Chapter 10.
Chemical Kinetics II.
Composite
Mechanisms
Physical Chem. 2
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Composite reactions are reactions that occur in more than one elementary step; they
also commonly referred to as complex reactions or as stepwise reactions.
Reactions can be classified as:
a. Elementary reactions
b. Composite reactions.
Composite reactions are reactions that occur in more than one elementary step
Type of Composite reactions
1. Simultaneous reactions
2. Opposing reactions
3. Consecutive reactions
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For a reaction: H2 + 2ICl → I2 + 2HCl
If this reaction occurred in a single elementary step, it would be third order, but
this reaction is a second order (experimentally).
𝑟𝑎𝑡𝑒 = 𝜐 = 𝑘 𝐻2 𝐼𝐶𝑙
This can be explained if there is initially a slow reaction between one molecule of
H2 and one of IC1,
1. H2 + ICl → HI + HCl
(slow)
2. HI + ICl → HCl + I2 (rapid)
The first step is said to be the rate-determining or rate-controlling step.
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Example
2Br − + H2O2 + 2H+ → Br2 + 2H2O
𝜐 = 𝑘 𝐻2 𝑂2 𝐻 + 𝐵𝑟 −
1. H+ + H2O2
⇋ H2O+ — OH (rapid equilibrium)
2. H2O+ — OH + Br- → HOBr + H2O (slow)
3. HOBr+ H+ + Br- → Br2 + H2O (fast)
The concentration of H2O+ — OH , which is a protonated H2O2 molecule, is proportional
to [H+][H2O2], and the rate of reaction 2 is proportional to [H2O+—OH ] and therefore to
[H2O2][H+] [Br-].
“kinetic evidence—or indeed any other kind of evidence—can never prove a reaction
mechanism, although the evidence may disprove a mechanism!”
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EXAMPLE 10.1 Suppose that a reaction of stoichiometry
A+B
→ Y +Z
is zero order in the reactant A, and first order in B. Suggest a mechanism that is
consistent with this result.
The obvious suggestion is that B forms some intermediate X by a slow step, and that X
then reacts rapidly with
B
→ X (slow)
X+A → Y +Z
(Fast)
The amount of A does not affect the rate, but there must be enough A present.
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EXAMPLE 10.2 The kinetics of the reaction
4−
−
2𝐹𝑒 𝐶𝑁 3−
6 + 2𝐼 ⟶ 2𝐹𝑒 𝐶𝑁 6 + 𝐼2
2 𝐼 − 2 𝐹𝑒 𝐶𝑁 4− −1 𝐼 𝑜
𝜐 = 𝑘 𝐹𝑒 𝐶𝑁 3−
2
6
6
What we notice at once about this rate equation is that the concentration of a product of the reaction
appears in it, and to a power of -1 . This can be explained if we assume that the product 𝐹𝑒 𝐶𝑁 4−
6 is
involved in a pre-equilibrium. The following is a possible scheme:
− ⇋ 𝐹𝑒 𝐶𝑁 4− + 𝐼 − (𝑟𝑎𝑝𝑖𝑑)
𝐹𝑒 𝐶𝑁 3−
+
2𝐼
6
6
2
The concentration of 𝐹𝑒 𝐶𝑁 3−
6 occurs to the second power suggests that this preequilibrium is followed by a slow reaction of 𝐼2− with another molecule of 𝐹𝑒 𝐶𝑁 3−
6
−
4−
𝐹𝑒 𝐶𝑁 3−
6 + 𝐼2 ⇋ 𝐹𝑒 𝐶𝑁 6 + 𝐼2
Physical Chem. 2
(𝑠𝑙𝑜𝑤)
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−
𝐹𝑒 𝐶𝑁 4−
𝐼
6
2
𝐾=
− 2
𝐹𝑒 𝐶𝑁 3−
𝐼
6
The rate of formation of the products
𝜐 = 𝑘2 𝐼2− 𝐹𝑒 𝐶𝑁 3−
6
3−
− 2
𝐾
𝐹𝑒
𝐶𝑁
𝐼
6
𝐼2− =
− 2
𝐹𝑒 𝐶𝑁 4−
𝐼
6
2 𝐼 − 2 𝐹𝑒 𝐶𝑁 4− −1
𝜐 = 𝑘2 𝐾 𝐹𝑒 𝐶𝑁 3−
6
6
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EXAMPLE 10.3
An investigation was made by M. J. Haugh and D. R. Dalton (J. Amer. Chem. Soc, 97,
5674(1975)) of the reaction of hydrogen chloride with propene at high pressures. They
found that under some circumstances the reaction was first order in propene and third order
in hydrogen chloride: = k[propene] [HC1].
