8.3 Factors Affecting RXN Rate

REACTION KINETICS
8.3 Factors Affecting Reaction Rate

Factors affecting Reaction Rate
Concentration or Pressure
Temperature
Catalyst
Particle Size
Explain the effect of temperature
Factors using Maxwell-Boltzman Curve
affecting Explain the effect of catalyst using
reaction energy profile diagram
rate
State Arrhenius equation
Relate temperature and Ea based on
Arrhenius Equation
Determine k,Ea, T and A using Arrhenius
Equation
Kindling (small sticks) Log
The kindling, with its greater surface area, allows

for more contact with oxygen and a faster reaction
REACTION RATE
 Each reaction has its own characteristic
rate
 Determined by the chemical nature of

the reactants
FACTORS AFFECTING
REACTION RATE
1. Concentration
 Concentration of reactant ↑
 Frequency of effective collision ↑
 Reaction rate↑
FACTORS AFFECTING
REACTION RATE
2. Pressure
 For gaseous reactants
 If pressure increased by reducing
volume
 [reactant] ↑, frequency of effective
collision ↑,
 rate of reaction ↑
FACTORS AFFECTING
REACTION RATE
3. Particle Size
 Size of reacting particles decrease, total

surface area exposed for reaction
increase
 Reaction rate increase

FACTORS AFFECTING
REACTION RATE
4. Temperature
The effect of temperature on the reaction rate
can be explained in terms of kinetic theory.
The average kinetic energy ∝ temperature
 The higher the temperature the higher the rate of

reaction.
 Maxwell-Boltzmann distribution shows the kinetic
energy distributions for a reaction mixture at two
different temperature.
Maxwell-Boltzmann distribution
•Only collisions with energy greater

than Ea are able to react.
•At lower temperature T1, the fraction

of energetic (and effective) collision
quite small
•As the temperature increases, the

distribution broadens and shifts
towards higher energy
•T ↑, average kinetic energy of the molecules ↑ ,

number of molecules with energy equal or greater than
Ea ↑ , effective collision ↑ , k ↑ .
FACTORS AFFECTING
REACTION RATE
5. Catalyst
A catalyst is a substance that increases the
rate of a chemical reaction without itself
being consumed.
A catalyst provides an alternative pathway

which has a lower activation energy
compared to the one without catalyst
Energy profile diagram showing the difference reaction
in the presence of catalyst and without catalyst.
Without catalyst With catalyst Combine diagram

ARRHENIUS EQUATION
k: rate constant
– Ea/RT R: universal gas constant
k = Ae
(8.314 Jmol–1 K–1 )
e: base of natural logarithm
T: absolute temperature (in K)
Ea: activation energy
A: frequency factor
Ea 1
ln k = ln A –
R T
k = Ae – Ea/RT
 Higher T  larger k  increased rate
 Temperature affects the rate by affecting

the rate constant (k)
TEMPERATURE AND RATE
Arrhenius equation k = Ae – Ea/RT
f = e – Ea/RT f: Fraction of molecular collisions

with energy greater or equal to Ea.
 T  average speed of particles 

collision frequency 
f = e – Ea/RT f: Fraction of molecular collisions

with energy greater or equal to Ea.
 Only those collisions with enough energy

to exceed Ea can lead to reaction
 Temperature rise enlarge fraction (f) of

collision with enough energy to exceed Ea
 Magnitudes of both Ea and T affect the

fraction of sufficiently energetic collisions
• Higher T  larger k  increased rate
• Larger Ea  smaller k  decreased rate

EXAMPLE: 1
The decomposition of N2O5 has an activation
energy of 103 kJ and a frequency factor of
4.3 x 1013 s–1. What is the rate constant for
this decomposition at
(a) 20oC
R = 8.314 Jmol–1 K–1
(b) 100oC
k = Ae– Ea/RT
CHAPTER 8.3
EXAMPLE: 1
(a)
k = Ae– Ea/RT
A = 4.3 x 1013 s–1 T = 20oC

Ea = 103 x 103 Jmol–1 = (20 + 273)K
= 293 K
R = 8.314 Jmol–1 K–1
103 x 103 Jmol–1

–
k = 4.3 x 1013 s–1 x e 8.314 Jmol–1 K–1 x 293 K
= 1.9 x 10–5 s–1

EXAMPLE: 1
(b)
k = Ae– Ea/RT
A = 4.3 x 1013 s–1 T = 100oC

Ea = 103 x 103 Jmol–1 = (100 + 273)K
= 373 K
R = 8.314 Jmol–1 K–1
103 x 103 Jmol–1

–
k = 4.3 x 1013 s–1 x e 8.314 Jmol–1 K–1 x 373 K
= 0.16 s–1
LEARNING OUTCOME
SK026 8.3 (f)
 Determine k, Ea, T and A using Arrhenius

equation by calculation and graphical method.
The equation:
 - E a  1 
ln k      ln A
 R  T 
compare with linear equation:
y = mx + c
ln k
y – axis : ln k
X – axis :  1 
ln A –
 
T
So,gradient, m = - E 
 a

 R 
 1  -1
c = ln A  K
T
Ea can be determine
from the graph!
Ea 1
ln k = ln A –
R T
see the linear relationship…?

