15
The Effects of
Temperature and
Catalyst on
Reaction Rate
15.1 Activation Energy and Arrhenius Equation
15.2 Interpretation of Rates of Gaseous
Reactions at Molecular Level
15.3 Energy Profile
15.4 Effect of Catalysts on Rates of Reactions
1
Activation Energy
and Arrhenius
Equation
2
Activation Energy
Exothermic reaction
related to the rate of reaction
Activation energy, Ea
= energy required to
start the reaction
3
Activation Energy
Endothermic reaction
related to the rate of reaction
4
Most reactions have positive Ea since energy is
absorbed to break bonds in reactant particles.
5
Arrhenius Equation
-Ea
k  Ae RT
Since rate = k[A]a[B]b...
At fixed concentrations, rate depends on k
which in turn depends on
temperature (T) and
the nature of the reaction (A and Ea)
A depends on the nature of the reaction and
varies with T
6
Q.23
R = 8.31 J K1 mol1
At 298K,k298  A298e
k308
k298
Ea
R(298K)
 Ea
R(308K)

 Ea
A298e
R(298K)
A308e
e
Ea  1
1 



R  298 308 
At 308K,k308  A308e
Assume A is a constant
= 1.92  2
A 10 K  in T doubles the rate
7
Ea
R(308K)
Arrhenius Equation
Rate of reaction 
exponentially with temperature
Rate  e
8
 Ea
RT
Arrhenius Equation
Rate  Ae
 Ea
RT
T ,
 A  A  T
 Rate  (Minor effect)
9
Arrhenius Equation
Rate  Ae
 Ea
RT
T ,
 Ea


RT
e
 Ea
RT
(less negative)
 (more positive)
 Rate  (Major effect)
10
Arrhenius Equation
Rate  Ae
 Ea
RT
T ,
 A  A  T
 Rate  (Minor effect)
11
Arrhenius Equation
Rate  Ae
 Ea
RT
T ,
 Ea


RT
e
 Ea
RT
(more negative)
 (less positive)
 Rate  (Major effect)
12
Determination of Activation Energy
k  Ae
 Ea
RT
logek  logeAe
 Ea
RT
log10k  log10Ae
13
 Ea  1
 logeA -  
 R T
 Ea
RT
 Ea  1
 log10A - 

 2.303R T
Determination of Ea by Graphical Method
logek
logeA
 Ea  1
logek  logeA -  
 R T
1/T (K1)
- Ea
slope 
R
14
Q.24
logek
14.9
9.4
6.8
3.2
15
1/T (K1)
1.80103
1.55103
1.43103
1.28103
logek
logeA  24.8
Ea = -slope  R  182 kJ mol1
< E(H – I)
H – H and I – I
bonds are formed
before the H – I bond is
completely broken (refer to p.32)
16
1/T (K1)
If logerate is plotted against 1/T,
since rate = k[A]x[B]y…
logerate = logek + loge[A]x[B]y…
= logek + constant
Ea
 logeA  constant 
RT
y-intercept
17
Determination of Activation
Energy Using Two Rate Constants
k1  A1e
Ea
RT1
k2  A2e
 Ea
RT2
k1 A1

