]a and [B]b (with the powers a and b corresponding to the
stoichiometric coefficients) and the product concentrations [Y]y and [Z]z does
not depend on a particular reaction mechanism or intermediate concentrations. This is different from the kinetics outside of equilibrium, where the rate law in
general, depends on the mechanism.
► FOWLER later augmented the microscopic reversibility principle by the detailed
balance principle that states that in a complex reversible reaction every elemen-tary step of the forward reaction must be mirrored by a reverse elementary step. In other words, equilibrium can be maintained by any of the following processes:
A →← B →← C or
while a cyclic process such as this: is not possible, since such a cyclic
process would mean different mechanisms for the A→C and C→A reactions: A transforms to C through the intermediate B, while the reverse process C→A is an elementary reaction, i.e., the forward and reverse processes would have diferent mechanisms, which is forbidden.
• It is not possible to prove the microscopic reversibility principle or the detailed ba-lance principle theoretically just using the phenomenological kinetics or thermo-dynamics. Their foundations come from the consideration of the potential energy
A B C
Vyboishchikov (106)
surface. The basic idea is as follows: if a pathway linking the reactant(s) A and the product B is passable (in terms of ∆‡Gº and ∆rGº) in the direction A→B, it should be also passable in the reverse direction B→A. Let us consider, for instance, the cyclic mechanism shown before. We will demonstrate that it is impossible.
Indeed, the irreversibility of the A→B steps means that Gº(B)<< Gº(A). Analogous-ly, Gº(C) << Gº(B). Hence, Gº(C) << Gº(A). If, nevertheless, the C→A reaction is kinetically affordable in terms of ∆‡Gº (as shown in the scheme above), the C→A transition state must lie fairly low with respect to C. But then it must be even lower with respect to A, and the direct A→C process must be even easier than the C→A process!
► Using purely phenomenological (“macroscopic”) arguments, it can be shown that without misroscopic reversibility, one could obtain effects that contradict to the classical thermodynamics.
For example, let’s imagine a reversible reaction A →← B that can occurs both with and without a catalyst K according to two following mechanisms:
1 2
1 2
A B A + K B + Kk k
k k− −
� �
At equilibrium, without taking into account the microscopic reversibility prin-ciple, we could write down v1 + v2 = v–1 + v–2. Thus,
k1[A] + k2[A][K] = k–1[B] + k–2[B][K]
Vyboishchikov (107)
Therefore,
1 2
eq1 2
[K][B][A] [K]
k kK
k k− −
+= =
+
which would mean that the equilibrium constant Keq would depend on the cata-lyst concentration. In other words, the catalyst would be able to shift the equilib-rium. This is impossible according to thermodynamics, since, if we actually had a catalyst that could shift the equlibrium, we would be able to build a perpetuum
mobile.
The paradox is resolved as follows: using the detailed-balance principle, we have to write down two separate equations for the rates: v1 = v–1 and v2 = v–2: k1[A] = k–1[B] and k2[A][K] = k–2[B][K], which follows that
1 2
1 2
[B][A]
k k
k k− −= =
.
► Sometimes we can draw direct conclusions about the reaction mechanism based solely on the microscopic reversibility principle. Let us consider the well-known
example of the N2O5 decomposition: 2 N2O5 → 4NO2 + O2. If the reaction were an
elementary one, the reverse reaction 4NO2 + O2 → 2 N2O5 must be elementary as
well. But it would be a pentamolecular reaction, which is impossible of course.
Thus, we conclude that 2 N2O5 → 4NO2 + O2 is not elementary, which is
confirmed by its first-order experimental kinetics.
Vyboishchikov (108)
Transition-state theory – practical calculation of rate constants
• The transition-state theory allows predicting the rate constant from molecular data (cf. the original name “absolute rate constant theory”).
• Unlike ∆rGº , ∆‡Gº is not available from any kind of thermochemical data.
• ∆‡Gº can be calculated from molecular partition functions Q of the reac-tants the transition state with the help of statistical thermodynamics. ► G is related to Q: G= E– RTlnQ. The Gaussian™ program calculates
Gº using the following equation: Gº = E –RTlnQº. The partition functions Qº are calculated as explained later (page 110). Qº means the partition function in the standard state, but we will drop the º notation.
• Therefore: unimolecular bimolecular
‡‡ ‡
Aº ln
QG E RT
Q∆ ∆= −
‡‡ ‡
A Bº ln
QG E RT
Q Q∆ ∆= −
‡B º /G RTk T
k eh
∆−= ‡B º /1
ºG RTk T
k ec h
∆−= ‡
B‡ E
RTA
k T Qk e
h Q
∆−
=
‡
B‡
A B
1º
E
RTk T Q
k ec h Q Q
∆−
=
Vyboishchikov (109)
Transition-state theory – various forms of the equation
• Both the reactant(s) and the transition state are needed:
– Molecular energies: ∆‡E.
– Atomic masses: Qtrans and Qrot.
– Molecular geometries: moment of inertia for Qrot (see next page 110).
– Vibrational frequencies: Qvib (see next page 110).
• The partition function is dimensionless and is defined as follows: B/i k T
i
Q e ε−= ∑ The summation runs over all the quantum states of the molecule; εi is the energy of the i-th state.
• The partition function can be separted into translations, rotational, and vibrational parts.
trans rot vibQ Q Q Q= There is also an electronic partition function Qel, but it is usually neglected. Exception: for spin-degenerate states, Qel = spin degeneration (2 for doublets (free radicals), 3 for triplets, etc.)
