View
5
Download
1
Category
Preview:
Citation preview
Chapter 20. Chemical KineticsConcentrations of Reactants and Products = functionα,β,γ,…(time)
The methods of monitoring the concentrations
1. Pressure change
2N2O5(g) → 4NO2(g) + O2(g) - constant-volume container
The pressure of the system increases during the course of reaction.
Inappropriate for the reactions that leave the overall pressure unchanged, and for reactions in solution.
2. Spectroscopy
H2(g) + Br2(g) → 2HBr(g)
By monitoring the intensity of absorption of light hν by the bromine, the progress of the reaction can be monitored.
http://bp.snu.ac.kr 1
toxic toxic
toxicodor
4. Electrochemical MethodsWhen a reaction changes the number or nature of ions present in a solution, its course may be followed by monitoring the electrical conductivity or pH of the solution.
5. Characterization Methodsmass spectroscopy charge/massemission spectroscopy hν, electron, ionnuclear-magnetic resonanceRaman spectroscopy hνinfra-red absorption x-ray absorption hνx-ray diffraction hνetc.
3. PolarimetryWhen the optical activity of a mixture changes in the course of reaction, it can be monitored by measuring the angle of optical rotation.
http://bp.snu.ac.kr 2
The Rates of Reactions
A+B → P
where [A], [B], [P]: the concentrations of the species, A, B, and P.
The rate of reaction
Methods of applying these analytical techniques1. Real-Time Analysis: the composition of the system is analyzed while
the reaction is in progress.
2. Quenching: the reaction is frozen after it has been allowed to proceed for a certain time, and then the composition is analyzed by any suitable techniques.
(Product)
http://bp.snu.ac.kr 3
[ ] [ ] [ ]dt
Bd
dt
Ad
dt
Pd−=−=
- - - - - - - - - - - - - - - - - - - - - -
Figure 20A.2. Real-Time AnalysisIn the stopped-flow technique, the reagents are driven quickly into the mixingchamber by the driving syringes, and then the time dependence of the concentrations is monitored.
http://bp.snu.ac.kr 4
P(t), T(t), V(t), Cα(t), Cβ(t), Cγ(t), …..
in situ in vivo
Steady State Quenching- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
Figure 20A.3The definition of (instantaneous) rate as the slope of the tangent drawn to the curve showing the variation of concentration with time. For negative slopes, the sign is changed when reporting the rate, so all reaction rates are positive.
Rate of Reaction
http://bp.snu.ac.kr 5
[ ] [ ] [ ] [ ]dt
ddt
ddt
ddt
dv DC31B
21AD3C2BA
==−=−=
+→+
___________________
_________
20A.2(c). Determination of the Rate LawA direct method for determining the rate law from the raw kinetic data giving the concentration as a function of time is from the measurement of slopes.*
The reaction between A and B:
A plot of the rate against various concentrations of A and B gives the orders a and b from the slopes, and log k from the intercept.
A + B → P (Product)
* Only when the rate constant k is independent of concentration.
__________________
http://bp.snu.ac.kr 6
[ ] [ ] [ ] [ ]
[ ] [ ] [ ]
[ ] [ ] [ ]BlogAloglogAlog
BAA
PBAA
bakdt
d
kdt
ddt
dkdt
dv
ba
ba
++=
−
=
−
==−= Reaction RateReaction Order
Figure 20A.4The plot of log v0 against (a) log[I]0 for a given [Ar]0, and (b) log[Ar]0 for a given [I]0.
The reaction between A and B
http://bp.snu.ac.kr 7
( ) ( ) ( ) ( )[ ] [ ]ArI
ArIArI22
2
kv
gggg
=
+→+
[ ] [ ] [ ]
[ ] [ ] [ ]
[ ] [ ]
[ ]Blog
AloglogAlog
BAA
BAA
b
akdt
d
kdt
d
kdt
d
ba
ba
+
+=
−
=
−
=−
I:slope 2
Ar:slope 1
A+2B → 3C+D
The ambiguity in the definition of Reaction Rate is avoided if we define the rate of reaction v as
where vJ is the stoichiometric constant of substance J.
