Kinetics Part IV:
Activation Energy
Jespersen Chapter 14 Sec 5 & 6
Dr. C. Yau
Spring 2014
1
The Collision Theory
The rate of a reaction is proportional to the
number of effective collisions per second
among the reactant molecules.
We already know concentration plays an
important part in rxn rate:
Conc
Freq Collision
Rxn Rate
Only EFFECTIVE collisions lead to products.
Only a small fraction of collisions lead to
products, based on two other factors:
1) Activation Energy
2
2) Molecular Orientation
Kinetic Energy Distribution
Ea = Activation energy = minimum energy needed
REMEMBER: This is the
for collision to be effective
Fig 14.11 p.666
Activation Energy is not affected by
increase in temperature.
graph for Kinetic Energy.
Don’t confuse with graph
for Potential Energy.
3
Collision Theory Of Reactions
For a reaction to occur, three conditions must be met:
1. Reactant particles must collide.
2. Collision energy must be enough to break bonds/initiate.
3. Particles must be oriented so that the new bonds can form.
e.g. NO2Cl + Cl
NO2 + Cl2
4
Eqn Summarizing 3 Factors in Collision Theory
Particulate Level:
Rxn Rate (molecules L-1 s-1) =
N x forientation x fKE
N = # collisions per second per liter of mixture
forientation = fraction of collisions with effective
orientation
fKE = fraction of collisions with sufficient
kinetic energy for effective collision (area
under the curve with KE  Ea
5
Mathematically, fKE
has been found to be related to Ea
and T in this equation:
-E a
ln f KE =
RT
Simplify by finding the anti ln of both sides of the equation...
antiln  ln f KE 
 -E a 
= anti ln 

RT


f KE =
e
 -E a 


 RT 
Still remember what fKE stands for?
6
Eqn Summarizing 3 Factors in Collision Theory
Macroscopic Level:
Equation has to be in terms of moles instead
of molecules.
1 mole of molecules
Conversion factor is
6.02x1023 molecules
So we divide the previous equation by
Avogadro’s number to get
reaction rate in units of mol L-1 s-1
Rxn Rate (molecules L-1 s -1 )
6.02x1023 (molecules mol-1 )
mol
= (M/s)
Ls
7
Temperature Effects
Changes in temperature affect the rate
constant, k, according to the
 Ea/RT
Arrhenius equation:
k  pZe
p is the steric factor
Z is the frequency of collisions.
Ea is the activation energy
R is the Ideal Gas Constant (8.314 J/(mol K)
T is the temperature (K)
We substitute A (the frequency factor) for (pZ))
This is an important
equation to remember!
k  Ae  Ea/RT
8
Graphical Determination of Ea
k  Ae  Ea/RT
ln k = ln A + ln (e -Ea/RT )
Ea
ln k = ln A RT
Ea  1 
ln k = ln A  
R T
Ea  1 
ln k =   + ln A
R T
y = m x + b
You are expected to be able to
derive this yourself.
How exactly do we
determine Ea?
What do we plot on
the x-axis? on the yaxis?
How do we find Ea
on the graph?
9
Example 14.12 p.670: Determine Ea in kJ/mol
What do we do with this data?
Ea  1 
ln k =   + ln A
R T
10
11
Ln k vs. 1/T
-0.70
slope =
5.0x10-5
= - 1.4x104 (units ?)
Ln k
Then what?...
-0.70
5.0x10-5
= -1.4x104 (units?)
slope =
How do we find Ea?
Ea  1 
ln k =    + ln A
R T
Ea =  slope x R
= 1.2x102 kJ/mol
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Determination of Ea from k at 2 temperatures
k  Ae  Ea/RT
Ratio form: Can be used when A isn’t known.
 k 2  -Ea  1 1 
ln   
 - 
 k 1  R  T2 T1 
You should be able to derive this equation
for yourself. We did a similar derivation
earlier this semester for Hvap, VP and T.
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Example
Given that k at 25°C is 4.61×10-1 M/s and
that at 50°C it is 4.64×10-1 M/s, what is the
activation energy for the reaction?
 k 2  -Ea  1 1 
ln   
 - 
 k 1  R  T2 T1 
 4.64 x10  1 M /s 
-E a
1 
 1
ln 