Vyboishchikov (110)
Transition-state theory – calculation of rate constant from molecular data
Partition functions:
• Vibrational partition function – depends on individual vibrational frequencies: ► From each vibrational coordinate (see Appendix 8)
B
B
/2
vib, /, 1,...,3 6(5)
1
m
m
h k T
m h k T
eq m N
e
ν
ν
−
−= = −
−
► The total one:
B
B
3 6 3 6 /2
vib vib, /1 1 1
m
m
N N h k T
m h k Tm m
eQ q
e
ν
ν
− − −
−= =
= =−
∏ ∏
► For a transition state, the reaction coordinate is not counted:
B
B
3 3 /2‡
vib,vib1 1
7
/
7
1
m
m
N N h k T
m h k Tm m
eQ q
e
ν
ν
− − −
−= =
= =−
∏ ∏
• Rotational partition function – depends on the molecule’s mass and size: ► For linear molecules: ► For non-linear molecules:
2
B2rot8 I
Q k Th
π=
( )
33/ 2
BA B C
rot 3
16 2 I I IQ k T
h
π π=
I is the moment of inertia IA, IB, IC are principal moments of inertia So-called “symmetry numbers” are ignored here – ask your professor or consult E.Besalú's book vol. 2 page 132 or Atkins page 595 and page 619 if you are interested.
Vyboishchikov (111)
Transition-state theory – calculation of rate constant from molecular data
• Translational partition function – depends on molecule’s mass and volume: ► From each translational coordinate (L is the length of the potential box):
B
trans
2 mk Tq L
h
π=
► The total one:
ɶB
3/2
trans trans3
(2 )º º
mk TQ V Q V
h
π== , where B
3/2
trans 3
(2 )mk TQ
h
π=ɶ
Thus, Qtrans depends on the standard volume Vº. This is the volume avail-able to a single molecule at the standard concentration.
1º
ºA
VN c
=
Qtrans is dimensionless, but Q̃trans has units of m–3.
• For unimolecular reactions:
‡ ‡ ‡
B B B‡ ‡ ‡ ‡‡vib rot vib rot
vib rot v
‡trans
tra i rotns bA
E E
RT RT RTA
E
A A A A
Q Q Q Qk T Q k T k Tk e e e
h Q h hQ Q Q Q
Q
Q
∆ ∆ ∆− − −
= = =
since mA = m‡ ⇒ QAtrans = Q‡
trans. Thus, the translational partition functions
(and the standard volume Vº) cancel and do not affect the result.
Vyboishchikov (112)
Transition-state theory – calculation of rate constant from molecular data
• For bimolecular reactions:
‡ ‡
‡
B B
B
B
‡‡ ‡‡trans rot vib
trans trans rot rot vib vib
‡‡ ‡trans rot vib
trans trans rot rot vib vib
‡ ‡trans rot
trans trans ro
1 1º
1
º
1º º
A
E E
RT RTA B A B A B A B
E
RTA B A B A B
A B
QQ Qk T Q k Tk e e
c h c hQ Q Q Q Q Q Q Q
QQ Qk Te
h Q Q Q Q Q Q
Q Qk T
h Q Q
V
Q
c
N
∆ ∆
∆
− −
−
= =
= =
=
ɶ
ɶ ɶ
ɶ
ɶ ɶ
‡‡vib
t rot vib vib
E
RTA B A B
Qe
Q Q Q
∆−
Of course, the result does not depend on cº. The units will be mol
–1·m3·s
–1.
► Application for a bimolecular reaction of two atoms: A + B → Pr.
2 2
B B2 2
‡ ‡2‡
rot rot rot
8 81;A B I d
Q Q Q k T k Th h
π π µ= = = =
‡
vib vib vib 1A BQ Q Q= = =
where µ = mAmB /(mA+mB) is the reduced mass and d‡ is the inter-atomic distance in the transition state.
Vyboishchikov (113)
Transition-state theory – calculation of rate constant from molecular data Now translational contributions:
B B
B
B B
3/2 3/2
trans trans3 3
3/2‡ trans 3
‡ 3/2 3trans 3
3/2 3/2trans trans
(2 ) (2 ); ;
(2 ( ) )
(( ) / )
(2 ) (2 )
A BA B
A B
A B A B
A B
m k T m k TQ Q
h h
m m k TQ
h
Q m m m m hh
k T k TQ Q
π π
π
π πµ
= =
+=
+= =
ɶ ɶ
ɶ
ɶ
ɶ ɶ
where µ = mAmB/(mA+mB) is again the reduced mass.
‡
‡
‡
B
B
1 / 2
B
2 B
2B
2
‡rot
rot r
‡tra
‡vibns
trans tr vibans
3
3 / 2
v
‡
t
2‡
ibo
8
(2 )
8
A
A
A
E
RTA B
E
RT
A
E
A BB
RT
Qk Tk N e
h
Q
Q Q
k TN e
h
k TN d
Q
Q Q Q Q
dk T
h
h
T
e
k
πµ
µ
π µ
π
∆
∆
∆
−
−
−
= =
= =
=
ɶ
ɶ ɶ
If d‡ is assumed to be the sum of atomic radii: d‡ = rA + rB , the above ex-pression coincides exactly with the collision theory (not a part of our course; see Appendix 6 on Page 148 if interested); for a reaction of two atoms only!
Vyboishchikov (114)
Transition-state theory – temperature dependence of k
• Temperature dependence of the pre-exponential factor can be analyzed based on partition functions:
‡
B‡1
º A B
E
RTk T Q
k ec h Q Q
∆−
=
► Each translational and rotational degree of freedom contribute ∝T1/2 to each partition function. Therefore:
Degrees of freedom Species
translational rotational Q
Single atom 3 0 ∝T3/2
Linear molecule 3 2 ∝T5/2
Non-linear molecule 3 3 ∝T6/2 = T
3
Note that vibrational degrees of freedom do not contribute significantly to the total Q at room temperature (and below), but can contribute at
higher temperatures ⇒ Qvib can be often (but not always) neglected.
► Example: a reaction of two atoms A and B (see Page 113 for more details): QA ∝ T 3/2 QB ∝ T 3/2
Q‡ = Q(A···B) ∝ T 5/2
‡ ‡ ‡5/2
B 1/2
A B 3/2 3/2
‡RT RT RT
E E Ek T Q T
k e T e T eh Q Q T T
∆ ∆ ∆− − −
= ∝ =
► See also problem 8 page 28 of the Llibret de problemes.