20B.1. The First-Order Reactions
(20A.3b)
http://bp.snu.ac.kr 8
[ ] [ ] [ ] [ ]dt
ddt
ddt
ddt
dv DC31B
21A
==−=−=
[ ]dt
dv
vJ
J1=
[ ] [ ][ ][ ]
[ ][ ][ ]
[ ]dtkd
dtkd
kdt
d
tt
∫∫ =
−
=−
=−
0 1
A
A
1
1
0 AA
AA
AA[ ] [ ]( )
[ ][ ][ ] [ ] ( )tk
tk
tk
t
t
t
10
10
10
expAAAAln
AlnAln
−=∴
−=
=−−
k1 = rate constant = independent of concentration
A → P____
______
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
The concentration of A falls exponentially with time with a rate determined by k1.
20B.2. The Second-Order Reactions A + A → P
the slope is second-order rate coefficient k2.
http://bp.snu.ac.kr 9
[ ] [ ] ( )tkt 10 expAA −=
[ ] [ ]
[ ][ ][ ]
[ ]dtkd
kdt
d
tt
∫∫ =
−
=−
0 2
A
A 2
22
0 AA
AA[ ] [ ]
[ ] [ ][ ]02
0
20
A1AA
A1
A1
tk
tk
t
t
+=
=−∴
[ ] and obained be should linestraight a ,against plotted isA1When t
t
First-Order Reactions
____- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
Figure 20B.1The exponential decay of the reactant in a first-order reaction. The larger the rate constant, the more rapid the decay: here klarge = 3ksmall.
First-Order Reaction
http://bp.snu.ac.kr 10
log [A] vs. time
[ ] [ ] ( )tkt 10 expAA −=A → P
(IQ 50)
Figure 20B.2The determination of the rate constant of a first-order reaction: a straight line is obtained when ln [A] (or, as here, ln p) is plotted against t, and the slope gives k.
First-Order Reaction
= First-Order Rate Constant
http://bp.snu.ac.kr 11
1k
[ ] [ ] ( )[ ] ( )τ/expA
expAA
0
10
ttkt
−=−=
= Time Constant= Life Time= Relaxation Time
1
1
k=τ
Figure 20B.3The variation with time of the concentration of a reactant in a second-order reaction. The dotted line is the corresponding decay in a first-order reaction with the same initial rate. For this illustration, klarge = 3ksmall.
Second-Order Reaction
First-Order (dotted lines)
Second-Order
http://bp.snu.ac.kr 12
[ ]A1
[ ] [ ]reactionorder 2nd
A1
A1
20
−
=− tk
t
A+A → P [ ] [ ]22 AA
kdt
d=−
(IQ 50)
Table 20B.3 Integrated rate laws
http://bp.snu.ac.kr 13
* Only when the rate constant k is independent of concentration.
http://bp.snu.ac.kr 14
* Only when the rate constant k is independent of concentration.
20B.1. Half-Life and Time ConstantA simple indication of the rate of a chemical reaction is the time it takes for the concentration of a reagent to fall to half of its initial value: this is called the half-life of the reaction, and is denoted t1/2.
First-Order Reaction:
- Time Constant- Life Time- Relaxation Time
15http://bp.snu.ac.kr
[ ] [ ][ ][ ]
[ ][ ]
12/1
100
021
2/11
2/100
2lnAAln 2ln
AAln
at A21A
kt
tktk
t
t
=∴
−=−==−
→
1
1k
=τ
For a first-order reaction, one-half of the reactant disappears in t1/2 independent of the initial concentration.
http://bp.snu.ac.kr 16
12/1
2ln k
t =
[ ] [ ] ( )[ ] ( )τ/expA
expAA
0
10
ttkt
−=−=
-161017(월)-161019(수) 휴강
where A and Ea are Pre-Exponential Factor: Independent of temperatureActivation Energy: Determined from a plot of ln kr against 1/T
Approximately independent of temperature
20D. Temperature Dependence of Reaction RatesIn many experiments, the temperature dependence of the rate has been found to fit the expression proposed by Arrhenius:
http://bp.snu.ac.kr 17
__________
−=
RTEAk a
r exp
RT
EAk a
r −= lnln
Rate Constant
Activation Enthalpy≈ Activation Energy
[ ] [ ]barkv BA−= A + B → P Reaction Rate
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
Rate Constant
Figure 20D.1A plot of ln k against 1/T is a straight line when the reaction follows the behavior described by the
Arrhenius-Type Temperature Dependence.
http://bp.snu.ac.kr 18
RTEAk a
r −= lnln
Figure 20D.3. A potential energyprofile for anexothermic (not always)reaction. The height of the barrierbetween the reactants and productsis the activation energy(activation enthalpy)of the reaction.