e
k2 A2
Ea  1 1 
  
R  T2 T1 
k1
ln  lne
k2
18
Ea  1 1
 
R  T2 T1
e



Ea  1 1 
  
R  T2 T1 
Ea

R
 1
1 

 
 T2 T1 
Interpretation of
Rates of Gaseous
Reactions at
Molecular Level
19
The Kinetic Theory of Gases
- Developed by Maxwell and Boltzman
1. Gas particles are in a state of constant and
random motion in all directions, undergoing
frequent collisions with one another and with
the walls of the container.
2. The pressure exerted on the container is due
to the collisions between gas particles and the
walls of the containers.
20
The Kinetic Theory of Gases
- Developed by Maxwell and Boltzman
3. Gas particles are treated as point masses
because their volumes are negligible when
compared with the volume of the container.
4. There is no interaction among gas particles
except collisions.
21
The Kinetic Theory of Gases
- Developed by Maxwell and Boltzman
5. Collisions between gas particles are perfectly
elastic,
i.e. the total kinetic energy is conserved.
22
Distribution of Molecular Speeds in a Gas
Consider a sample of gas:
Transfer of K.E.
among molecules
 Distribution of
molecular speeds
23
Distribution of Molecular Speeds in a Gas
Area under curve
= total no. of gas molecules
24
The Kinetic Theory of Gases
- Developed by Maxwell and Boltzman
The mean kinetic energy of a sample of gas
particles is proportional to its absolute
temperature (T).
1
2
mc  T
2
25
The Kinetic Theory of Gases
- Developed by Maxwell and Boltzman
1
mc2  T
2
2
2
2
n1c1  n2c2  n3c3  ...
c  mean square velocity 
n1  n2  n3  ...
2
n1 = no. of molecules with velocity c1 and
n1 + n2 + n3 + ... = n (total no. of molecules)
c2  root mean square velocity
26
The distribution of velocity is not symmetrical
 the average velocity of a gas sample is best
represented by the root mean square velocity c 2
c2
27
The Kinetic Theory of Gases
- Developed by Maxwell and Boltzman
1
mc2  T
2
For a sample of gas containing n molecules,
1
2
PV  nRT  mn c
3
where m is the absolute mass of the gas
molecule
28
Variation in the distribution of molecular Speeds with T
As T ,
• Molecular speeds 
• Curve becomes
flattened
• Wider distribution of
molecular speeds at a
higher temp
• Area under the curve
remains unchanged.
29
1
PV  mn c2  nRT
3
If n = 1
3RT
3RT

c 
M
mn
2
M : molar mass of gas in Kg
M   r.m.s velocity 
30
3RT
c 
M
2
The areas underneath the curves are the same
The lighter molecules are more spread out in
molecular speeds.
31
Q.25
H2
3RT
3  8.314 273
c 

M
2.0 10-3
2
= 1845 ms1
CO2
3RT
3  8.314 273
c 

M
44.0 10-3
2
= 393 ms1
The lightest gases (H2, He) can escape from the
gravitational pull of small planets
 Very rare in the Earth’s atmosphere
32
Simple Collision Theory
For a reaction to occur, the reactant
particles must collide with
(1) kinetic energy  Ea
(2) proper orientation.
33
HCl(g) + NH3(g)  NH4Cl(s)
Proper Orientation
34
Improper Orientation
35
Improper Orientation
36
Simple Collision Theory
For a reaction to occur, the reactant
particles must collide with
(1) kinetic energy  Ea
(2) proper orientation.
Effective
collision
No. of effective collisions = Z  e
Z = collision frequency
37
 Ea
RT
p
Simple Collision Theory
For a reaction to occur, the reactant
particles must collide with
(1) kinetic energy  Ea
(2) proper orientation.
Effective
collision
No. of effective collisions = Z  e
e
38
Ea
RT
 Ea
RT
p
 fractionof collisionswith K.E.  Ea
Simple Collision Theory
For a reaction to occur, the reactant
particles must collide with
(1) kinetic energy  Ea
(2) proper orientation.
Effective
collision
No. of effective collisions = Z  e
 Ea
RT
p = fraction of collisions with proper
orientation
39
p
Theoretically, (from collision theory and
kinetic theory)
No. of effective collisions = Z  e
rate  Z  e
Ea
RT
 Ea
RT
p
p
Experimentally,
rate  k[X] [Y] ...  Ae
a
40
b
 Ea
RT
[X] [Y] ...
a
b
rate  Z  e
Ea
RT
p
rate  k[X]a [Y]b ...  Ae
 Ea
RT
[X]a [Y]b ...
If molarities of X, Y,… are fixed
A  Zp
A T
41
= no. of collisions with
proper orientation
Interpretation of the Effect of
Temperature Change on Rate of Reaction
rate  k[X]a [Y]b ...  Ae
T 
 Ea
RT
[X]a [Y]b ...
A  Zp  T
 speed of reactant particles 
 collision frequency (Z) 
A
 rate  (minor effect)
42
Interpretation of the Effect of
Temperature Change on Rate of Reaction
rate  k[X]a [Y]b ...  Ae
 Ea
RT
[X]a [Y]b ...
T 
 K.E. of reactant particles 
 fraction of collisions with K.E.  Ea 
i.e. e
-Ea
RT

