Demo today ???
Colloids
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Hydrophilic and Hydrophobic Colloids
Focus on colloids in water.
“Water loving” colloids: hydrophilic.
“Water hating” colloids: hydrophobic.
Molecules arrange themselves so that hydrophobic
portions are oriented towards each other.
If a large hydrophobic macromolecule (giant molecule)
needs to exist in water (e.g. in a biological cell),
hydrophobic molecules embed themselves into the
macromolecule leaving the hydrophilic ends to interact
with water.
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Hydrophilic and Hydrophobic Colloids……..
Typical hydrophilic groups are polar (containing C-O,
O-H, N-H bonds) or charged.
Hydrophobic colloids need to be stabilized in water.
Adsorption: when something sticks to a surface we say
that it is adsorbed.
If ions are adsorbed onto the surface of a colloid, the
colloids appears hydrophilic and is stabilized in water.
Consider a small drop of oil in water.
Add to the water sodium stearate.
Hydrophilic and Hydrophobic Colloids
Hydrophilic and Hydrophobic Colloids
• Sodium stearate has a long hydrophobic tail
(CH3(CH2)16-) and a small hydrophobic head (-CO2-Na+).
• The hydrophobic tail can be absorbed into the oil drop,
leaving the hydrophilic head on the surface.
• The hydrophilic heads then interact with the water and
the oil drop is stabilized in water.
Removal of Colloidal Particles
• Colloid particles are too small to be separated by
physical means (e.g. filtration).
• Colloid particles may be coagulated (enlarged) until
they can be removed by filtration.
• Methods of coagulation:
– heating (colloid particles move and are attracted to each other
when they collide);
– adding an electrolyte (neutralize the surface charges on the
colloid particles).
– Dialysis: using a semipermeable membranes separate ions
from colloidal particles
Chapter 14
14.1
14.2
14.3
14.4
14.5
14.5
14.7
Chemical Kinetics
Factors that Affect Reaction Rates
Reaction Rates
Changes of Rate with Time
Reaction Rates and Stoichiometry
Concentration and Rate
Exponents in the Rate Law
Units of Rate Constants
Using Initial Rates to Determine Rate Laws
The Change of Concentration with Time
First-Order Reactions
Second-Order Reactions
Half-Life
Temperature and Rate
Reaction Mechanisms
Catalysis
Following the Progress of the Reaction A  B
C4H9Cl(aq) + H2O (l)  C4H9OH (aq) + HCl (aq)
Rate  
[C4 H 9Cl ]
[C4 H 9OH ]

t
t
Note the signs!
In fact, the instantaneous rate corresponds to d[A]/dt
2 HI(g)  H2(g) + I2(g)
Consider the reaction
It’s convenient to define the rate as
1 [ HI ]
[ H 2 ]
[ I 2 ]
rate  


2 t
t
t
And, in general for
aA
+
bB

cC + dD
1 [ A]
1 [ B] 1 [C ] 1 [ D]
Rate  



a t
b t
c t
d t
Sample exercise 14.2
The decomposition of N2O5 proceeds according to the equation
2 N2O5 (g)  4 NO2 (g) + O2 (g)
If the rate of decomposition of of N2O5 at a particular instant in a vessel
is 4.2 X 10-7 M/s, what is the rate of appearance of (a) NO2; (b) O2 ?
1 [ N 2O5 ]
1 [ NO2 ]
1 [O2 ]
Rate  