∆G = Gactivated – Greactant= ∆H – T ∆S= ∆E – T ∆S + P ∆V
http://bp.snu.ac.kr 19
***∆G
20C.1. First-Order Reactions Approaching Equilibrium- considering Backward Reaction
If the initial amount of A is [A]0, and if initially there is no B present, then [A]+[B]=[A]0 at all times:
The solution of this first-order differential equation:
If (no reverse reaction), the equation becomes
(20C.2)
(20C.4)
http://bp.snu.ac.kr 20
[ ] [ ] [ ]BAAbf kk
dtd
+−=
[ ] [ ] [ ] [ ]( ) ( )[ ] [ ]00 AAAAAAbbfbf kkkkk
dtd
++−=−+−=
[ ] [ ] ( )[ ]
++−+
=bf
bffbt
kk
tkkkkAA
exp0
0=bk[ ] [ ] ( )tkAA ft −= exp0
→ fk
←bk
B A
__________________________________________
ppt 20-9
Figure 20C.1The approach of concentrations to their equilibrium values as predicted by the equation for a reaction that is first-order in each direction, and for which .
http://bp.snu.ac.kr 21
A B
Sec. 20C.1
bf kk 2=
B A → fk←
bk
[A]+[B]=[A]0
What is the final state of the system?
(20C.6)Equilibrium Constant
http://bp.snu.ac.kr 22
________
Sec. 20C.1
∞→t[ ] [ ] [ ] [ ] [ ] [ ]
[ ][ ] b
f
bf
f
bf
b
kk
K
kkk
kkk
==
+=−=
+=
∞
∞
∞∞∞
AB
AAAB ,AA 0
00
(20C.4)
→ fk←
bk
A
B
[ ] [ ] ( )[ ]
++−+
=bf
bffbt kk
tkkkkAA
exp0
First-Order Reactions Approaching Equilibrium
The term relaxation denotes the return of a system to equilibrium. It is used in chemical kinetics to indicate that an externally applied influence has shifted the equilibrium position of a reaction, normally suddenly, and that the reaction is adjusting to the equilibrium composition characteristic of new conditions.
where x0 is the departure from equilibrium immediately after the temperature jump, and x is the departure from equilibrium at the new temperature after a time t.
20C.2. Relaxation Method
Consider
http://bp.snu.ac.kr 23
(First-Order)Temperature jump Suddenly from T1 to T2
at T2
→ fk
←bk
B A
bf
t
kkexx +==−
ττ 1 ,0
[ ] [ ] ( )[ ]
++−+
=bf
bffbt kk
tkkkkAA
exp0at T1
We write the deviation of [A] from its new equilibrium value as x, so and at T2. The concentration of A then changes as:
Since , its value may be combined with the relaxation-time measurement to find the individual kf and kb.http://bp.snu.ac.kr 24
[ ] [ ]eqbeqf kk BA =
[ ] [ ] xeq += AA [ ] [ ] xeq −= BB
[ ] [ ] [ ][ ]( ) [ ]( )
( )xkkxkxk
kkdt
d
bf
eqbeqf
bf
+−=
−++−=
+−=
BA
BAA
[ ]( )
( ) ( )
( )bf
ttkk
bfbf
bf
kkexexx
tkkx
xdtkk
x
dx
xkkdt
dx
dtdxdtAd
bf +===
+−=⇒+−=
+−=
=
−+−
ττ 1
where ,
ln
,// of Because
00
0
bf kkK ≈
[ ][ ] b
f
kk
K ==∞
∞
AB
Equilibrium Constantat T1
at T2 kf (T2) kb (T2)perturbation
________________________________________________________________________________at T2
Figure 20C.2The relaxation to the new equilibrium composition
when a reaction initially at equilibrium at a temperature T1 is subjected to a sudden change to temperature T2.
http://bp.snu.ac.kr 25
[A][A]eq at T2
at T1
Temperature Jumping Up
20E.2. Consecutive Reactions and the Steady State
23.5 min 2.35dayshalf-lives
We suppose that only A is present initially, and that its concentration is then [A]0.