 rate  exponentially (major effect)
43
No. of
molecules convex
concave
speed / K.E.
44
The shaded area = no. of
particles with K.E. > E
n
e
n0
E
RT
E
shaded area
Fraction of particles with K.E. > E =
total area
45
n
e
n0
If E = Ea
Ea
RT
Ea
shaded area
Fraction of particles with K.E. > Ea =
total area
46
n
e
n0
Ea
RT
As T , the fraction of particles with K.E. > Ea
increases exponentially.
 Rate increases exponentially with T
47
Limitations of Collision Theory
Collision theory is based on the calculations from
kinetic theory of ideal gases. Thus, it is ONLY
applicable to reactions in gas phase.
In aqueous phase, the interactions between the
reactant particles and the solvent molecules
have to be considered.
The fraction of collisions with proper orientation
(the steric factor, p) cannot be predicted.
It can only be determined experimentally.
48
Q.26
Consider the 2nd order single-step gas phase rx
R(g) + R(g)  products
Given : k = 1.00102 mol1 dm3 s1 at 473 K,
[R(g)]initial = 1102 mol dm3, L = 6.021023 mol1
Gas constant = 8.31 J K1 mol1, Ea = 100 kJ mol1
Initial collision frequency(Z) = 7.771032 s1
(a) Estimate
(i) The no. of effective collisions per m3 per second.
(ii) The no. of collisions with K.E.  Ea per m3 per second.
(b) Hence, deduce the steric factor, p of the reaction.
49
(a)(i)
Initialrate  k[R]
2
initial
= (1.0010-2 mol1 dm3 s1)(1.0010-2 mol dm3)2
= 1.0010-6 mol dm3 s-1
= 1.0010-6 mol  6.021023 mol-1 dm-3 s-1
= 6.021017 molecules dm-3 s-1
= 6.021020 molecules m-3 s-1
6.021020 molecules of R are decomposed
per cubic meter per second
50
(a)(i)
Consider the 2nd order single-step gas phase rx
R(g) + R(g)  products
Rate = 6.021020 molecules m-3 s-1
One effective collision leads to
decomposition of Two molecules of R.
Thus, no. of effective collisions per cubic
meter per second
= 3.011020
51
(a)(ii)
No. of effective collisions = Z  e
 Ea
RT
p
= 3.011020 m-3 s-1
No. of collisions with K.E.  Ea
 Ze
 Ea
RT
1001000 J mol-1
 7.7710 m s  e
32
3 1
(8.31 J K -1 mol-1 )(473 K)
= 7.771032 m-3 s-1  (8.9310-12)
= 6.941021 m-3 s-1
52
(b)
 Ea
No. of effective collisions = Z  e RT  p
= 3.011020 m-3 s-1
No. of collisions with K.E.  Ea
= 6.941021 m-3 s-1
no. of effectivecollisions
p
no. of collisionswith K.E.  Ea
3.01 1020 m 3s 1
p
 0.0434 = 4.34 %
21
3 1
6.94  10 m s
53
Q.27
rate  k[X]a [Y]b ...  Ae
 Ea
RT
[X]a [Y]b ...
If Ea  0
e
Ea
RT
1
rate  Ae
Ea
RT
[X]a [Y]b ...  A[X]a [Y]b
Rate is independent of T (A-level)
54
No. of
molecules
Ea
55
K.E.
No. of effective collisions = Z  e
56
p
Zp
No. of
molecules
Ea
 Ea
RT
Ea
Ea
K.E.
Energy Profile
Transition State
Theory
57
Transition State Theory
- focuses on what happens after the
collisions have started.
58
Energy profile
- shows the variation of the potential
energy of the reaction mixture as the
reaction proceeds.
P.E.
59
reaction
coordinate
Consider the one-step reaction,
A–B + X  A + B–X
P.E. of the reaction mixture
are calculated for any A–B
and B–X distances, and the
results are plotted on a
contour diagram
60
Consider the one-step reaction,
A–B + X  A + B–X
At R,
A-B distance is short
B-X distance is long
 before reaction
61
Consider the one-step reaction,
A–B + X  A + B–X
The valley at R represents
the potential energy for the
initial state of the system,
i.e. A–B and X
62
Consider the one-step reaction,
A–B + X  A + B–X
At P,
A-B distance is long
B-X distance is short
 after reaction
63
Consider the one-step reaction,
A–B + X  A + B–X
The valley at P represents
the potential energy for the