Suggest a mechanism that is consistent with this result.
Solution
2𝐻𝐶𝑙 ⇌ 𝐻𝐶𝑙 2
𝐾1
𝐻𝐶𝑙 + 𝑝𝑟𝑜𝑝𝑒𝑛𝑒 ⇌ 𝐻𝐶𝑙. 𝑝𝑟𝑜𝑝𝑒𝑛𝑒
𝐾2
𝐻𝐶𝑙 2 + 𝐻𝐶𝑙. 𝑝𝑟𝑜𝑝𝑒𝑛𝑒 → 𝐶𝐻3 𝐶𝐻𝐶𝑙𝐶𝐻3 + 2𝐻𝐶𝑙
The rate of this reaction is 𝜐 = 𝑘3 𝐻𝐶𝑙 2 𝐻𝐶𝑙. 𝑝𝑟𝑜𝑝𝑒𝑛𝑒
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𝐻𝐶𝑙 2 = 𝐾1 𝐻𝐶𝑙 2
𝐻𝐶𝑙. 𝑝𝑟𝑜𝑝𝑒𝑛𝑒 = 𝐾2 𝐻𝐶𝑙 𝑝𝑟𝑜𝑝𝑒𝑛𝑒
𝜐 = 𝑘3 𝐻𝐶𝑙 2 𝐻𝐶𝑙. 𝑝𝑟𝑜𝑝𝑒𝑛𝑒
= 𝑘3 𝐾1 𝐻𝐶𝑙 2 𝐾2 𝐻𝐶𝑙 𝑝𝑟𝑜𝑝𝑒𝑛𝑒
𝜐 = 𝑘3 𝐾1 𝐾2 𝐻𝐶𝑙 3 𝑝𝑟𝑜𝑝𝑒𝑛𝑒
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10.1 EVIDENCE FOR A COMPOSITE MECHANISM
1.The kinetic equation does not correspond to the stoichiometric
equation.
2. The detection of reaction intermediates (by chemical or other
means).
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Type of Composite reactions
1. Simultaneous reactions
A+B → Y
Where B and C compete one another for a (called sometimes:
A+C
→
parallel or side reactions)
Z
k1 is the rate constant for forward reaction
𝑘1
2. Opposing reactions
𝐴+𝐵 ⇌ 𝑍
K-1 is the rate constant for reverse reaction
𝑘−1
3. Consecutive reactions: these are reactions in which the product of one reaction becomes
the reactant for the other reaction.
k1
k2
k3
A → X →Y → Z
The intermediate Y may catalyze (positive feedback) or
Inhibit (negative feedback)
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10.3 RATE EQUATIONS FOR COMPOSITE MECHANISMS
𝑘1 𝑘2
𝐴 ⇌𝑋⇌ 𝑍
𝑘−1 𝑘−2
There are four elementary reactions
k1
A → X
k-1
X → A
k2
X → Z
k-2
Z → X
𝜐1 = 𝑘1 𝐴
𝜐−1 = 𝑘−1 𝑋
෍ 𝜐𝑥 = 𝜐1 + 𝜐−2
the total rate into X (produce X)
෍ 𝜐−𝑥 = 𝜐−1 + 𝜐2
the total rate out of X (remove X)
෍ 𝜐𝑥 = 𝑘1 𝐴 + 𝑘−2 𝑍
𝜐2 = 𝑘2 𝑋
𝜐−2 = 𝑘−2 𝑍
෍ 𝜐−𝑥 = 𝑘−1 + 𝑘2 𝑋
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At complete equilibrium
෍ 𝜐𝑥 = ෍ 𝜐−𝑥
Consecutive reactions:
The simplest consecutive mechanism is
k1
k2
A → X → Z
A reaction of this type was first investigated in 1865 by the chemist Augustus George
Vernon Harcourt (1834-1919), and the kinetic equations were solved by the
mathematician William Esson (1839-1916).