If this result is inserted into the equation for B and the condition [B]0=0 imposed, we arrived at
http://bp.snu.ac.kr 26
(Assume no backward reaction)
(20E.4a)
(20E.4b)
(20E.4c)
CBA'11 →→ kk
PuNpU e.g. 23994
23993
23992 →→ ββ
[ ] [ ] [ ] [ ] [ ] [ ] [ ]BC ,B -AB ,AA '1
'111 k
dtdkk
dtdk
dtd
==−=
[ ] [ ] ( )tkt 10 expAA −=
[ ] [ ] ( )[ ] [ ] [ ] [ ]
[ ] [ ] [ ] [ ] [ ] ( )tktkttt
ttt
tktkt
ekekkk
eekk
k
'11
'11
1'1'
1100
0
1'1
10
11ABAAC
ACBA
AB
−−
−−
−
−
+=−−=
=++
−
−=
The solution for [B]
(1)
(2)
(3)
(4)
(4) → (2)
(5)
http://bp.snu.ac.kr 27
CBA'11 →→ kk
[ ] [ ][ ] [ ] [ ][ ] [ ]
[ ] [ ] ( )tk
kdt
d
kkdt
d
kdt
d
10
'1
'11
expAA
BC
BAB
AA
−=
=
−=
−=
[ ] [ ] ( ) [ ]BexpAB '1101 ktkk
dtd
−−=
(skip)
Rearranging eq (5)
(7)
From eqs (6) and (7)
(6)
http://bp.snu.ac.kr 28
[ ][ ] [ ]
[ ] [ ]BB
B
BLet
'1
'1
'1
'1
'1
'1
kdt
ddtdYe
ekdtBde
dtdY
eY
tk
tktk
tk
+=∴
+=
=
−
[ ] [ ] [ ] ( )tkkkdt
d101
'1 expABB
−=+
[ ] ( )
[ ] ( )
[ ] ( ) dtekdY
ekdtdY
tkkdtdYe
tkk
tkk
tk
1'1
1'1
'1
01
01
101
A
A
expA
−
−
−
=
=
−=[ ] ( ) [ ]
[ ] [ ] ( ) [ ]
[ ] [ ] ( )tktk
tkkk
tkkt
eekk
kkk
kekk
ke
kkke
kkkYY
'11
1'1
'1
1'1
1'1
01
1'1
01
1'1
01
1'1
01
1'1
010
AB
AA0B
AA
−−
−
−
−−
=
−−
−=−
−−
−=−
(skip)
(i) The three equations for [A], [B] and [C] indicate how to analyze a reaction scheme consisting of two consecutive first-order reactions.
(ii) “Rate-Determining Step” {Bottle-Neck Effect}If k1
’>>k1, whenever B molecules is formed, it decays quickly into C.
The formation of C depends only on the smaller rate coefficient, as anticipated. For this reason, the step with the slowest rate is called the rate-determining step of the reaction.
http://bp.snu.ac.kr 29
[ ] [ ] ( )
[ ] [ ] ( )tktk
tktkt
eekk
k
ekekkk
C
11
'11
1A 1A
1
1A
0'11
'1
0
1'1'
11
0
−−
−−
−≈
−+≅
−
−+=
[ ] [ ] ( )tkt e
kk'11AC
, If
0
1'1
−−≈
<<
CBA'11 →→ kk
병목현상
Sec. 20E.4
________________________________________________________________________________
____________________________________________________________________
Figure 20E.4(a) The first step is rate determining.(b) The second step is rate-determining.(c) Although one step is slow, it is not
rate-determining step because there is a fast route that circumvents it. Figure 20E.5. Reaction profile for a
mechanism in which the first step is rate-determining step.
http://bp.snu.ac.kr
SlowFast
FasterG
낙하산인사
x-axis:High (1023) Dimensional Space- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
in Series
The steady-state approximation assumes that during the major part of the reaction,
the concentrations of all reaction intermediates are approximately constant, and
the rates of changes of all reaction intermediates are negligibly small.
20E.3. Steady-State Approximation
http://bp.snu.ac.kr 31
A couple of kinetic steps can inducemathematical complexity.