2
t
4 t
1 t
i.e. the rate of disappearance of N2O5 is 4.2 x 10-7 M/s
the rate of the reaction is 2.1 x 10-7 M/s
the rate of appearance of NO2 is 8.4 x 10-7 M/s
and the rate of appearance of O2 is 2.1 x 10-7 M/s
2 N2O5 = 4 NO2 + O2 (g)
at T = 45 oC in carbon tetrachloride as a solvent
Time
min
∆t
min
[N2O5]
mol/L
0
2.33
184
2.08
319
1.91
526
1.67
867
1.35
1198
1.11
1877
0.72
∆[N2O5]
mol/L
- ∆[N2O5]/ ∆t
mol/L-min
2 N2O5 = 4 NO2 + O2 (g)
at T = 45 oC in carbon tetrachloride as a solvent
Time
min
∆t
min
0
[N2O5]
mol/L
- ∆[N2O5]/ ∆t
mol/L-min
0.25
1.36 x 10-3
0.17
1.26 x 10-3
0.24
1.16 x 10-3
0.32
0.94 x 10-3
0.24
0.72 x 10-3
0.39
0.57 x 10-3
2.33
184
184
2.08
135
319
1.91
207
526
1.67
341
867
1.35
331
1198
1.11
679
1877
∆[N2O5]
mol/L
0.72
The information on the previous slide is a bit of a
nuisance, since the instantaneous rate keeps changing
—and you know how much we like constant values
or linear relationships!
So let’s try something rather arbitrary at this point.
Let’s divide the instantaneous, average rate by
[N2O5]
and/or
[N2O5]2
2 N2O5 = 4 NO2 + O2 (g)
at T = 45 oC in carbon tetrachloride as a solvent
[N2O5]
mol/L
[N2O5]av ∆[N2O5]
avg
mol/L
- ∆[N2O5]/ ∆t
mol/L-min
Avg rate
/[N2O5]av
Avg rate
/[N2O5]av2
2.21
0.25
1.36 x 10-3
6.2 x 10-4
2.8 x 10-4
2.00
0.17
1.26 x 10-3
6.3 x 10-4
3.2 x 10-4
1.79
0.24
1.16 x 10-3
6.5 x 10-4
3.6 x 10-4
1.51
0.32
0.94 x 10-3
6.2 x 10-4
4.1 x 10-4
1.23
0.24
0.72 x 10-3
5.9 x 10-4
4.8 x 10-4
0.92
0.39
0.57 x 10-3
6.2 x 10-4
6.7 x 10-4
2.33
2.08
1.91
1.67
1.35
1.11
0.72
Notice the nice
constant value!!!
which we’ll call ‘k’
It’s convenient to write this result in symbolic form:
Rate = k [N2O5]
where the value of k is about 6.2 x 10-4
so that when [N2O5] = 0.221,
Rate = (6.2 x 10-4 )(0.221)
= 1.37 x 10-4 which is the ‘average rate’
we started with
In fact, we really should take into account the 2 in front of the
N2O5, in accordance with the rule we developed earlier.
This leads us to the general concept of Reaction Order
When Rate = k [reactant 1]m [reactant 2]n
we say the reaction is
m-th order in reactant 1
n-th order in reactant 2
and (m + n)-th order overall.
Be careful—because these orders are NOT related necessarily
to the stoichiometry of the reaction!!!
Other reactions and their observed reaction orders
2 N2O5 = 4 NO2 + O2 (g)
Rate = k [N2O5]
!!!
CHCl3 (g) + Cl2 (g)  CCl4 (g) + HCl(g) Rate = k[CHCl3][Cl2]1/2
H2 (g) + I2 (g)
 2 HI (g)
Rate = k[H2][I2]
The order must be determined experimentally!!!
We’ll see later that it depends on the Reaction Mechanism,
rather than the overall stoichiometry.
Be careful: the measurement of the rate will always depend on
observations of the reactants or products and involves stoichiometry,
but the part on the right, the order, does not depend on the stoichiometry.
Let’s explore the results when
Rate = k [N2O5]
This can be expressed as
Rate = - (Δ[N2O5] / Δ t = - d[N2O5] / dt = k [N2O5]
or, in general for
A  products
Rate = - Δ[A] / Δt = - d[A] / dt = k [A]
rearrangement and integration from time = 0 to t = t gives the result
ln[A]t - ln[A]o = - kt
or
ln [A]t = - kt + ln [A]o
or
ln ([A]t/[A]o = - kt
where [A]o = conc. at t = 0
This is the expression of concentration vs time
for a First-Order Reaction
Same information in a better format:
[ A]
d [ A]
Rate  

 k[ A]
t
dt

[ A]t
[ A]0
t
d [ A]
 k  dt
0
[ A]
ln[ A]t   kt  ln[ A]0
To give
these forms
of the
“integrated
rate law”:
or ln[ A]t  ln[ A]0   kt
 [ A]t 
or ln 
   kt
[ A]0 
Consider
First-Order Reactions
[ A]
d [ A]
Rate  