𝑑𝐴
𝜐1 = −
= 𝑘1 𝐴
𝑑𝑡
𝐴 = 𝐴 𝑜 𝑒 −𝑘1 𝑡
The net rate of formation of X is
Physical Chem. 2
𝐴 = 𝐴 𝑜 𝑎𝑡 𝑡 = 0
𝑑𝑋
= 𝑘1 𝐴 − 𝑘2 𝑋
𝑑𝑡
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𝑑𝑋
𝑘1
−𝑘
𝑡
= 𝑘1 𝐴 𝑜 𝑒 1 − 𝑘2 𝑋
𝑋 = 𝐴𝑜
𝑒 −𝑘1 𝑡 − 𝑒 −𝑘2 𝑡
𝑑𝑡
𝑘2 − 𝑘1
𝑍 = 𝐴𝑜− 𝐴 − 𝑋
𝐴𝑜= 𝐴 + 𝑋 + 𝑍
𝐴𝑜
𝑍 =
𝑘2 1 − 𝑒 −𝑘1 𝑡 − 𝑘1 1 − 𝑒 −𝑘2 𝑡
𝑘2 − 𝑘1
We see that [A] falls exponentially, while [X] goes through a
maximum. Since the rate of formation of Z is proportional to
the concentration of X, the rate is initially zero and is a
maximum when [X] reaches its maximum value.
For an initial period of time it may be impossible to detect
any of Induction Period the product Z, and the reaction is
said to have an induction period.
An induction period in chemical kinetics is an initial slow stage of a chemical reaction; after
the induction period, the reaction accelerates.
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Two limiting cases
a. k1 >> k2
The reactant A is then rapidly converted into the intermediate X, which
slowly forms Z.
k 2 is small, e−k2t shows a very slow decay, while e−k1t shows a vey rapid fall.
𝑘1
𝑋 = 𝐴𝑜
𝑒 −𝑘1 𝑡 − 𝑒 −𝑘2 𝑡 ≈ 𝐴 𝑜 𝑒 −𝑘2 𝑡 − 𝑒 −𝑘1 𝑡
𝑘2 − 𝑘1
𝑋 rises rapidly to the value 𝐴 𝑜 ,
and then slowly declines.
k1 =1; k2 = 0.001
Physical Chem. 2
time
𝒆−𝒌𝟐 𝒕 − 𝒆−𝒌𝟏 𝒕
1
0.6915
2
0.8716
3
0.9486
5
0.9883
10
0.99
100
0.9048
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1.1
e-k2t - e-k1t
1
0.9
0.8
0.7
0.6
0.5
0
50
100
150
200
250
Time
Variations with time of the exponentials,
when k1 >> k2
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Steady state treatment
b. k2 >> k1
𝑒 ~0 − 𝑒 ~∞ = 1 − 0 ~1
𝑒 −𝑘1 𝑡 − 𝑒 −𝑘2 𝑡 ~1
𝑘1 −𝑘 𝑡
𝑋 = 𝐴𝑜
𝑒 1 − 𝑒 −𝑘2 𝑡
𝑘2
At t=0, [X] =0, after a very short time
𝑘1
𝑋 = 𝐴𝑜
,
𝑘2
𝑋 ≪ 𝐴𝑜
𝑘2 ≫ 𝑘1
After this short time, [X] remains constant.
Steady state approximation:
𝒅𝑿
=𝟎
𝒅𝒕
𝒅𝑿
= ෍ 𝜐𝑥 − ෍ 𝜐−𝑥
𝒅𝒕
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𝑘1
𝐴+𝐵 ⇌ 𝑋
𝑘−1
𝑘2
𝑋→𝑍
𝑑𝐴
𝑑𝐵
−
=
= 𝑘1 𝐴 𝐵 − 𝑘−1 𝑋
𝑑𝑡
𝑑𝑡
𝑑𝑍
= 𝑘2 𝑋
𝑑𝑡
𝑑𝑋
= 𝑘1 𝐴 𝐵 − 𝑘−1 𝑋 − 𝑘2 𝑋
𝑑𝑡
According to SSA,
𝑑𝑋
𝑑𝑡
=0
𝑑𝑋
𝑘1 𝐴 𝐵
= 𝑘1 𝐴 𝐵 − 𝑘−1 𝑋 − 𝑘2 𝑋 = 0 ⇒ 𝑋 =
𝑑𝑡
𝑘−1 + 𝑘2
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Example
+
k1
I
CH2
I-
+
k-1
(Equilibrium, fast)
+
CH2
+
H2O
k2
OH
+
H+
(slow)
𝑑 𝑃ℎ𝐶𝐻2 𝑂𝐻
𝜐=
= 𝑘2 𝑃ℎ𝐶𝐻2+ 𝐻2 𝑂
𝑑𝑡
𝑃ℎ𝐶𝐻2+
𝑃ℎ𝐶𝐻2 𝐼
= 𝐾𝑒𝑞
𝐼−
But 𝑃ℎ𝐶𝐻2+ is not directly measured.