Vyboishchikov (115)
Reactions in solution
• Solvents have a strong influence on a chemical reaction:
– Much higher concentrations are available than in the gas phase.
– Ionic substances dissociate into ions, which has a profound effect on the reaction mechanism.
– Solvent can stabilize the reactants, transition state, and products diffe-rently – this is the usual solvent effect, which is the most important.
– For bimolecular reactions, ∆‡Sº is less negative in the solvent than in the gas phase.
– “Cage effect” (= efecte de gàbia) (Franck–Rabinowitch effect) (example of Norrish reaction – photochemical decomposition of ketones):
R1–C(O)–R2 → ·R1 + ·R2 + CO → R1–R2 + CO
In the gas phase, ·R1 and ·R2 undergo recombination randomly, giving R1R1, R1R2, and giving R2R2 in statistic fractions 1:2:1. In solvent, ·R1 and ·R2 recombine in the same solvent cage, where they were formed, giving predominantly the R1R2 product.
– Some solvents (notably H2O, NH3, alcohols,...) can form hydrogen bonds with the solute.
Vyboishchikov (116)
Solvent effect
• The prediction of the solvent effect is done by comparison of the effect of the solvent on the reactant(s) and on the transition state. The solvent effect is predominantly of energetic (not entropic) nature. That is, ∆‡E, ∆‡Hº, and, hence, ∆‡Gº change. Entropic effects do play a role, but are usually less important.
►The solvent almost always stabilizes both the reactant(s) and the transi-tion state, but possibly not to the same extent, leading to a change in the reaction barrier. Thus, ∆‡E, ∆‡Hº, and ∆‡Gº can be either larger or smaller in the solvent than in the gas phase, depending on whether the reactants or the transition state is stabilized more:
reaction coordinate
E
solvent
gas phase
reaction coordinate
E
solvent
gas phase
solvent favors the reaction solvent disfavors the reaction
Vyboishchikov (117)
Solvent effect – example
• Menshutkin reaction
R3N + R' X → [R3R'N]+X–
The product is ionic; the transition state [R3N··· R' ···X]‡ (albeit not ionic) is
strongly polar. Therefore, such a reaction proceeds much faster in more polar solvents.
► Menshutkin reaction rate constants:
Solvent k ·105, M–1·s–1 dielectric constant ε
dipole mo-ment µ (D)
Hexane 0.5 1.9 0
Toluene 25.3 2.4 0.43
Bromobenzene 166.0 5.2 1.55
Acetone 265.0 20.7 2.85
Nitrobenzene 1380.0 34.8 4.02
Vyboishchikov (118)
Primary kinetic salt effect – qualitative explanation
► With increasing ionic strength, each ion gets a more dense ionic atmosphere (consisting of other ions). This atmos-phere provides a shielding (=apantalla-
ment) for the given ion, thus weakening the interaction between the ions. This is favorable for ions of the same charge (less repulsion), but disfavorable for ions of opposite charge (less attraction). Cf. Prof. Blancafort's lectures.
► Note that the ionic strength (and thus the salt effect) depends on all ions present in the solutions, not only on those participating in the reaction. Thus, even adding a inert salt will affect the rate constant!
► Examples:
– Reaction rate increases with increasing I: [Co(NH3)5Br]
2+ + Hg
2+ → HgBr
+ + [Co(NH3)5Br]
3+
– Reaction rate decreases with increasing I: [Co(NH3)5Br]
2+ + OH
– → Br
– + [Co(NH3)5(OH)]
2+
Ionic strength (força iònica): I = ½Σi cizi
2
Vyboishchikov (119)
Primary kinetic salt effect
• Let us go back to the derivation of the Eyring equation, but now consider that the solutions are not ideal:
‡ A
AA
B
BB
‡
‡
‡(A···B) ‡ ‡ ‡[(A···B) ]/ º 1
[(A···B) ] [A][B]( [A]/ º ) ( [B]/ º ) º
a cK K
a a c c c
γ γ γγ γ γ
= = ⇒ =
where γA, γB, γ‡ are the activity coefficients of the reactants and activated complex, correspondingly; 0 < γ < 1.
► For diluted solutions of non-electrolytes, γ is close to 1 and does not affect the rate constant substantially.
► For electrolytes γ must be taken into account.
• The rate constant in an electrolyte solution is given by:
‡ oB A B A B A B/solv ‡ ‡ ‡
( )( ) (0)
(0)G RTk T k I
k I e kh k
γ γ γ γ γ γ
γ γ γ∆−= = ⇒ =
where k(0) is the rate constant at zero ionic strengh.
Vyboishchikov (120)
Primary kinetic salt effect – Brønsted–Bjerrum equation
• The rate constant in an electrolyte solution is given by:
A B
‡
( )(0)
k I
k
γ γ
γ=
► The activity coefficient γ depends on the ionic strength I as described by the Debye–Hückel theory:
2
10log i iAz Iγ = − (Debye–Hückel limiting law)
where A is a constant (≈0.51 M–1/2 in aqueous solutions at 25 °C ); zi is the charge of i-th ion;
I is the ionic strength: I = ½Σi cizi2
A B 210 10 10 A 10 B 10 A B
2A B
‡ 2‡
( )log log log log log
(0)( )
k IAz I Az I
kA z z I
γ γγ γ γ
γ= = + − = − − +
+ +
10 A B
( )log 2
(0)k I
Az z Ik
=
Brønsted–Bjerrum equation
• The Brønsted–Bjerrum equation describes the primary kinetic salt effect: the rate constant for an ionic reaction increases with ionic strength if the reacting ions are of the same charge (zAzB >0) and decreases if the ions are of opposite charge (zAzB <0).
Vyboishchikov (121)
Secondary salt effect
• Ionic strength can affect the concentration of a reactant or a catalysts due to changing in ionization constant.
► Example: Catalysis of sucrose inversion with acetic acid.
Addition of NaCl accelerates the reaction due to enhanced dissociation of acetic acid due to presence of Na+ and Cl–.
► Reactions in which the ionic form of a reactant reacts, get faster in the presence of an electrolyte.