 k[ A]
t
dt

[ A]t
[ A]0
t
d [ A]
 k  dt
0
[ A]
ln[ A]t   kt  ln[ A]0
To give
these forms
of the
“integrated
rate law”:
or ln[ A]t  ln[ A]0   kt
 [ A]t 
or ln 
   kt
[ A]0 
An example of the plots of concentration vs time
for a First-Order Reaction
The Change of Concentration with Time
Half-Life
• Half-life is the time taken for the
concentration of a reactant to drop to half its
original value.
• That is, half life, t1/2 is the time taken for [A]0
to reach ½[A]0.
• Mathematically,
t1  
2
ln 1
k
2  0.693
k
The Change of Concentration with Time
For a First-Order Reaction
The identical length of the
first and second half-life
is a SPECIFIC characteristic
of First-Order reactions
Consider now
Second-Order Reactions
[ A]
d [ A]
Rate  

 k[ A]2
t
dt

[ A]t
[ A]0
t
d [ A]
 k  dt
2
0
[ A]
1
1
 kt 
[ A]t
[ A]0
Not
e
the
2!
Second-Order Reactions
• We can show that the half life
1
t1 
2 k A 0
• A reaction also can have rate constant
expression of the form
rate = k[A][B],
i.e., be second order overall, but be first order
in A and in B.
Recall for
1st order:
t1  
ln 1
k
2
2  0.693
k
And for 2nd order:
t1 
2
Is this first or second order in the reactant? What is k?
1
k A 0
t1/2 = 1.73 sec
(‘2nd half-life’)
t1/2 = 0.693/0.4 =
1.73 sec
t1/2 = [(0.4)(0.5)] -1
= 5.0 sec
(‘2nd half life’)
t1/2 = (k[A]0)-1 = [(0.4)(1.0)] -1
= 2.5 sec
(‘1st half-life’)
MQ-1 122, WI 07:
THE MEAN WAS
THE MODE WAS 95.00 WITH 10 STUDENTS.
THE MEDIAN WAS 112.50
THE STANDARD DEVIATION WAS 29.35
112.00 (64.00%)
High: 175
Low 38
N = 472
THE POSSIBLE RANGE WAS FROM 0.00 TO 175.00
THE ACTUAL RANGE WAS FROM 38.00 TO 175.00 (21.7% TO 100.0%)
CUTS TAKEN AT 72.88 (APROX. 72.50) AND 26.86 (APROX. 27.50) PERCENTILES.
THE UPPER CUT CONTAINED 130 STUDENTS; THE LOWER, 130 STUDENTS.
TOTAL NUMBER OF DETECTED ERRORS: 0.
DISTRIBUTION OF SCORES
0.00 - 9.99
0
10.00 - 19.99
0
20.00 - 29.99
0
30.00 - 39.99
1
40.00 - 49.99
8
50.00 - 59.99
6
60.00 - 69.99
22
70.00 - 79.99
32
80.00 - 89.99
45
90.00 - 99.99
50
100.00 - 109.99 51
110.00 - 119.99 59
120.00 - 129.99 50
130.00 - 139.99 53
140.00 - 149.99 38
150.00 - 159.99 42
160.00 - 169.99 12
170.00 - 175.00
3
These students are in
danger of failing!
Please See Me!!!
Correct answers:
BCDAE CCDDE CADBE BDABD AAADC ACDBA A
Chapter 14
14.1
14.2
14.3
14.4
14.5
14.5
14.7
Chemical Kinetics
Factors that Affect Reaction Rates
Reaction Rates
Changes of Rate with Time
Reaction Rates and Stoichiometry
Concentration and Rate
Exponents in the Rate Law
Units of Rate Constants
Using Initial Rates to Determine Rate Laws
The Change of Concentration with Time
First-Order Reactions
Second-Order Reactions
Half-Life
Temperature and Rate
Reaction Mechanisms
Catalysis
First Order Reactions
d [ A]
rate  
  k [ A]
dt
ln[ A]t   kt  ln[ A]0
or ln[ A]t  ln[ A]0   kt
Second Order Reactions
d [ A]
rate  
 k [ A]2
dt
1
1
 kt 
[ A]t
[ A]0
 [ A]t 
or ln 
  kt