𝑃ℎ𝐶𝐻2 𝐼
𝜐 = 𝑘2 𝐾𝑒𝑞
𝐻2 𝑂
−
𝐼
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𝐵𝑦 𝑆𝑆𝐴,
𝑑 𝑃ℎ𝐶𝐻2+
=0
𝑑𝑡
𝑘1 𝑃ℎ𝐶𝐻2 𝐼 −𝑘−1 𝑃ℎ𝐶𝐻2+ 𝐼 − − 𝑘2 𝑃ℎ𝐶𝐻2+ 𝐻2 𝑂 = 0
𝑃ℎ𝐶𝐻2+
𝑘1 𝑃ℎ𝐶𝐻2 𝐼
=
𝑘−1 𝐼 − + 𝑘2 𝐻2 𝑂
𝜐 = 𝑘2
𝑘1 𝑃ℎ𝐶𝐻2 𝐼
𝑘−1 𝐼 − + 𝑘2 𝐻2 𝑂
Physical Chem. 2
𝐻2 𝑂
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Rate-Controlling (Rate-Determining) Steps
Previously in this chapter slow reaction steps followed by rapid ones have been said to be
rate-determining or rate-controlling, the rate of the overall reaction being the same as the
rate of the initial step.
A rate controlling step
a. a step which has a strong influence on the overall rate of the reaction.
b. if there are rapid pre-equilibria followed by a slow step
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𝑘1
𝐴+𝐵 ⇌ 𝑋
𝑘−1
𝐶𝑎𝑠𝑒 1,
𝑘2
X →𝑍
𝑘2 ≫ 𝑘−1
𝑘1 𝐴 𝐵
𝜐=
= 𝑘1 𝐴 𝐵
𝑘−1 + 𝑘2
𝐶𝑎𝑠𝑒 2,
𝜐 = 𝑘2 𝑋 ,
𝑘2 ≪ 𝑘−1
𝑘1
𝑋 =
𝐴 𝐵
𝑘−1
𝑘1 𝑘2
𝜐=
𝐴 𝐵 = 𝑘2 𝐾1 𝐴 𝐵
𝑘−1
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EXAMPLE 10.4 The reaction between iodide ions and the cobalt complex Co 𝐶𝑁 5 𝑂𝐻22−
for which the stoichiometric equation
Co 𝐶𝑁 5 𝑂𝐻22− + 𝐼 − ⟶ Co 𝐶𝑁 5 𝐼 3− + 𝐻2 𝑂
is believed to go by the mechanism
𝑘1
Co 𝐶𝑁 5 𝑂𝐻22− ⇌ Co(𝐶𝑁)2−
5 +𝐻2 𝑂
𝑘−1
𝑘2
−
3−
Co(𝐶𝑁)2−
+𝐼
Co
𝐶𝑁
𝐼
→
5
5
Assume that the intermediate exists in a steady state and derive the general rate equation.
Write the rate equation for the special cases of low and high iodide concentrations and
decide which is the rate-controlling step in each case.
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The steady-state equation for Co(𝐶𝑁)2−
5
2−
𝑘1 Co 𝐶𝑁 5 𝑂𝐻22− − 𝑘−1 Co 𝐶𝑁 2−
−
𝑘
Co
𝐶𝑁
𝐼− = 0
2
5
5
2−
𝑘
Co
𝐶𝑁
𝑂𝐻
1
5
2
Co 𝐶𝑁 2−
=
5
𝑘−1 + 𝑘2 𝐼 −
𝜐 = 𝑘2 Co 𝐶𝑁 2−
5
2−
−
𝑘
𝑘
Co
𝐶𝑁
𝑂𝐻
𝐼
1 2
5
2
𝐼− =
𝑘−1 + 𝑘2 𝐼 −
𝐶𝑎𝑠𝑒 1, 𝑘−1 ≫ 𝑘2 𝐼 − (𝑖. 𝑒. 𝐼 − is very low)
𝑘1 𝑘2
𝜐=
Co 𝐶𝑁 5 𝑂𝐻22− 𝐼 −
𝑘−1
The rate is thus proportional to the rate constant k2 for the
second step, and to the equilibrium constant k1/k-1 for the
pre-equilibrium.