Vyboishchikov (122)
Meaning of ∆‡E in the Eyring equation
• The Eyring equation can be expressed through ∆‡Ee and partition func-tions (for a unimolecular reaction):
B B
‡ ‡
B B
A A A A
‡ ‡ ‡‡trans rot vib
trans rot vib
e e
k T T
E E
kQ Q Qk T Q k Tk e e
h hQ Q Q Q
∆ ∆− −
= =
Here ∆‡Ee = E‡ –EA is the barrier on the potential energy surface (energy difference between the transition state and the reactant minimum). ► Let us consider the vibrational partion function Qvib:
B
B
B B
B
Z
B
P
B
E
3 6 3 6 3 6 3
3 6
13 6
v
6/2/2
vib vib, / /1 1 1 1
ib/1
1·
1 1
exp1
/ 2· exp ·
1
m
m
m
m m
N N N N
N
h k T
h k Th k T
m h k T h k Tm m m
N
m
m
m
mh
eQ q e
e
Qe
e
k T k T
E
νν
ν ν
ν
ν
− − − −−−
− −= =
−
−=
= =−
=
= = = =− −
= − = −
−
−
∑
∏ ∏
∏
∏ ∏
ɶ
where EZP = ∑mhνm /2 is the so-called zero-point vibrational energy.
Sergei
Rectangle
Sergei
Rectangle
Vyboishchikov (123)
Meaning of ∆‡E in the Eyring equation
• Consider
0
BZP
B BAZP B
ZPZP
B B
‡ ‡ A‡
‡‡ A A ‡
A A
A A
‡ ‡ /vib vib
/vib vib
) ( )‡ ‡vib vib
vib vib
(
e E k Tk T k T
E k T
E E
k T
E E
E
k T
E
E E
Q Q ee e
Q Q e
Q Qe e
Q Q
∆
∆
−−− −
−
−− −
+ +
= =
= =
ɶ
ɶ
ɶ ɶ
ɶ ɶ
► Thus,
B
0
B
‡ ‡
B BA A
‡‡ e
k T k
E E
TQk T Q k T
k e eh hQ Q
∆ ∆− −
= =ɶ
ɶ
This means that ∆‡E in the Eyring equations can have two meanings: · with ∆‡Ee = E‡ –EA is the pure energy barrier
· with ∆‡E0 = (E‡+E‡ZP)– (EA+EA
ZP) = ∆‡Ee + ∆EZP
∆‡E0 is the energy barrier including zero-point vibrational energies.
► E‡ZP has 1 degree of freedom less than EA
ZP ,
thus usually E‡ZP < EA
ZP and ∆‡E0 < ∆‡Ee.
E e
E 0
E ‡ e
E ‡ 0
reaction coordinate
E
Vyboishchikov (124)
Kinetic isotope effect
• The kinetic isotope effect (=efecte isotòpic cinètic = efecte cinètic d’isòtop) is the change of reaction rate with isotopic substitution.
► Usually plays a role for hydrogen, when the common 1H isotope (“pro-tium”) is substituted by deuterium (D) or tritium (T).
– Quantitatively, the kinetic isotope effect is defined as KIE = kH /kD.
► is much smaller, but still measurable, for other elements (C,N,O,...)
• The potential energy surface does not depend on the isotope ⇒ ∆‡Ee does
not depend on the isotope either.
• However, ∆‡E0 , as well as Qtrans, Qrot, and Q~
vib do depend on the mass. Thus, the Eyring equation will predict different rates for different masses (and isotopes):
0
B
‡
BA
‡T
E
kQk T
k eh Q
∆−
=ɶ
ɶ
Thus, the main origin (but not the only one, see below) of the primary isotope effect is the dif-ference in the ZPE energies. Practically, the dissociation energy for an X–D bond is slightly higher than for an X–H bond.
Ee E0(D)
E‡e
E‡0
reaction coordinate
E
E0(H)
Vyboishchikov (125)
Kinetic isotope effect
► For simplicity, let us consider a unimolecular reaction in which a X–H (X–D) bond is broken:
B B
0‡ ‡
0
B B‡‡ ‡ ‡
trans rot vib
trans rot vib
k T k TA AA A
E EQk T Q k T Q Q
k e eh h Q QQ Q
∆ ∆− −
= =ɶɶ
ɶ ɶ
‡ ‡vib vib
H Dvib vib
(H) / (D)/
(H) / (D)A A
Q Qk k
Q Q≈
► Neglecting Qtrans and Qrot:
QAtrans and Q‡
trans cancel exactly;
Q Arot(H) ≈ Q‡
rot(D).
0‡‡ ‡
vib vibH D
vib vib
(H) / (D)/
(H) / (D)T
E
RA A
Q Qk k e
Q Q
∆∆−
=ɶ ɶ
ɶ ɶ , where ∆∆‡E0 = ∆‡E0(H)– ∆‡E0(D) =
= ∆EZP(H)– ∆EZP (D) and is negative.
• Limiting case 1 (high frequencies, low temperatures):
BB vib vib/
(H1
/ 1 )1
) (1 Dmh k T
h k T Q Qe ν
ν−
⇒ ≈ ⇒ ≈−
ɶ ɶ≫
H D
B B
0‡
H D
( )2/
h
k T
E
k Tk k e e
ν ν∆∆ −−
= =
Here we used the approximation e–x≈0 ⇒ 1– e–x≈1 for x >>1.
Vyboishchikov (126)
Kinetic isotope effect • Limiting case 2 (low frequencies, high temperatures): hν/kBT << 1
( ) ( )
B
BD
( ) /B
vib vib ( )/B
H ( ) /1 ( )H / D
( ) / ( )D DH H
1
h k T
h k T
h k TeQ Q
h k Te
ν
ν
ν νν ν
−
−
−≈ ==
−ɶ ɶ
Here we used the approximation e–x ≈1– x ⇒ 1–e–x≈x for x <<1.