 [ A]0 
ln 2 0.693
t 12 

k
k
1
t1 
2
k A0
Example Exercise 14.8
NO2 (g)  NO (g) + ½ O2 (g) at 300 oC Page 540
Time/s[NO2]
0.0
50.0
100.0
200.0
300.0
ln[NO2]
0.0100
0.00787
0.00649
0.00481
0.00380
-4.610
-4.845
-5.038
-5.337
-5.573
Is the reaction first or second order in NO2 ?
What is the rate constant for the reaction?
1/[NO2]
100
127
154
208
263
Second-Order Processes
The decomposition of NO2 at 300°C is described by
the equation
NO2 (g)
NO (g) + 1/2 O2 (g)
and yields data comparable to this:
Time (s)
0.0
50.0
100.0
200.0
300.0
[NO2], M
0.01000
0.00787
0.00649
0.00481
0.00380
Second-Order Processes
• Graphing ln [NO2] vs. t
yields:
• The plot is not a straight
line, so the process is not
first-order in [A].
Time (s)
0.0
50.0
[NO2], M
0.01000
0.00787
ln [NO2]
−4.610
−4.845
100.0
200.0
300.0
0.00649
0.00481
0.00380
−5.038
−5.337
−5.573
Second-Order Processes
• Graphing ln
1/[NO2] vs. t,
however, gives this
plot.
Time (s)
0.0
50.0
100.0
200.0
300.0
[NO2], M
0.01000
0.00787
1/[NO2]
100
127
0.00649
0.00481
0.00380
154
208
263
• Because this is a
straight line, the
process is secondorder in [A].
Example of Second-Order Plots of conc vs time
NO2 (g)  NO (g) + ½ O2 (g) at 300 oC (See page 591)
Since graph (b) gives a straight line, it is a second-order reaction in NO2 .
i.e. rate = k[NO2]2 .
And the slope is k, where k = 0.534 M -1 s -1 .
Similar problem based on reaction
2 C2F4  C4F8
Similar problem based on reaction
C2F4  1/2 C4F8
which we are told is 2nd order in the reactant,
with a rate constant of 0.0448 M -1 s-1
or 0.0448 L mol -1 s -1 at 450 K.
If the initial concentration is 0.100 M, what
will be the concentration after 205 s ?
1/Ct = 1/Co + kt
= 1/(0.100 M) + (0.0448 M -1 s-1 )(205 s)
= 19.2 M -1 or Ct = 5.21 x 10 -2 M
General Order of reaction
First Order reactions
Second Order reactions
Integrated form of each
Half lives of each
Take out a clean sheet of paper and print on it
your name and section number
and an indication it is “pop quiz # 2”.
Mon Rec, 2:30, 3:30
Wed Rec, 2:30, 3:30
Josh Dettman
Brian Culp
Venuka Durani
Xiangke Chen
115
120
116
119
117
118
Jason Hoy
Luyuan Zhang
109
112
110
113
111
113
Take out a clean sheet of paper and print on it
your name and section number
and an indication it is “pop quiz # 2”.
Tue Rec
Thur Rec
Dan Poole
122
Dan Poole
126
Nick Selner
123
Nick Selner
127
Jim Bowsher
124
Jim Bowsher
125
Yu-Kay Law
Nadia Casillas
129
128
Yu-Kay Law
121
A certain reaction has the form A  B. At 400 K and
[A]0 = 2.8 x 10 -3 M, data for a plot of [A] vs t were collected.
It was then found that a plot of
1/[A] vs t yielded a
straight line with a slope of 3.60 x 10-2 L∙mol -1 ∙s -1 .
a) Write an expression for the rate law.
rate = k[A]2
b) Write an expression for the integrated rate law.
1/[A] = 1/[A]0 + kt
c) What is the rate constant for the reaction?
3.60 x 10-2 L∙mol -1 ∙s -1
d) What is the half-life for the reaction, as given?
1
1
10 5
3
t1 



9
.
9
x
10
2
k [ A]0
(3.60 x102 )( 2.8 x10 3 ) (3.6)( 2.8)
A certain reaction has the form A  B. At 400 K and
[A]0 = 5.6 x 10 -3 M, data for a plot of [A] vs t were collected.
It was then found that a plot of
1/[A] vs t yielded a
straight line with a slope of 1.80 x 10-2 L∙mol -1 ∙s -1 .
a) Write an expression for the rate law.
rate = k[A]2
b) Write an expression for the integrated rate law.
1/[A] = 1/[A]0 + kt
c) What is the rate constant for the reaction?
1.80 x 10-2 L∙mol -1 ∙s -1
d) What is the half-life for the reaction, as given?
1
1
10 5
3
t1 



