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1
Reaction
Kinetics (3) Xuan Cheng
Xiamen University
PhysicalChemistry
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Determination of theRate Law
PhysicalChemistry
λ β α ][][][ L B Ak r =The rate law (17.48)
1. Half-life method
o A
n
Ank n
t ][log)1()1(
12loglog 10
1
102/110 −−−
−=
−(17.49)
[ ] Ano
n
k Ant 1
1
2/1 )1(
12
−
−
−
−
= (17.29)For n ≠ 1
o A
n
Ank n
t ][log)1()1(
12loglog 10
1
102/110 −+−
−=
−(17.49)
ReactionKinetics
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半衰期法确定反应级数
用半衰期法求除一级反应以外的其它反应的级数。
以 lnt 1/2 ~ ln[A]o 作图从直线斜率求 n 值。从多个实验数据
用作图法求出的 n 值更加准确。
根据 n 级反应的半衰期通式: 取两个
不同起始浓度 [A]o , [A]o’ 作实验,分别测定半衰期为
t 1/2 和 ,因为同一反应,常数相同,所以:
1
1
2/1][
1
)1(
12−
−
−
−=
no A
n
Ak nt
1/ 2't
PhysicalChemistry
1
2/1
2/1
][
'][
'
−
=
n
o
o
A
A
t
t
)]/[']ln([
)'/ln (1 2/12/1
oo A A
t t n +=
o An K t ]ln [)1(lnln 2/1 −+=
ReactionKinetics
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PhysicalChemistryDetermination of the
Rate Law2. Powell-plot method
o A A ]/[][≡α
n Ak r ][=
(17.50)
the fraction of A unreacted
t Ak no A
1][ −≡φ
[ ][ ]
[ ] t k n A A
A A
no
n
o
)1(1 11−+=
−−
For n ≠ 1 (17.28)
φ α )1(11 −=−− nn For n ≠ 1
(17.13)
[ ]
[ ]t k
A
A A
o
−=ln
For n = 1
φ α −=ln For n = 1 (17.51)
ReactionKinetics
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PhysicalChemistry
Determination of theRate Law
3. Initial-rate method
λ β α ][][][ L B Ak r =The rate law (17.48)
The ratio of initial rates for runs 1 and 2
α
=
1,0
2,01,02,0
][
][/ A
Ar r
Measure r 0 for two different initial concentrations [A]0,1 and [A]0,2
while keeping [B]0, [C]0, …fixed.
α can be found
The orders β ,…λ can be found similarly
ReactionKinetics
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PhysicalChemistry
Determination of theRate Law4. Isolation method
λ β α ][][][ L B Ak r =The rate law (17.48)
Make initial concentrations of reactant A much less than the
concentrations of all other species
[B]0 >> [A]0, [C]0 >> [A]0, …
The rate law becomes
α λ β α ][][][][ 00
A j L B Ak r == (17.52)λ β 00 ][][ L Bk jwhere ≡
Where j is essentially constant.
The reaction has the pseudo-order α .
The orders β ,…λ can be found
similarly.
ReactionKinetics
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PhysicalChemistryRate Laws and EquilibriumConstants for Elementary Reactions
DC B A +⇔+
]][[]][[][
11 DC k B Ak dt
Ad −+−=
Show that for a reaction that takes place in a sequence of steps, the
overall equilibrium constant is a product of ratios of the rate
constants for each step.
It is sufficient to consider a reasonably general but simple two-stepreaction sequence, such as
F E C +⇔
F E D B A ++⇔+
(second-order in each direction, k 1, k -1)
(first-order forwarded, second-order reverse, k 2, k -2)
(overall)
]][[][][
22 F E k C k dt
C d −+−=
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PhysicalChemistry Reaction Mechanisms
p r o d u c t s A →
The Rate-Determining-Step Approximation
The reaction mechanism is assumed to consist of one or morereversible reactions that stay close to equilibrium during most of the
reaction, followed by a relatively slow rate-determining step, which
in turn is followed by one or more rapid reactions.
products B A →+ products A →2
productsC B A →++ products B A →+2 products A →3
unimolecular
The number of molecules that react in an elementary stepThe molecularity of the elementary reaction
bimolecular
trimolecular (termolecular)
ReactionKinetics
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PhysicalChemistry
12 k k >>
Reaction Mechanisms
C B Ak k → → 21
t k t k ee 12 −− << 212 k k k ≈−
Suppose now that
Then whenever a B molecule is formed it decays rapidly into C.