( )( )
H D
B B
‡0
H D
( )vib 2
vib
/D
H (H)(D)
h
k
E
T k TQk k e e
Q
ν ννν
∆∆ −−
= ≈ɶ
ɶ
►The frequency ratio:
D
H
1 (H)2
2 (D)f m
m m
νν
π ν= ⇒ = ≈
• Not necessarily any of the two limiting cases take place. • Typical primary isotope effect for H/D is between 1 and 8.
• For other elements, it is much smaller (between 1.02 and 1.1)
Vyboishchikov (129)
Appendix 1 – Guggenheim relation
The Guggenheim relation is enormously useful for solving kinetics problems, as it links any measurable quantity Y that characterizes the advance of a chemical reaction with the con-centration of the reactant [A]:
0 0
[ ] [ ][ ] [ ]
t tY Y A A
Y Y A A∞ ∞
∞ ∞
− −=
− −
The only condition of applicability of the Guggenheim relation is that the quantity Y depends linearly on the concentration of all components of the reaction mixture:
M [ ]i i
i
Y Y Aλ= + ∑
Thus, Y can be optical absorbance in spectrophotometry, electrical conductivity, total pressure of a gas mixture, angle of rotation of polarized light, a concentration of a species other than the reactant, and many other quantities. However, it cannot be optical transmittance, electrical resistance, or pH, as they are not linear functions of the concentrations. A rigorous derivation of the Guggenheim relation is presented on Pages 35–36 of the Llibret
de problemes. We will now discuss how it can be used for solving chemical kinetics problems. First of all, the Guggenheim relation expresses laws of stoichiometry. On its own, it does not carry any information about the kinetics of the process under study. Therefore, it must be combined with a valid kinetic rate law in the integrated form. We will consider three distinct cases: (a) first-order irreversible reaction; (b) first-order reversible reaction; (c) second-order irreversible reaction with equal reactant concentrations: [A]0 = [B]0; (d) zeroth-order irrevers-ible reaction.
Vyboishchikov (130)
(a) first-order irreversible reaction: Rate law: Guggenheim relation (taking into account that [A]∞ = 0):
00
[ ][ ] [ ]
[ ]kt ktt
t
AA A e e
A− −= ⇒ =
0 0 0
[ ] [ ] [ ][ ] [ ] [ ]
t t tY Y A A A
Y Y A A A∞ ∞
∞ ∞
− −= =
− −
Therefore,
0 00
( ) ln( ) ln( )kt kttt t
Y Ye Y Y Y Y e Y Y Y Y kt
Y Y− −∞
∞ ∞ ∞ ∞∞
−= ⇒ − = − ⇒ − = − −
−
If Yt < Y∞, the latter equation should be written as follows: ln(Y∞– Yt) = ln(Y∞–Y0) – kt In general, the final equation will be:
0ln lntY Y Y Y kt∞ ∞− = − −
This one is directly suitable for doing a linear regression in the coordinates |Yt – Y∞|(t). The
slope will give the rate constant k.
(b) first-order reversible reaction:
Rate law: 1 1 1 1( ) ( )0
0
[ ] [ ][ ] [ ] [ ] [ ]
[ ] [ ]( ) k k t k k tt
t
A AA A A A e e
A A− −− + − +∞
∞ ∞∞
−= + − ⇒ =
−
Guggenheim relation: 0 0
[ ] [ ][ ] [ ]
t tY Y A A
Y Y A A∞ ∞
∞ ∞
− −=
− − (this time [A]∞ =/ 0)
Therefore,
1 1( )
0
k k ttY Ye
Y Y−− +∞
∞
−= ⇒
− 0 1 1ln ln| | ( )tY Y Y Y k k t∞ ∞ −− = − − +
This one is directly suitable for a linear regression in the coordinates |Yt – Y∞|(t). The slope
will give the sum of the rate constants k1+k–1.
Vyboishchikov (131)
(c) second-order irreversible reaction, equal initial concentrations ([A]0 = [B]0): Rate law: Guggenheim relation (taking into account that [A]∞ = 0):
0
1 1[ ] [ ]t
ktA A
− = 0 0 0
[A] [A] [A][A] [A] [A]
t t tY Y
Y Y∞ ∞
∞ ∞
− −= =
− −
Multiplying the above equation by [A]0 and substituting [A]t /[A]0 in the denominator from the Guggenheim relation:
00 0 0
0
0
0 0 00 0 0
0
0 0 0
[ ]1 1 11 [ ] [ ] 1 [ ]
[ ] [ ] [ ]
( ) ( )1 [ ] [ ] [ ]
1 1[ ] [ ]
tt t
t t
t t t
t tt
t
Akt k A t k A t k A t
Y YA A A
Y Y
Y Y Y Y Y Y Y Yk A t k A t k A t
Y Y Y Y Y Y
Y Y Y YY Y
Y Y k A t k A t
∞
∞
∞ ∞ ∞
∞ ∞ ∞
∞∞
− = ⇒ − = = ⇒ − = ⇒−−
− − − − −⇒ − = ⇒ = ⇒ = ⇒
− − −
− − ⇒ = ⇒ = + −
This is suitable for a linear regression in the coordinates Yt((Y0 –Yt) / t). The slope will give 1/(k[A]0), which is equal to the half-life (see page 22.) Note that in this case you do not need to know Y∞, but you have to know [A]0.
Vyboishchikov (132)
(d) zeroth-order irreversible reaction: Rate law: Guggenheim relation (taking into account that [A]∞ = 0):
0[ ] [ ]tA A kt= − 0
0 0 0 0
[ ] [ ] [ ][ ] [ ]
[ ] [ ] [ ]t t t t
t
Y Y A A A Y YA A
Y Y A A A Y Y∞ ∞ ∞
∞ ∞ ∞
− − −= = ⇒ =
− − −
( )
0 00
0 0
00
0
[ ] [ ]
1[ ]
[ ]
t
t
t
Y YA A kt
Y Y
Y Y kt
Y Y A
k Y YY Y t
A
∞
∞
∞
∞
∞
−= −
−−
= −−
−= +
This is suitable for a linear regression in the coordinates Yt(t). The slope is k(Y∞– Y0)/[A]0, from which k can be obtained.