STEP 2 is the rate-controlling step.
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𝐶𝑎𝑠𝑒 2, 𝑘−1 ≪ 𝑘2 𝐼 − (𝑖. 𝑒. 𝐼 − is high enough)
𝑘1 𝑘2 Co 𝐶𝑁 5 𝑂𝐻22− 𝐼 −
𝜐=
𝑘−1 + 𝑘2 𝐼 −
25
𝜐 = 𝑘1 Co 𝐶𝑁 5 𝑂𝐻22−
STEP 1 is the rate-controlling step.
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RATE CONSTANTS, RATE COEFFICIENTS, AND EQUILIBRIUM CONSTANTS
For any mechanism, involving any number of elementary steps, the overall equilibrium
constant is the product of the equilibrium constants for the individual steps and is the
product of the rate constants for the reactions in the forward direction divided by the
product of those for the reverse reactions:
𝑘1 𝑘2 𝑘3 …
𝐾𝑐 = 𝐾1 𝐾2 𝐾3 … =
𝑘−1 𝑘−2 𝑘−3 …
This conclusion is related to an important principle, the principle of microscopic
reversibility, which was stated in 1938 by the American physical chemist Richard Chase
Tolman (1881-1948).
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𝑘1
𝐴+𝐵 ⇌ 𝑌+ 𝑋
𝑘−1
𝑘1
If the reaction is elementary, 𝐾𝑐 =
𝑘−1
k1
But, if the reaction is composite, K c is NOT necessarily equal to
k −1
k1 is the RATE CONSTANT (elementary reaction)
k1 is the RATE COEFFICIENT (composite reaction)
If a reaction occurs by a composite mechanism, and we measure a rate coefficient k1 for
the overall reaction from left to right and also measure a rate coefficient k-1, for the
overall reaction from right to left, at the same temperature, the ratio k1/k-1 is not
necessarily the equilibrium constant for the overall reaction.
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10.5 FREE-RADICAL REACTIONS
A+ + B−
(ions)
A ‫ ــــــــ‬B
A. + B. (radicals)
The breaking of covalent bonds, homolytically can be done by supplying energy in two ways:
a. Heating
O
b. Irradiation with light
H
Physical Chem. 2
H-O. + H.
H
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Chain Reactions
A chain mechanism consists of three steps:
a. Initiation-step: radicals are generated