−
+−
−= −− t k t k o e
k k
k e
k k
k AC 21
12
1
12
21][][ (17.41)
reduces to ( )t k o e AC 11][][ −−=
The formation of C depends on only the smaller of the two rate constants
B Ak → 1 is called the rate-determining step of the
reaction.
For the consecutive unimolecular reactions
ReactionKinetics
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PhysicalChemistry
The Steady-State Approximation
Assumes that during the major part of the reaction, the rates of
change of concentrations of all reaction intermediates are
negligibly small
0][ ≈dt
te Intermediad
Reaction Mechanisms
Reactants
Products
Intermediates
Time
Conce
ntration
C B Ak k → → 21 (17.35)
0][][][
21 =−=
Bk Ak dt
Bd
][][
2
1 A
k
k B = ][][
][12 Ak Bk
dt
C d ==
ReactionKinetics
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PhysicalChemistry Reaction Mechanisms
0001 ])[1(][][ 11 Aee Ak C t k t t k −− −=∫ =
C is formed by a first-order decay of A, with a rate constant k 1,
the rate constant of the slower, rate-determining step.
][][][
12 Ak Bk dt
C d ==
[ ] [ ]t k
o e A A 1−
= (17.38)
( )t k o e AC 11][][ −−=
The same result as before, butobtained much more quickly.
ReactionKinetics
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PhysicalChemistry
C B →
21 k k >>−
A B →
(the rate-
determining step)
Reaction Mechanisms
DC B A k k k → → → 321
←−1k
←−2k
←−3k
Consider the following mechanism composed of unimolecular reactions
is slower than
B A ⇔ remains close to equilibrium
23 k k >> 23 −>> k k
C B ⇔ is not in equilibrium
D is rapidly formed
from C
ReactionKinetics
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PhysicalChemistry Reaction Mechanisms
O H N H C NH H C HNO H Br
22562562 2+ → ++ ++ −
]][][[ 2−+= Br HNO H k r
Example 17.4
is observed to be
A proposed mechanism is
(17.55)
The rate law for the Br --catalyzed aqueous reaction
O H ONBr Br HNOk
222 + → + −
−+ ++ → + Br O H N H C NH H C ONBr k
22562563
++ → + 2221 NO H HNO H
k
←−1k
rapid equilib.
(17.56)slow
fast
ReactionKinetics
h i l
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PhysicalChemistry Reaction MechanismsExample 17.4
]][][[ 2−+= Br HNO H k r (17.55)
Deduce the rate law for this mechanism and relate the observed rate
constant k in (17.55) to the rate constants in the assumed mechanism(17.56)
++ → + 2221 NO H HNO H
k
← −1k
rapid equilib.
(17.56)O H ONBr Br HNOk
222 + → + − slow
−+ ++ → + Br O H N H C NH H C ONBr k
22562563 fast
the rate-determiningstep
(1)
(2)
(3)
The formation of ONBr in (2) ]][[ 222−+= Br NO H k r (17.57)
Step (1) is near equilibrium. Equation (17.53) gives
ReactionKinetics
Ph i l
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PhysicalChemistry Reaction MechanismsExample 17.4
b
f
c k
k
K = elementary reaction (17.53)*
]][[
][
2
22
1
11,
HNO H
NO H
k
k K c +
+
−== ]][[][ 2
1
122 HNO H
k
k NO H +
−
+ =
]][[ 222−+= Br NO H k r (17.57)
]][][)[/( 2121−+
−= Br HNO H k k k r
21,121 )/( k K k k k k c== −
]][][[ 2−+= Br HNO H k r (17.55)
Example 17.5
ReactionKinetics
Ph i l
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PhysicalChemistry Reaction Mechanisms
More examples in using the steady-state approximation
)()(4)(2 2252 g O g NO g O N +→ ][ 52O N k r =
on the basis of the following mechanism:
Account for the rate law for the decomposition of N2O5
3252 NO NOO N ak + →
5232
'
O N NO NO ak → +
NOO NO NO NO bk ++ → +2232