Vyboishchikov (133)
Appendix 2 • Reversible reaction:
( ) 1 01 1
[A[[A A] ]
]dk k
dtk+ += −
− −
Separating variables [A] and t:
1 1 1 0
[A][1
( [A]) A]d dt
k k k− −=
− + +
Taking a definite integral:
( )
0
0
[A]
1 1 1 0[A] 0
1 1 1 01 1
1 1 1 0
1 1 1 1 1 0
1 1 1 0
1 1 1 0
1
1 1
0
[A]
[A] 0
[A][A]
[A]
[A
1( ) [A]
1ln ( ) [A]
1 ( ) [A]ln
( ) [A]
1 ( ) [A]ln
[
][A
A]
1 (ln
]
[A]
t
t
d dtk k k
k k k tk k
k k kt
k k k k k
k k kt
k k k
k k
k k
− −
− −−
− −
− − −
− −
−
−
−
=− + +
− − + + =+
− + + − = + − + + − + + − = + −
+−
+
∫ ∫
1 1
1 1
1 1 0
1 0
1 1 1 0 ( )
1 0
( )1 1 1 0 1 0
) [A][A]
( ) [A][A]
( ) [A] [A
[A]
[A]
[ ]A]
k k t
k k t
kt
k
k k ke
k
k k k k e
−
−
−
− − − +
− +− −
=
−
+
=
+ = +
−
Therefore
1 11 0 1 0 ( )
1 1 1 1
[A] [A][A] k k tk k
ek k k k
−− − +
− −= +
+ +
Vyboishchikov (134)
Appendix 3a
• Solution by a substitution [B](t) = K(t)e–k2t:
11 0 2
[B][A] [B]k td
k e kdt
−= −
Let us make the substitution [B](t) = K(t)e–k2t, where K(t) is a new (auxiliary) unknown function. Then:
( )2 2 2 22
[B]( ) ( )k t k t k t k td dK d dK
e K t e e k K t edt dt dt dt
− − − −= + = −
Inserting this into then left-hand side of equation
1
1 0 2[B]
[A] [B]k tdk e k
dt−= −
we obtain a differential equation with respect to the new (auxiliary) unknown function
2 2 1 2
2 1
1 2
2 1 0 2
1 0
( )1 0
( ) [A] ( )
[A]
[A]
k t k t k t k t
k t k t
k k t
dKe k K t e k e k K t e
dt
dKe k e
dt
dKk e
dt
− − − −
− −
− −
− = −
=
=
Integration of the latter equation yields:
2 1 2 11 0( ) ( )1 0
2 1 2 1
1 [A]( ) [A] ·k k t k k tk
K t k e C e Ck k k k
− −= + = +− −
and
2 11 0 ( )
2 1
[A]( ) k k tk
K t e Ck k
−= +−
Vyboishchikov (135)
Therefore,
2 2 1 2 1 21 0 1 0( )
2 1 2 1
[A] [A][B]( ) ( ) k t k k t k t k t k tk k
t K t e e C e e Cek k k k
− − − − −= = + = + − −
To determine the value of C, we need the initial condition [B](t = 0) = 0, since the concentra-tion of the intermediate is 0 at the beginning. Thus, we obtain C = –k1[A]0/(k2 –k1)
( )1 2 1 2 1 21 0 1 0 1 0 1 0
2 1 2 1 2 1 2 1
[A] [A] [A] [A][B]( ) k t k t k t k t k t k tk k k k
t e Ce e e e ek k k k k k k k
− − − − − −= + = − = −− − − −
In a special case k1 = k2 = k the derivation (and the resulting equation) is different:
1 2( )1 0 0
0
0 0
[A] [A]
( ) [A]
[B]( ) ( ) ( [A] ) [A]
k k t
kt kt kt kt
dKk e k
dt
K t k t C
t K t e k t C e k te Ce
− −
− − − −
= =
= +
= = + = +
Using the initial condition [B](t=0) = 0, we obtain C = 0. Thus, [B](t) = k[A]0te–kt
Vyboishchikov (136)
Appendix 3b
• Solution by an integrating factor:
1
1 0 2[B]
[A] [B]k tdk e k
dt−= −
Let us multiply both parts by ek2t:
2 2 1
2
2
2 2 1
1 0 2
(2
)1 0
[B][A] [B]
[B][B] [A]
k t k t k t k t
k t k kt tk
de k e e k e
dt
de k e
dtk e
−
−
= −
+ =
Note that k2ek2t is the derivative of ek2t and, thus, the entire left-hand side is the derivative of
ek2t[B]:
( )2 2 1( )
1 0[B] [A]k t k k tde k e
dt−=
Now we should simply integrate the above equation with respect to t:
2 2 1
2 2 1
2 1 2 2
1 2
( )1 0
( )1 0
2 1
( )1 0
2 1
1 0
2 1
[B] [A]
[A][B]
[A][B]
[A][B]
k t k k t
k t k k t
k k t k t k t
k t k t
e k e dt
ke e C
k k
ke e Ce
k k
ke Ce
k k
−
−
− − −
− −
=
= +−
= +−
= +−
∫
Vyboishchikov (137)
where C is an integration constant. To determine the value of C, we need initial condition [B](t = 0) = 0, since the concentration of the intermediate is 0 at the beginning. Thus, we obtain C = –k1[A]0/(k2–k1)
( )− − − − − −= + = − = −
− − − −1 2 1 2 1 21 0 1 0 1 0 1 0
2 1 2 1 2 1 2 1
[A] [A] [A] [A][B] k t k t k t k t k t k t
t
k k k ke Ce e e e e
k k k k k k k k
In a special case k1 = k2 = k the derivation (and the resulting equation) is different:
( )
1 0
1 0
1 0
1 0
1 0
1 0
1 0
[B][A] [B]
[B][A] [B]
[B][B] [A]
[B] [A]
[B] [A]
[B] [A]
[B] [A]
kt
kt kt
kt kt
kt
kt
kt
kt kt
dk e k
dt
de k k e
dt
de ke k
dt
de k
dt
e k dt
e k t C
k te Ce
−
− −
= −
= −
+ =
=
=
= +
= +
∫
Using the initial condition [B]t=0 = 0, we obtain C = 0. Thus, [B]t = k[A]0te–kt
Vyboishchikov (138)
Appendix 4