(y = mx + c)
R must be 8.314 Jmol–1K–1
not 0.08206 atmLmol–1K–1
because the unit of Ea
is in J or kJ !
CHAPTER 8.3
EXAMPLE: 2
The rate constant for the decomposition of
nitrous oxide (N2O) into N2 molecule and O atom
has been measured at different temperatures:
k (1/M•s) T (oC)
0.00187 600
0.01130 650
0.05690 700
0.24400 750
Determine graphically the activation energy

for the reaction.
EXAMPLE: 2
k (1/M•s) ln k T (oC) T (K) 1/T (K–1)
0.00187 – 6.281 600 873 1.15 x 10–3
0.01130 – 4.483 650 923 1.08 x 10–3
0.05690 – 2.866 700 973 1.03 x 10–3
0.24400 – 1.411 750 1023 9.78 x 10–4
 Plot graph: ln k versus 1/T

EXAMPLE: 2
ln k Ea 1
ln k = ln A –
R T
ln A
Slope = – Ea/R = – 2.89 x 104 K
Ea = ( 8.314 J/K•mol )( 2.89 x 104 K)

= 2.40 x 105 J/mol
= 240.2 kJ/mol
Ea
slope = –
R
1/T (K–1)
Ea 1
ln k = ln A –
R T
 If we know the rate constant at

two different temperature:
Ea 1 1
ln k2 = ln A – ln k1 = ln A – Ea
R T2 R T1
k2 Ea 1 1
ln = – –
k1 R T2 T1
CHAPTER 8.3
EXAMPLE: 3
The decomposition of HI has rate constants
k = 0.079 Lmol–1 s–1 at 508oC
k = 0.24 Lmol–1 s–1 at 540oC.
What is the activation energy of this reaction in
kJmol–1?
k2 Ea 1 1
ln = – –
k1 R T2 T1
EXAMPLE: 3
k2 Ea 1 1
ln = – –
k1 R T2 T1
k1 = 0.079 Lmol–1 s–1 T1 = 508oC

= (508 + 273)K = 781 K
k2 = 0.24 Lmol–1 s–1 T2 = 540oC

= (540 + 273)K = 813 K
Ea = ?
EXAMPLE: 3
k2 Ea 1 1
ln = – –
k1 R T2 T1
0.24 Lmol–1 s–1 Ea 1 1

ln = – –
0.079 Lmol–1 s–1 8.314 Jmol–1K–1 813 K 781 K
Ea = 183311 Jmol–1
= 183.3 kJmol–1
CHAPTER 8.3
EXAMPLE: 4
The rate constant of a first–order reaction is
4.60 x 10–4 s–1 at 350oC.
If the activation energy is 104 kJ/mol,
calculate the temperature at which its rate
constant is 8.80 x 10–4 s–1.
EXAMPLE: 4
k1 = 4.60 x 10–4 s–1 T1 = 350oC = (350 + 273) K

= 623 K
k2 = 8.80 x 10–4 s–1 T2 = ?
Ea = 104 x 103 Jmol–1
k2 Ea 1 1
ln = – –
k1 R T2 T1
EXAMPLE: 4
k2 Ea 1 1
ln = – –
k1 R T2 T1
8.80 x 10–4 s–1 104 x 103 Jmol–1 1 1

ln =– –
4.60 x 10–9 s–1 8.314 Jmol–1K–1 T2 623 K
1 1
– = – 5.186 x 10–5
T2 623 K
1
= 1.553 x 10–5 K–1
T2
T2 = 644 K = 371oC
END OF
SLIDE SHOW

8.3 Factors Affecting RXN Rate

Uploaded by

Copyright:

Available Formats

8.3 Factors Affecting RXN Rate

Uploaded by

Document Information

Original Description:

Original Title

Copyright

Available Formats

Share this document

Share or Embed Document

Sharing Options

Did you find this document useful?

Is this content inappropriate?

Copyright:

Available Formats

8.3 Factors Affecting RXN Rate

Uploaded by

Copyright:

Available Formats

REACTION KINETICS

8.3 Factors Affecting Reaction Rate

The kindling, with its greater surface area, allows

 Determined by the chemical nature of

 Size of reacting particles decrease, total

 Reaction rate increase

 The higher the temperature the higher the rate of

•Only collisions with energy greater

•At lower temperature T1, the fraction

•As the temperature increases, the

•T ↑, average kinetic energy of the molecules ↑ ,

A catalyst provides an alternative pathway

Without catalyst With catalyst Combine diagram

 Higher T  larger k  increased rate

 Temperature affects the rate by affecting

Arrhenius equation k = Ae – Ea/RT

f = e – Ea/RT f: Fraction of molecular collisions

 T  average speed of particles 

f = e – Ea/RT f: Fraction of molecular collisions

 Only those collisions with enough energy

 Temperature rise enlarge fraction (f) of

 Magnitudes of both Ea and T affect the

• Higher T  larger k  increased rate

• Larger Ea  smaller k  decreased rate

A = 4.3 x 1013 s–1 T = 20oC

103 x 103 Jmol–1

= 1.9 x 10–5 s–1

A = 4.3 x 1013 s–1 T = 100oC

103 x 103 Jmol–1

 Determine k, Ea, T and A using Arrhenius

see the linear relationship…?

Determine graphically the activation energy

k (1/M•s) ln k T (oC) T (K) 1/T (K–1)

0.00187 – 6.281 600 873 1.15 x 10–3

0.01130 – 4.483 650 923 1.08 x 10–3

0.05690 – 2.866 700 973 1.03 x 10–3

0.24400 – 1.411 750 1023 9.78 x 10–4

 Plot graph: ln k versus 1/T

Slope = – Ea/R = – 2.89 x 104 K

Ea = ( 8.314 J/K•mol )( 2.89 x 104 K)

 If we know the rate constant at

k1 = 0.079 Lmol–1 s–1 T1 = 508oC

k2 = 0.24 Lmol–1 s–1 T2 = 540oC

0.24 Lmol–1 s–1 Ea 1 1

k1 = 4.60 x 10–4 s–1 T1 = 350oC = (350 + 273) K

k2 = 8.80 x 10–4 s–1 T2 = ?

Ea = 104 x 103 Jmol–1

8.80 x 10–4 s–1 104 x 103 Jmol–1 1 1

You might also like