252 3 NOO N NO ck → +
First identify the intermediates NO and NO3
ReactionKinetics
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PhysicalChemistryReaction Mechanisms
]][[]][[][
5232 O N NOk NO NOk dt
NOd cb −=
]][[]][[][][
3232'
523 NO NOk NO NOk O N k
dt
NOd baa −−=
3252 NO NOO N
ak
+ →
5232
'
O N NO NO ak → +
NOO NO NO NO bk ++ → + 2232
252 3 NOO N NO ck → +
]][[]][[][][
5232'
5252 O N NOk NO NOk O N k
dt
O N d caa −+−=
ReactionKinetics
Ph i l
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PhysicalChemistry Reaction Mechanisms
0]][[]][[ 5232 =− O N NOk NO NOk cb
][
][
)(][][
]][[
2
52
'32
52
NO
O N
k k
k
NO NO
O N NO
k
k
ba
a
b
c
+==
According to the steady-state approximation, set both rates equal
to zero 0][ =dt NOd 0][ 3 =
dt NOd
)(][ 'bac
ba
k k k
k k
NO +=
][
]][[][
2
523
NO
O N NO
k
k NO
b
c=
0]][[]][[][ 3232
'
52 =−− NO NOk NO NOk O N k baa
][
][
)(][
][
)(][
2
52'
2
52'3
NO
O N
k k
k
NO
O N
k k k
k k
k
k NO
ba
a
bac
ba
b
c
+=
+=
ReactionKinetics
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PhysicalChemistry Reaction Mechanisms
]][[]][[][][
5232'
5252 O N NOk NO NOk O N k
dt
O N d caa −+−=
The net rate of change of concentration of N2
O5
is
][
)(][
][
)(
][][][
52'2
52
'2
'52
52 O N
k k k
k k k
NO
O N
k k
k NOk O N k
dt
O N d
bac
bac
ba
aaa
+−
++−=
)(
]['
bac
ba
k k k
k k NO
+
=
][
][
)(][
][
)(][
2
52'
2
52'3
NO
O N
k k
k
NO
O N
k k k
k k
k
k NO
ba
a
bac
ba
b
c
+=
+=
][)(
][)(
][52'52'
''52 O N
k k
k k O N
k k
k k k k k k
dt
O N d
ba
ba
ba
aabaaa
+−
+
+−−=
ReactionKinetics
Ph i l
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PhysicalChemistry Reaction Mechanisms
][
2][
52'
52
O N k k
k k
dt
O N d
ba
ba
+−=
because 2][ 52−=O N υ
][][ 5252' O N k O N k k
k k r
baba =+=
][ 52O N k r =
ba
ba
k k
k k k
+=
'
It follows that the reaction rate is
where
ReactionKinetics
Ph sical
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PhysicalChemistry Reaction MechanismsPre-equilibria
From a simple sequence of consecutive reactions we now turn to a
slightly more complicated mechanism:
Where C denote the intermediate.
P C B A ba k k → → + ←
'ak
This scheme involves a pre-equilibrium, in which an intermediates
is in equilibrium with the reactants.
A pre-equilibrium arises when the rates of formation of the
intermediate and its decay back into reactants are much faster than
its rate of formation of products; thus, the condition is possible
when k’ a>>k b but not when k b >>k’ a. Because we assume that A, B,
and C are in equilibrium.
ReactionKinetics
Physical i
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PhysicalChemistry Reaction MechanismsPre-equilibria
We can write:
c K B A
C =]][[
][ 'a
ack
k K =
In writing these equations, we are presuming that the rate of
reaction of C to form P is too slow to affect the maintenance of the
pre-equilibrium (see the following example). The rate of formation
of P may now be written:
]][[][][
B A K k C k dt
P d cbb ==
This rate law has the form of a second-order rate law with a
composite rate constant:]][[
][ B Ak
dt
P d = '
a
bacb
k
k k K k k ==where
ReactionKinetics
Physical R i
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PhysicalChemistry Reaction MechanismsPre-equilibria
Example: Analyzing a pre-equilibrium
Repeat the pre-equilibrium calculation but without ignoring the
fact that C is slowly leaking away as it forms P.