• H2 + Br2 → 2HBr (Bodenstein) (cf. Llibret de problemes, problem 26, page 14)
Br2 → 2Br k1 Br + H2 → HBr + H k2 H + Br2 → HBr + Br k3 H + HBr → H2 + Br k4
2Br → Br2 k5
► Product formation rate:
2 2 3 2 4
[HBr][ ][H ] [ ][Br ] [ ][HHr H rB B ]
dk k k
dt= + −
► Steady-state equations for active intermediates Br and H:
2 2 3 2 4 2 2 4 3 2
3 2
2
1 2 2 2 3 2 4 5
1 2 5
[ ][ ][H ] [ ][Br ] [ ][HBr] 0 [ ][H ] [ ][HBr] [ ][Br Br
BrBr Br
HH H H Br ]
[HBr]2 [ ][Br ]
[ ]2 [Br ] [ ][H ] [ ][Br ] [ ][HBr] 2 [ ]
2 [Br ] [
H
H H
0
2
H
dk k k k k k
dtd
kdt
dk k k k k
dt
k k
= − − ≈ ⇒ − =
=
= − + + −
−
=���������������������������������
( )1/2
2 1/2
1 5 2] 0 [ ] / [Br ]
[ ][H ]
Br B
Br
r k k
k
= ⇒ =
( )
( )
1/21/2
2 1 5 2 2
3 2 41/2
1/2
2 1 5 2 2
2 2
3 2 4
3 2
3 4
3
2
2
/ [Br ] [[ ][H ][ ]
[Br ] [HBr]
[HBr]
H ]H
[Br ] [HBr]
/ [Br2 [ ][Br ] 2 [Br ]
2[HBr
] [H ]H
[Br ] [
]
HBr]
Brk
k k
k k k
k k
k k k
k k
dk k
dt
d
dt
+= =
+
= =
=
+
( )1/2
1/2
3 2 1 5 2 2
3 4 2
/ [Br ] [H ]
[HBr]/ [Br ]
k k k k
k k+
Experimental equation:
1/2
2 2
2
[H ][Br ]
1 [HBr]/ [Br ]
kv
k
′=
′′+
Vyboishchikov (140)
Appendix 6 – Collision theory
• In the collision theory, we consider reacting molecules as hard spheres, pay-ing attention to the collision between them but not to what happens within the reacting molecule.
• The idea is to calculate the number of collisions between molecules of reac-tants A and B per unit time per unit volume (collision frequency) and derive the rate of the reaction from it.
• Collision classification: ► Elastic collisions:
– Exchange of kinetic energy only;
– Velocity and the direction of translational motion change; – Translational energy of each molecule change; the total translational energy
does not; – Rotational and vibrational energy do not change.
– No chemical changes (bond formation or breaking) occur.
► Inelastic collisions:
– A part of kinetic energy is transferred to vibrational and rotational energy; – The molecule(s) become(s) vibrationally and rotationally excited;
– No chemical changes (bond formation or breaking) occur.
► Reactive collisions:
– Chemical changes leading to formation of product(s) from the reactants (bond breaking and/or formation).
Vyboishchikov (141)
Collision theory
• Let’s consider a collision of two hard spheres A and B moving at a relative velocity
v:
b
A
B
v
► b (impact parameter) is defined as minimum distance between the sphere
centers.
► Kinetic energy of two particles moving at velocities vA and vB:
2 2
c.m.kin
( )2 2
A Bv m vmE
µ += +
where µ =mAmB /(mA+mB) is the reduced mass; v =|v1– v2| is the relative velocity and vc.m. is the velocity of the center of mass.
Thus, the relative kinetic energy is given by Ekin= µv2/2.
Vyboishchikov (142)
Collision theory – collision cross-section
• Let’s consider again a collision of two hard spheres A and B:
► The collision occurs if b ≤rA + rB.
V
b
A
B
• The collision cross-section (secció de col·lisió) for non-interacting spheres is
defined as
2 2( )c A Br r dσ π π= + =
d = rA+rB is referred to as collision diameter. • A real collision of two spheres with velocities vA and vB and masses mA and mB will be re-
placed by shooting of a point mass (massa puntual) µ and velocity v on a fictitious fixed target with area σc.
• The collision cross-section for interacting molecules is influenced by intermolecular interaction:
rA+rB
r A +r B
rA+rB
no interaction attraction repulsion
• σceff is referred to as effective collision cross-section (secció eficaç de col·lisió)
σceff
σceff
σceff
Vyboishchikov (143)
Collision theory – collision frequency
• We want to calculate the collision frequency Z (number of collisions per unit time per unit volume) of molecules moving with given velocities vA and vB.
► The collision frequency ZAB is then calculated as
A B·effAB cZ v n nσ= ,
where nA and nB is the number of molecules A and B per unit volume. A detailed
derivation is give in the other file.