b. Propagation-step: one radical generates another
c. Termination-step: use up one or both of the reactive intermediates.
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The hydrogen-bromine reaction
𝐻2 + 𝐵𝑟2 ⟶ 2𝐻𝐵𝑟
The rate law established by M. Bodenstein and S. C. Lind in 1906 is
𝑑 𝐻𝐵𝑟
𝑑𝑡
=
𝑘 𝐻2 𝐵𝑟2 1/2
𝑚 𝐻𝐵𝑟
1+
𝐵𝑟2
Suggested Mechanism
𝑘1
1. 𝐵𝑟2 →2Br
Initiation
𝑘3
3. 𝐻 + 𝐵𝑟2 →HBr + Br
𝑘4
4. 𝐻 + 𝐻𝐵𝑟 →𝐻2 + Br
Chain Propagation
𝑘2
2. 𝐵𝑟 + 𝐻2 →HBr + H
k and m are constants
Physical Chem. 2
𝑘−1
5. 2𝐵𝑟 → 𝐵𝑟2
Termination or Chain ending
31
K. Al-Sou'od
𝑑 𝐻𝐵𝑟
𝜐𝐻𝐵𝑟 =
= 𝑘2 𝐵𝑟 𝐻2 + 𝑘3 𝐻 𝐵𝑟2 − 𝑘4 𝐻 𝐻𝐵𝑟
𝑑𝑡
Using SS method for both H and Br
𝑑[𝐻]
= 𝑘2 𝐵𝑟 𝐻2 − 𝑘3 𝐻 𝐵𝑟2 − 𝑘4 𝐻 𝐻𝐵𝑟 = 0 … . (1)
𝑑𝑡
𝑘2 𝐵𝑟 𝐻2 − 𝑘4 𝐻 𝐻𝐵𝑟 = 𝑘3 𝐻 𝐵𝑟2
𝝊𝑯𝑩𝒓 = 𝟐𝒌𝟑 𝑯 𝑩𝒓𝟐
𝑑[𝐵𝑟]
= 𝑘1 𝐵𝑟2 − 𝑘2 𝐵𝑟 𝐻2 + 𝑘3 𝐻 𝐵𝑟2 + 𝑘4 𝐻 𝐻𝐵𝑟 −𝑘−1 𝐵𝑟 2 = 0 … . (2)
𝑑𝑡
Equ. (1) + Equ (2)
𝑘1 𝐵𝑟2 −𝑘−1 𝐵𝑟 2 = 0
Physical Chem. 2
𝑘1
𝐵𝑟 =
𝑘−1
K. Al-Sou'od
1/2
𝐵𝑟2 1/2
32
Substitute [Br] into equation (1)
𝑘2
𝑘1
𝑘−1
𝑘2
𝑘1
𝑘−1
1/2
𝐵𝑟2 1/2 𝐻2 − 𝑘3 𝐻 𝐵𝑟2 − 𝑘4 [𝐻] 𝐻𝐵𝑟 = 0
1/2
𝐵𝑟2 1/2 𝐻2 = 𝑘3 𝐵𝑟2 + 𝑘4 𝐻𝐵𝑟
𝑘1 1/2
𝑘2
𝐻2 𝐵𝑟2 1/2
𝑘−1
𝐻 =
𝑘3 𝐵𝑟2 + 𝑘4 𝐻𝐵𝑟
𝐻
𝝊𝑯𝑩𝒓 = 𝟐𝒌𝟑 𝑯 𝑩𝒓𝟐
insertion of the expression for [H] into 𝜐𝐻𝐵𝑟 leads to
𝑘1 1/2
𝟐𝑘2 𝒌𝟑
𝐻2 𝐵𝑟2 3/2
𝑘−1
𝝊𝑯𝑩𝒓 =
𝑘3 𝐵𝑟2 + 𝑘4 𝐻𝐵𝑟
Physical Chem. 2
𝑘1 1/2
𝟐𝑘2
𝐻2 𝐵𝑟2 1/2
𝑘−1
𝝊𝑯𝑩𝒓 =
1 + 𝑘4 /𝑘3 𝐻𝐵𝑟 / 𝐵𝑟2
K. Al-Sou'od
33
𝑘1 1/2
𝟐𝑘2
𝐻2 𝐵𝑟2 1/2
𝑘−1
𝝊𝑯𝑩𝒓 =
1 + 𝑘4 /𝑘3 𝐻𝐵𝑟 / 𝐵𝑟2
𝑑 𝐻𝐵𝑟
𝑑𝑡
=
Physical Chem. 2
𝑘 𝐻2 𝐵𝑟2 1/2
1+
𝑚 𝐻𝐵𝑟
𝐵𝑟2
K. Al-Sou'od
34
Organic Decompositions
A typical organic free-radical chain reaction is the decomposition of ethane,
𝐶2 𝐻6 ⟶ 𝐶2 𝐻4 + 𝐻2
Physical Chem. 2
K. Al-Sou'od
35
𝑘1
1. 𝐶2 𝐻6 →2𝐶𝐻3
𝑘2
2. 𝐶𝐻3 + 𝐶2 𝐻6 →𝐶𝐻4 + 𝐶2 𝐻5
Initiation
𝑘3
3. 𝐶2 𝐻5 →𝐶2 𝐻4 + 𝐻
𝑘4
4. 𝐻 + 𝐶2 𝐻6 →𝐻2 + 𝐶2 𝐻5
𝑘5
5. 2 𝐶2 𝐻5 →𝐶2 𝐻4 + 𝐻
Chain Propagation
Termination or Chain ending
Apply SSA for CH3, C2H5, H
Physical Chem. 2
K. Al-Sou'od
36
For CH3: k1[C2H6] – k2[CH3][C2H6] = 0 …. (1)
For C2H5: k2[CH3][C2H6]– k3[C2H5] + k4[H][C2H6]-k5[C2H5]2 = 0 …. (2)
For H: k3[C2H5] - k4[H][C2H6] = 0 …. (3)