9
.
9
x
10
2
k [ A]0
(1.80 x102 )(5.6 x10 3 ) (1.8)(5.6)
Remember this figure? Let’s use it for another point.
The ‘initial rate’
is very useful in
determining the
order of a reaction.
It is the only time
when there are no
products present!
SAMPLE EXERCISE 14.6 Determining a Rate Law from Initial Rate Data
The initial rate of a reaction
B, and the results are as follows:
was measured for several different starting concentrations of A and
Using these data, determine (a) the rate law for the reaction, (b) the magnitude of the rate constant, (c) the rate
of the reaction when [A] = 0.050 M and [B] = 0.100 M.
Solution
Analyze: We are given a table of data that relates concentrations of reactants with initial rates of reaction and
asked to determine
(a) the rate law, (b) the rate constant, and (c) the rate of reaction for a set of concentrations not listed in the table.
Plan: (a) We assume that the rate law has the following form: Rate = k[A]m[B]n, so we must use the given data
to deduce the reaction orders m and n. We do so by determining how changes in the concentration change the
rate. (b) Once we know m and n, we can use the rate law and one of the sets of data to determine the rate
constant k. (c) Now that we know both the rate constant and the reaction orders, we can use the rate law with the
given concentrations to calculate rate.
Solve: (a) As we move from experiment 1 to experiment 2, [A] is held constant and [B] is doubled. Thus, this
pair of experiments shows how [B] affects the rate, allowing us to deduce the order of the rate law with respect
to B. Because the rate remains the same when [B] is doubled, the concentration of B has no effect on the
reaction rate. The rate law is therefore zero order in B (that is, n = 0).
SAMPLE EXERCISE 14.6 continued
In experiments 1 and 3, [B] is held constant so these data show how [A] affects rate. Holding [B] constant while
doubling [A] increases the rate fourfold. This result indicates that rate is proportional to [A] 2 (that is, the reaction
is second order in A). Hence, the rate law is
This rate law could be reached in a more formal way by taking the ratio of the rates from two experiments:
Using the rate law, we have
2n equals 1 under only one condition:
We can deduce the value of m in a similar fashion
Using the rate law gives
SAMPLE EXERCISE 14.6 continued
Because 2m = 4, we conclude that
(b) Using the rate law and the data from experiment 1, we have
(c) Using the rate law from part (a) and the rate constant from part (b), we have
Because [B] is not part of the rate law, it is irrelevant to the rate, provided that there is at least some B present to
react with A.
Check: A good way to check our rate law is to use the concentrations in experiment 2 or 3 and see
if we can correctly calculate the rate. Using data from experiment 3, we have
Thus, the rate law correctly reproduces the data, giving both the correct number and the correct
units for the rate.
SAMPLE EXERCISE 14.7 Using the Integrated First-Order Rate Law
The decomposition of a certain insecticide in water follows first-order kinetics with a rate constant of 1.45 yr–1
at 12°C. A quantity of this insecticide is washed into a lake on June 1, leading to a concentration of 5.0  10–7
g/cm3. Assume that the average temperature of the lake is 12°C. (a) What is the concentration of the insecticide
on June 1 of the following year? (b) How long will it take for the concentration of the insecticide to drop to
3.0  10–7 g/cm3?
Solution
Analyze: We are given the rate constant for a reaction that obeys first-order kinetics, as well as information
about concentrations and times, and asked to calculate how much reactant (insecticide) remains after one year.
We must also determine the time interval needed to reach a particular insecticide concentration. Because the
exercise gives time in (a) and asks for time in (b), we know that the integrated rate law, Equation 14.13, is
required.
Plan: (a) We are given k = 1.45 yr–1, t = 1.00 yr, and [insecticide]0 = 5.0  10–7 g/cm3, and so Equation 14.13
can be solved for 1n[insecticide]t. (b) We have k = 1.45yr–1, [insecticide]0 = 5.0  10–7 g/cm3, and [insecticide]t
= 3.0  10–7 g/cm3, and so we can solve Equation 14.13 for t.
Solve: (a) Substituting the known quantities into Equation 14.13, we have