Units SI: m2·(m·s–1)·(m–3)·(m–3)= m–3·s–1
► The collision frequency ZAA for identical molecules:
21·
2eff
AA c AZ v nσ=
Vyboishchikov (144)
Collision theory – Maxwell distribution
• To use the formula for the collision frequency we have to know the mean relative velocity v,
which is obtain from the Maxwell distribution. • Mean relative velocity from the Maxwell distribution:
B
0
8( )
k Tv vf v dv
πµ
∞
= =∫
v1
v2
v
f(v)
high temperature
or low molecular mass
intermediate temperature
or molecular mass
v
f(v) low temperature
or high molecular mass
Vyboishchikov (145)
Mean molecular velocities at various temperatures, m·s–1
298 K 1273 K He 1256 2596 N2 475 981 CO2 379 783 C6H6 284 589
Vyboishchikov (146)
Collision theory – collision frequency
• Mean relative velocity (µ is the reduced mass):
B8k T
vπµ
=
• Collision frequency in general case:
AB A BeffcZ v n nσ=
B
A B
8effc
k TZ n nσ
πµ=
► Collision frequency in the case of identical molecules (µ = mAmA /(mA+mA)=mA /2):
B B2
A28
2( / 2)
12 A
eff effAA c c
k T k TZ n n
m mσ σ
π π= =
Vyboishchikov (147)
Collision theory – requirement of minimum energy
• Not all the collisions (by far not all!) are reactive ones (cf. elastic and inelastic ones). It is reasonable to assume that a collision is reactive when the relative kinetic energy is
greater than a certain threshold energy (= energia llindar) E0. Then the frequency of
reactive collisions will be as follows:
B B
0 0
BAB A Breact
8E E
effk T k Tc
k TZ Z e e n nσ
πµ
− −= =
► Let’s express Zreact in terms of molar concentrations:
0
BB2react
8[A][B]A
E
eff k Tc
k TZ N eσ
πµ
−=
Now Zreact is essentially the reaction rate, but expressed in terms of number of reacting
molecules per unit volume per unit time (m–3
·s–1
). To get the reaction rate in molar units
(mol·m–3
·s–1
), we have to divide Zreact by NA:
0
Breact B8[A][B]
E
eff k TA c
A
Z k Tv N e
Nσ
πµ
−= =
Here v is the reaction rate, not the molecular velocity!
Vyboishchikov (148)
Collision theory – steric factor and Trautz–Lewis equation
• Still, not all the collisions (even with a kinetic energy above the threshold) are reactive, because the reaction only occurs when the molecules are favorably oriented. Thus, the reaction rate must be multiplied by a steric factor P (0<P<1):
0 0
B BB B8 8[A][B] [A][B]
E E
k T k TeffreactA
fA
e fc
k T k Tv N e NP e
π πµσσ
µ
− −= =
This is the Trautz–Lewis equation.
► For a one-substrate reaction:
B B
0 0
B B2 22 2[A] [A]E E
effeff k T k TA c A reactm m
k T k Tv N P e N eσ σ
π π
− −= =
• Relation between the threshold energy E0 in the Trautz–Lewis equation and the Arrhenius
activation energy Ea is as follows (see page 56):
012aE E RT= −
This is because in the Trautz–Lewis equation the rate constant temperature dependence is
of type 0
1/2( )E
RTk k T aT e−
= = .
Vyboishchikov (149)
Collision theory – opacity function
• When we considered a collision of two hard spheres A and B:
V
b
A
B
we assumed that the collision occurs if and only if b ≤rA + rB.
► We can now take a more general view, assuming that for any impact parameter b there
is some probability P(b) that the collision takes place:
2
0 0
( ) ( ) 2 ( )effc P b d b bP b dbσ π π
∞ ∞
= =∫ ∫
• The function P(b) is referred to as opacity function. ► In the above example of non-interacting hard spheres, the opacity function is
1, if ( )
0, if A B
A B
b r rP b
b r r
≤ +=
> +
Then 2
0
2 ( )A B
A B
r r
effc bdb r rσ π π
+
= = +∫
Vyboishchikov (152)
Appendix 8 – Vibrational partition function and zero-point energy • The Eyring equation can be expressed through ∆‡E and partition functions
(for a unimolecular reaction):
B B
‡ ‡
B BA A A A
‡ ‡ ‡‡trans rot vib
trans rot vib
k
E
k T
E
T Q Q Qk T Q k Tk e e
h hQ Q Q Q
∆ ∆− −
= =
Here ∆‡E = E‡ –EA is the barrier on the potential energy surface (energy difference between the transition state and the reactant minimum) ► Separating the numerators:
B BB
B B B
B B
B
ZPE ZPE
/2 /2/2
vib / / /
/ /vib/
1 1· ·
1 1 1
1· ·
1
m mm m
m m m
m
h k T h k Th k T
h k T h k T h k Tm m m m
E k T E k Th k T
m
eQ e e
e e e
e e Qe
ν νν
ν ν ν
ν
− −−
− − −
− −−
∑= = =
− − −
= =−
∏ ∏ ∏ ∏
∏ ɶ
where ZPE / 2mmE hν= ∑ is the so-called zero-point vibrational energy. This is
the energy of the vibrational ground state (the energetically lowest state) with respect to the energy of the minimum.
The product ∏m is running over all the vibrational modes m, of which the are 3N–6 for a non-linear molecule, 3N–5 for a linear molecule; for a transition state, there are 3N–7 (non-linear) or 3N–6 (linear).
• Consider a part of the expression:
ZPE ZPEZPE B
B B BAZP B
0
B
E
‡‡ ‡ A ‡ A A ‡‡
A A AA
) ( )‡ ‡ / ‡ ‡vib vib vib vib
/vib vib ivib
(
v b
E EE k Tk T k T
E E
k T k T
E E E
k T
E
E
Q Q e Q Qe e e e
Q Q QQ e
∆ ∆− −−− − − −
−
+ +
= = =ɶ ɶ ɶ
ɶ ɶɶ
Vyboishchikov (153)
► Thus,
B
0
B
‡ ‡
B BA A
‡‡ e
k T k
E E
TQk T Q k T
k e eh hQ Q
∆ ∆− −
= =ɶ
ɶ
This means that ∆‡E in the Eyring equations can have two meanings: · with ∆‡E = E‡ –EA is the pure energy barrier
· with ∆‡E0 = (E‡+ E‡ZPE) – (EA + EA
ZPE) = ∆‡Ee + ∆EZPE
The former quantity (that we usually call ∆‡E) is often denoted as ∆‡Ee in order to distinguish it
from the ∆‡E0. ∆‡E0 is the energy barrier
including zero-point vibrational energies
E e
E 0
E ‡ e
E ‡ 0
reaction coordinate
E
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