Addition of all three equations gives
k1[C2H6]- k5[C2H5]2 =0
𝐶2 𝐻5 = 𝑘1 /𝑘5 1/2 𝐶2 𝐻6 1/2
The rate of formation of ethylene is
𝜐 = 𝑘3 𝐶2 𝐻5 = 𝑘3 𝑘1 /𝑘5 1/2 𝐶2 𝐻6 1/2
Physical Chem. 2
K. Al-Sou'od
37
EXAMPLE 10.6 The mechanism originally proposed in 1934 by Rice and Herzfeld for
the ethane decomposition was
𝑘1
1. 𝐶2 𝐻6 →2𝐶𝐻3
𝑘2
2. 𝐶𝐻3 + 𝐶2 𝐻6 →𝐶𝐻4 + 𝐶2 𝐻5
Initiation
𝑘3
3. 𝐶2 𝐻5 →𝐶2 𝐻4 + 𝐻
𝑘4
4. 𝐻 + 𝐶2 𝐻6 →𝐻2 + 𝐶2 𝐻5
𝑘5
5. 𝐻 + 𝐶2 𝐻5 →𝐶2 𝐻6
Chain Propagation
Termination or Chain ending
Physical Chem. 2
K. Al-Sou'od
38
For CH3: k1[C2H6] – k2[CH3][C2H6] = 0 …. (1)
For C2H5: k2[CH3][C2H6]– k3[C2H5] + k4[H][C2H6]-k5[H][C2H5] = 0 …. (2)
For H: k3[C2H5] - k4[H][C2H6]- k5[H][C2H5] = 0 …. (3)
Addition of all three equations gives
𝑘1 𝐶2 𝐻6
𝐻 =
𝑘5 𝐶2 𝐻5
k1[C2H6]- k3[H][C2H5] = 0
Insertion of this expression into the SS equation for H gives, after some rearrangement,
The general solution of this quadratic equation is
𝑘3 𝑘5 𝐶2 𝐻5 2 − 𝑘1 𝑘5 𝐶2 𝐻6 𝐶2 𝐻5 − 𝑘1 𝑘4 𝐶2 𝐻6 2 = 0
The general solution of this
quadratic equation is
𝑘1 𝑘5 𝐶2 𝐻6 ± 𝑘12 𝑘52 𝐶2 𝐻6 2 + 4𝑘1 𝑘3 𝑘4 𝑘5 𝐶2 𝐻6 2
𝐶2 𝐻5 =
2𝑘3 𝑘5
Physical Chem. 2
K. Al-Sou'od
39
𝑘1 𝐶2 𝐻6
𝐻 =
𝑘5 𝐶2 𝐻5
k3[C2H5] - k4[H][C2H6]- k5[H][C2H5] = 0
𝑘1 𝐶2 𝐻6
𝑘1 𝐶2 𝐻6
k3[C2H5] - k4
[C2H6]- k5
[C2H5] = 0
𝑘5 𝐶2 𝐻5
𝑘5 𝐶2 𝐻5
𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑦 𝑏𝑦 𝑘5 𝐶2 𝐻5
𝑘3 𝑘5 𝐶2 𝐻5 2 − 𝑘1 𝑘5 𝐶2 𝐻6 𝐶2 𝐻5 − 𝑘1 𝑘4 𝐶2 𝐻6 2 = 0
Physical Chem. 2
K. Al-Sou'od
40
𝑘3 𝑘5 𝐶2 𝐻5 2 − 𝑘1 𝑘5 𝐶2 𝐻6 𝐶2 𝐻5 − 𝑘1 𝑘4 𝐶2 𝐻6 2 = 0
−𝑏 ± 𝑏 2 − 4𝑎𝑐
𝑥=
2𝑎
𝑥 = 𝐶2 𝐻5
𝑎 = 𝑘3 𝑘5
𝑏 = −𝑘1 𝑘5 𝐶2 𝐻6
𝑐 = −𝑘1 𝑘4 𝐶2 𝐻6 2
𝑘1 𝑘5 𝐶2 𝐻6 ± 𝑘12 𝑘52 𝐶2 𝐻6 2 + 4𝑘1 𝑘3 𝑘4 𝑘5 𝐶2 𝐻6 2
𝐶2 𝐻5 =
2𝑘3 𝑘5
Physical Chem. 2
K. Al-Sou'od
41
𝐶2 𝐻5 = 𝐶2 𝐻6
= 𝐶2 𝐻6
𝑘1
+
2𝑘3
𝑘1 𝑘5
+
2𝑘3 𝑘5
𝑘1
2𝑘3
The terms involving
𝐶2 𝐻5 =
𝑘1 𝑘4
𝑘3 𝑘5
2
𝑘12 𝑘52 + 4𝑘1 𝑘3 𝑘4 𝑘5
2𝑘3 𝑘5
𝑘1 𝑘4
+
𝑘3 𝑘5
A satisfactory approximation to the solution can be
obtained by noting that the rate constant k1 is very
small, since it involves the largest activation energy.