-1
-1 -1 
 4.61x10 M /s  8.314 J m ol K  323K 298K 
Ea= 188 J/mol = 2 x102 J/mol
Can you think of a reason why the graphical method
would give a more accurate value for Ea?
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 1 1
 

 T2 T1 
-E a
4.64x10-1
1 
 1
ln
=


-1
-1 -1 
4.61x10
8.314 J mol K  323 K 298 K 
-E a
-1
-1
ln 1.006
=
0.00309
5
K
0.00335
5
K

-1 -1 
8.314 J mol K
-E a
-1
0.0059
=
(0.00026
0
K
)
-1 -1
8.314 J mol K
 0.0059  8.314 J mol-1 
Ea
=
 188.66...  2x10 2 J/mol
0.000260
k2
ln
k1
-E a
=
R
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Working With The Arrhenius Equation
Given the following
data, predict k at 75°C
using the graphical
approach.
k  Ae  Ea/RT
T °C
k (M/s)
0.000886
25
0.000894
50
0.000908
100
0.000918
150
Ea  1 
ln k =   + ln A
R T
Trendline: y = -36.025x – 6.908
k = ? at T = 75oC
ANS k=9.01×10-4M/s
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T deg C
25
50
k
0.000886
0.000894
1/T (in K-1)
0.003355705
0.003095975
ln k
-7.028793607
-7.019804783
150
0.000918
0.002364066
-6.993313167
100
0.000908
0.002680965
-7.004266179
ln k vs. 1/T
y = -36.025x - 6.908
R² = 0.9997
-6.99
-6.995
-7
-7.005
-7.01
-7.015
-7.02
-7.025
-7.03
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In the reaction 2N2O5(g) 4 NO2(g) + O2(g)
the following temperature and rate
constant information is obtained.
What is the activation energy of the
reaction?
T (K)
k (s-1)
A. 102 kJ mol-1
B. -102 kJ mol-1
C. 1004 kJ mol-1
D. -1004 kJ mol-1
E. none of these
338
328
318
4.87(10-3)
1.50(10-3)
4.98(10-4)
Ea  1 
ln k =    + ln A
R T
If we are to determine Ea graphically, what do we graph?
Slope = -1.224x104 K and y-intercept = 30.9, what are the
units? What is Ea?
Practice with Example 14.12 p.672, Exer.26,27,28
18
Potential Energy Diagrams
The product is said to be “thermodynamically favored”
over the reactant. LEARN THIS TERMINOLOGY.
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Potential Energy Diagrams
• demonstrate the energy needs and
products as a reaction proceeds
• tell us whether a reaction is exothermic or
endothermic
• tell us if a reaction occurs in one step or
several steps
• show us which step is the slowest
Do not confuse PE diagram with KE
diagram! Learn the terminology!
So, remember which is the KE diagram?
20
Potential Energy Diagram
What would the potential energy diagram
look like for an endothermic reaction?
Make a sketch of a PE diagram for an
endothermic reaction.
Where do we look to find the activation
energy?
Where do we look to find the heat absorbed
during the reaction?
What is thermodynamically favored?
21
Catalysts
• speed a reaction, but
are not consumed by
the reaction
• may appear in the rate
law
• lower the Ea for the
reaction
• may be heterogeneous
or homogeneous
22
PE
PE Graph
Ea of uncatalyzed rxn
Ea of
catalyzed
rxn
Reaction Coordinate
Fraction of Molecules
KE Graph:
Effect of Catalyst
Ea with catalyst
Ea without catalyst
000
000
Kinetic Energy
23
Catalytic Actions
• may serve to weaken bonds through
induction
• may serve to change polarity through
amphipathic/surfactant effects
• may reduce geometric orientation effects
• Heterogeneous catalyst: reactant and
product exist in different states.
• Homogeneous catalyst: reactants and
catalyst exist in the same physical state
24
Example of a heterogeneous catalyst
Well-known “The Haber Process.”
3H2 (g) + N2 (g)
Fe
2NH3 (g)
Fe
Note: Fe is never consumed.
Catalysts do not have to be in large amounts.
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