][][ C k dt P d b=
The net rates of change of P and C are
0][][]][[][ ' =−−= C k C k B Ak
dt
C d baa
ba
a
k k
B Ak C
+=
'
]][[][
where
P C B A ba k k → → +
← 'ak
]][[][
B Ak dt
P d =
ba
ba
k k
k k k
+=
'
ReactionKinetics
Physical R ti
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PhysicalChemistry Reaction Mechanisms
Pre-equilibria
where]][[][
B Ak dt
P d =
ba
ba
k k
k k k
+=
'
]][[][
B Ak dt
P d =
'a
bacb
k
k k K k k ==where
When the rate constant for the decay of C into products is muchsmaller than that for its decay into reactants '
ab k k <<
ReactionKinetics
Physical R ti
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Homework Physical
Chemistry
Page 592
Prob. 17.28
Prob. 17.29
Prob. 17.33
Page 593
Prob. 17.39
Prob. 17.52
ReactionKinetics
Physical
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速率决定步骤
在连续反应中,如果有某步很慢,该步
的速率基本上等于整个反应的速率,则该慢步骤称为速率决定步骤,简称速决步或速
控步。利用速决步近似,可以使复杂反应的
动力学方程推导步骤简化。
PhysicalChemistry
Physical
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速率决定步骤
A][B][1k r ≈慢步骤后面的快步骤可以不考虑。
只需用平衡态近似法求出第 1 , 2 步的速
率。虽然第二步是速决步,但中间产物 C
的浓度要从第一步快平衡求。
例 1. 慢 快 快A B C D E+ → → →
PhysicalChemistry
例 2.
快 慢 快 快 F E DC B A →→→→+
Physical
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稳态近似
从反应机理导出速率方程必须作适当近似,
稳态近似是方法之一。
假定反应进行一段时间后,体系基本上处于稳态,这时,各中间产物的浓度可认为保持不
变,这种近似处理的方法称为稳态近似,一般
活泼的中间产物可以采用稳态近似。
PhysicalChemistry
Physical
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氢与碘的反应
2 2
2 2
H I 2HI
1 d[HI][H ][I ]
2 dtr k
+ →= =
总包反应
实验测定的速率方程
分别用稳态近似和平衡假设来求中间产物 [I]
的表达式,并比较两种方法的适用范围。
2
2 2 [I]1 d[HI]
[H ]2 dt k =
PhysicalChemistry
2
2
(1) I M 2I M
(2) H 2I 2HI
+ +
+ →
反应机理:
快平衡
慢
M I M I +→+ 22
Physical
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用稳态近似法求碘原子浓度
因为 (1) 是快平衡, k-1 很大; (2) 是慢反应, k2 很小,
分母中略去 2k2[H2] 项,得:1 2
2 22 2
1
[H ][ [H ]I [] I ]r k k k
k −= =
与实验测定的速率方程一致。
2 2
1 2 -1 2 2
1 d[I]
[I ][M]- [I] [M]- [H ][I] 02 dk k k
t = =
2 1 2
1 2 2
[I ][M][I]
[M] 2 [H ]
k
k k −
=+
2 1 2 2 22 2
1 2 2
[H ][I ][M][H ][I]
[M] 2 [H ]
k k r k
k k −
= = +
PhysicalChemistry
2
2
(1) I M 2I M
(2) H 2I 2HI
+ +
+ →
反应机理:
快平衡慢
M I M I +→+ 22
Physical
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显然这个方法简单,但这个方法只适用于快平衡下面是慢反应的机理 , 即 k -1>>k 2 。
反应 (1) 达到平衡时:2
1 2 -1[I ][M] [I] [M]k k =
2 1
21
[I] [I ]k
k −=
2 1 22 2 2 2 2 2
1
[H ][I] [H ][I ] [H ][I ]k k
r k k k
−
= = =
用平衡假设法求碘原子浓度Physical
Chemistry
2
2
(1) I M 2I M
(2) H 2I 2HI
+ ++ →
反应机理:
快平衡
慢
M I M I +→+ 22
Physical
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稳态近似法与平衡态近似法的比
较
稳态近似法 — 优点:所得最终动力学方程中包含了复合
反应中的全部动力学参数 ( 如 k 1 , k -1 ,
k 2)优缺点
—平衡态近似法 缺点:所得最终动力学方程中只有一个
动力学参数(k 2)
,而且包含在k 2 K c
的
乘积中
优点:所得动力学方程的形式简单
缺点:所得动力学方程的形式复杂
PhysicalChemistry