Figure 20E.1The curves are plots with k1 = 10k1`. If the intermediate B is in fact the desired product, it is important to be able to predict when its concentration is greatest.
http://bp.snu.ac.kr 32
CBA '11 →→ kk
[A]
[B]
[C]
intermediates
Fig. 20E.2Steady-State ApproximationThe concentrations of intermediates remain small and hardly change during most of the course of the reaction.
http://bp.snu.ac.kr 33
Rate-Limiting Step↓
(20E.5)
____
CBA'11 →→ kk
[ ] [ ] [ ] [ ] ( )[ ][ ]
( )( )
[ ][ ]
( )[ ][ ] [ ] [ ] [ ]
[ ] [ ] [ ]
[ ] 0B
AAB
AAB ,AB ,1
1AB , If
1AB
AB ,AA
'1
1
'1
1'1
1'1
'1
11
'1
1'1
1
1'1
010
'1
'11
'111
≈∴
<<=
<<=∴=>>
−=>>
−−
=
−−
==
−
−
−−−
dtd
dtd
dtd
kk
dtd
kk
kktk
kekkk
ekk
k
eekk
ke
tk
tkk
tktktk
(ppt 20-27) [A]
[C]
[B]
-161024(월)-161026(수) 학회
20E.5. Pre-Equilibrium
Steady-state approximation
Therefore, the reaction has overall second-order kinetics.http://bp.snu.ac.kr 34
much faster too slow
The intermediate is in equilibrium with the reactants (approximately).
intermediate
(20E.12)
( ) C AB BA + → 2k←
−1k→ 1k
( )[ ] [ ][ ] ( )[ ]ABBAAB12 −−= kk
dtd
[ ][ ] ( )[ ]
( )[ ] [ ][ ]
( )[ ] [ ][ ][ ] [ ][ ]BA ]AB[ C
BAAB
BAAB
0ABBA
11
1
2
12
Kkkdt
dKkk
kk
==
=
=
≈−
−
− [ ][ ]
[ ][ ] b
f
k
kK ===
∞
∞
A
B
Reactant
Product
20F.1. UnimolecularReactions
Lindemann-Hinshelwood mechanism
If the unimolecular step is slow enough to be the rate-determining step, the overall reaction will have the first-order kinetics (by steady-state).
Rate=k[cyclo-propane]
(*)http://bp.snu.ac.kr 35
(not first-order yet) (20F.4)
22
2
CHCHCH
23 CH CHCH
]A[]A[ ],A[[P] PA
]A][A[]A[ 2A A A
]A[]A[ A AAA
'
2
∗∗
∗∗
∗∗
∗
∗∗
−==→
−=→+
=+→+
bbk
ak
ak
kdt
dkdt
d
kdt
d
kdt
d
b
'a
a
[ ]]A[
]A[]A[P]A[
]A[]A[ 0]A[]A][A[]A[]A[
'
2
'
2'2
ab
bab
ab
abaa
kkkkk
dtd
kkkkkk
dtd
+==∴
+=∴≅−−=
∗
∗∗∗∗
반응속도가단일반응물의 1차반응으로표시되는반응. 그러나실제로는타분자와의충돌에인한에너지이동을포함한복잡한과정이며,외견상간단한속도식으로표시되는경우가많다.________________________________________________________________________________
(skip---)
Figure 20F.1Unimolecular Reactions.
The species A is excited by collision with A, and the excited A molecule (A*) may either be deactivated by a collision with A, or go on to decay by a unimolecular process to form products.
http://bp.snu.ac.kr 36
(skip)
Lindemann-Hinshelwood mechanism:
As the concentration (and therefore the partial pressure) of A is reduced, the reaction should switch to overall second-order kinetics. The physical reason for the change of order is that at low pressures the rate-determining step is the bimolecular formation of A*.
(*)
http://bp.snu.ac.kr 37
_______________
______
(20F.7)
(20F.6)
]A[]A[]P[
'
2
ab
ba
kkkk
dtd
+=
],A[]P[ then ,]A[or ,]A[]A][A[ If '' kdt
dkkkk baba =>>>> ∗∗
' wherea
ba
kkkk =
( )∗=∴
<<
eq from ]A[]P[,]A[ If
2
'
a
ba
kdt
dkk
(skip)
38http://bp.snu.ac.kr
Problems from Chap. 20
E 20A.2(b) 20A.5(b)
D 20B.2
E 20B.4(b)
P 20B.16
D 20D.2
P 20E.3 (2017-)
Recommended