𝑘 𝑘
𝑘1
can therefore be neglected in comparison with 1 4
2𝑘3
𝑘3 𝑘5
𝐶2 𝐻6
The rate of formation of ethylene is
Physical Chem. 2
𝜐 = 𝑘3 𝐶2 𝐻5 = 𝑘3
K. Al-Sou'od
𝑘1 𝑘4
𝑘3 𝑘5
𝐶2 𝐻6
42
𝜐 = 𝑘3 𝐶2 𝐻5 = 𝑘3
𝑘1 𝑘4
𝑘3 𝑘5
𝐶2 𝐻6
The Rice-Herzfeld mechanism thus gives first-order kinetics, in agreement with the
experiment. Nevertheless it turned out not to be the correct mechanism. One
difficulty is that it was found that the ethyl radical concentration is much higher than
the hydrogen atom concentration, so that the termination process C2H5 + C2H5 must
be more important than C2H5 + H.
Physical Chem. 2
K. Al-Sou'od
43
Decomposition of acetaldehyde
CH3CHO → CH4 + CO
Under usual conditions the order of reaction is 1.5.
𝑘1
1. 𝐶𝐻3 𝐶𝐻𝑂 ⟶ 𝐶𝐻3 + 𝐶𝐻𝑂
Suggested mechanism
𝑘2
2. 𝐶𝐻3 + 𝐶𝐻3 𝐶𝐻𝑂 ⟶ 𝐶𝐻4 + 𝐶𝐻3 𝐶𝑂
𝑘3
3. 𝐶𝐻3 𝐶𝑂 ⟶ 𝐶𝐻3 + 𝐶𝑂
𝑘4
4. 𝐶𝐻3 + 𝐶𝐻3 ⟶ 𝐶2 𝐻6
Physical Chem. 2
K. Al-Sou'od
44
The steady-state equation for CH3 is
𝑘1 [𝐶𝐻3 𝐶𝐻𝑂] − 𝑘2 [𝐶𝐻3 ] 𝐶𝐻3 𝐶𝐻𝑂 + 𝑘3 [𝐶𝐻3 𝐶𝑂] − 𝑘4 𝐶𝐻3 2 = 0
(1)
The steady-state equation for CH3CO is
𝑘2 [𝐶𝐻3 ] 𝐶𝐻3 𝐶𝐻𝑂 − 𝑘3 [𝐶𝐻3 𝐶𝑂] = 0 (2)
Addition of these two equations gives
𝑘1 [𝐶𝐻3 𝐶𝐻𝑂] − 𝑘4 𝐶𝐻3 2 = 0 (3)
𝒌𝟏
𝑪𝑯𝟑 =
𝒌𝟒
The rate of change of the concentration of acetaldehyde
𝜐 = 𝑘2
𝑘1
𝑘4
1/2
𝐶𝐻3 𝐶𝐻𝑂 3/2
𝟏/𝟐
𝑪𝑯𝟑 𝑪𝑯𝑶 𝟏/𝟐
(𝟒)
𝝊 = 𝒌𝟐 [𝑪𝑯𝟑 ] 𝑪𝑯𝟑 𝑪𝑯𝑶
This agrees with the experimental fact that the order of
the acetaldehyde decomposition is three-halves.
Physical Chem. 2
K. Al-Sou'od
45
HW
Physical Chem. 2
K. Al-Sou'od
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Physical Chem. 2
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Physical Chem. 2
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Physical Chem. 2
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Physical Chem. 2
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Physical Chem. 2
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Physical Chem. 2
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Physical Chem. 2
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