We use the ln function on a calculator to evaluate the second term on the right, giving
To obtain [insecticide]t = 1 yr, we use the inverse natural logarithm, or ex, function on the calculator:
Note that the concentration units for [A]t and [A]0 must be the same.
SAMPLE EXERCISE 14.7 continued
(b) Again substituting into Equation 14.13, with [insecticide] t = 3.0  10–7 g/cm3, gives
Solving for t gives
Check: In part (a) the concentration remaining after 1.00 yr (that is,1.2  10–7 g/cm3) is less than the original
concentration (5.0  10–7 g/cm3), as it should be. In (b) the given concentration (3.0  10–7 g/cm3) is greater
than that remaining after 1.00 yr, indicating that the time must be less than a year. Thus, t = 0.35 yr is a
reasonable answer.
PRACTICE EXERCISE
The decomposition of dimethyl ether, (CH3)2O, at 510°C is a first-order process with a rate constant of
6.8  10–4s–1:
If the initial pressure of (CH3)2O is 135 torr, what is its partial pressure after 1420 s?
Answer: 51 torr
SAMPLE EXERCISE 14.8 Determining Reaction Order from the Integrated Rate Law
The following data were obtained for the gas-phase decomposition of nitrogen dioxide at 300°C,
Is the reaction first or second order in NO2?
Solution
Analyze: We are given the concentrations of a reactant at various times during a reaction and asked to
determine whether the reaction is first or second order.
Plan: We can plot ln[NO2] and 1/[NO2] against time. One or the other will be linear, indicating whether the
reaction is first or second order.
SAMPLE EXERCISE 14.8 continued
Solve: In order to graph ln[NO2] and 1/[NO2] against time, we will first prepare the following table from
the data given:
SAMPLE EXERCISE 14.8 continued
As Figure 14.8 shows, only the plot of 1/[NO2] versus time is linear. Thus, the reaction obeys a second-order rate
law: Rate = k[NO2]2. From the slope of this straight-line graph, we determine that k = 0.543 M–1 s–1 for the
disappearance of NO2.
PRACTICE EXERCISE
Figure 14.8 Kinetic
data for
decomposition of NO2.
The reaction is NO2(g)
NO(g)
+ 1/2O2(g), and the data
were collected at 300°C.
(a) A plot of [NO2] versus
time is not linear,
indicating that the
reaction is not first order
in NO2. (b) A plot of
1/[NO2] versus time is
linear, indicating that the
reaction is second order
in NO2.
Consider again the decomposition of NO2 discussed in the Sample Exercise. The reaction is second order in NO 2
with k = 0.543 M–1s–1. If the initial concentration of NO2 in a closed vessel is 0.0500 M, what is the remaining
concentration after 0.500 h?
Answer: Using Equation 14.14, we find [NO2] = 1.00  10–3 M
Using ‘Initial Rates’ to determine order of reaction.
Chemical Kinetics (cont)
14.3
14.4
14.5
The Change of Concentration with Time
First-Order Reactions
Half-Life
Second-Order Reactions
Temperature and Rate
The Collision Model
Activation Energy
The Orientation Factor
The Arrhenius Equation
Reaction Mechanisms
Elementary Steps
Multistep Mechanisms
Rate Laws of Elementary Steps
Rate Laws of Multistep Mechanisms
Mechanisms with and Initial Fast Step
Note the DRAMATIC effect of temperature on k
Temperature and Rate
The Collision Model eg H2 + I2
The Collision Model
• The more molecules present, the greater the
probability of collision and the faster the rate.
• Complication: not all collisions lead to products. In
fact, only a small fraction of collisions lead to
product.
• The higher the temperature, the more energy
available to the molecules and the faster the rate.
• In order for reaction to occur the reactant molecules
must collide in the correct orientation and with
enough energy to form products.
Activation Energy
• Arrhenius: molecules must posses a
minimum amount of energy to react. Why?
– In order to form products, bonds must be broken
in the reactants.
– Bond breakage requires energy.
• Activation energy, Ea, is the minimum energy
required to initiate a chemical reaction.
Activation Energy
Activation Energy
• Consider the rearrangement of acetonitrile:
H3C N C
H3C
N
C
H3C C N
– In H3C-NC, the C-NC bond bends until the C-N
bond breaks and the NC portion is
perpendicular to the H3C portion. This structure
is called the activated complex or transition state.
– The energy required for the above twist and
break is the activation energy, Ea.
– Once the C-N bond is broken, the NC portion
can continue to rotate forming a C-CN bond.