final state of the system, i.e.
A and B–X
64
Consider the one-step reaction,
A–B + X  A + B–X
The energy contours rise in all
directions from the valleys at R
and P,
but the ‘easiest’ path is shown
by the bold line RTP
65
The transition state is like a col (山坳) in
a mountain region
66
T
R
67
P
A-B + X  ABX  A + B-X
In the transition state,
• Bond between A and B is partially broken
• Bond between B and X is partially formed
Thus, Ea is lower than E(A-B)
68
Transition state (Activated complex) is the
least stable arrangement of the system in the
most probable reaction pathway.
T
R
69
P
Advantages of Transition State Theory
1. Ea and A can be calculated
A  Zp
 the steric factor p can be predicted
2. It explains why the reaction pathway is
specific.
3. It is applicable to gaseous and aqueous
reactions.
70
Energy Profile : One-step Mechanism
A-B + X  ABX  A + B-X
71
Example of One-step Mechanism
Rate = k[CH3Cl][OH]
Bimolecular One-step 2nd Order
Nucleophilic Substitution Reaction
SN2
72
CH3Cl + OH  CH3OH + Cl
73
Energy Profile : Multi-step Mechanism
E1 > E 2
 Step 1 is the rate determining step
74
Energy Profile : Multi-step Mechanism
Rate  k1[A  B]  A1e
75
E1
RT
[A  B]
Multi-step Mechanism
• Chemical reactions take place in two or
more steps
• Formation of an intermediate
Step 1:
A ─ B  A + B (intermediate)
Step 2:
A + B + X  A + B─ X
Overall reaction: A ─ B + X  A + B ─ X
76
Example of Multi-step Mechanism
• Hydrolysis of 2-chloro-2-methylpropane
(1)
(2)
77
carbocation
Rate = k[C(CH3)3Cl]
Unimolecular Two-step 1st Order
Nucleophilic Substitution Reaction
SN 1
78
E1 > E 2
E1
E2
+ OH
Rate = k[C(CH3)3Cl]
79
Step 1 is r.d.s.
Reaction Mechanism and Rate Law
Reaction mechanisms are theoretical proposals
used to explain the experimentally determined
rate laws.
Reaction mechanism is the detailed sequence
of steps that occur in a reaction .
80
Reaction Mechanism and Rate Law
Each of the steps in a mechanism is called an
elementary step.
The number of reactant particles that takes
part in each elementary step is called the
molecularity of that step.
81
Reaction Mechanism and Rate Law
The number of reactant particles that takes
part in each elementary step is called the
molecularity of that step.
Unimolecular – one particle collides with the wall
of the vessel or the excess solvent
82
Pseudo-1st order reaction
CH3COOCH3 + H2O  CH3COOH + CH3OH
Rate = k[CH3COOCH3][H2O]
If H2O is used as solvent (in large excess)
[H2O]  constant throughout the reaction
Rate = k’[CH3COOCH3]
Unimolecular reaction
83
Reaction Mechanism and Rate Law
The number of reactant particles that takes
part in each elementary step is called the
molecularity of that step.
Unimolecular – one particle collides with the wall
of the vessel or the excess solvent
Bimolecular – two particles collide together
Termolecular – three particles collide together
simultaneously (very rare)
84
The slowest step in a particular mechanism is
called the rate-determining step
Requirements for writing reaction mechanisms :
1.
The sum of elementary steps must give the
overall balanced equation for the reaction.
2. The mechanism must agree with the
experimentally determined rate law.
85
Consider the reaction
A + B + C
D
Rate = k[A][B]
Only one intermediate
Proposed mechanism : -
86
A + B
R
(slow)
R + C
D
(fast)
r.d.s
Q.28
A + B
X
Y + C
87
X
Y
(slow)
(fast)
D
(fast)
r.d.s.
Q.28
E1 > E 2  E3
Step one is the r.d.s
Rate = k[A][B]
P.E.
E1
E2
X+C
A+B+C
E3
Y+C
D
Reaction coordinate
88
Q.28
A + B
X
k1
k
Y + C
k2
Y
X
(fast)
(slow) r.d.s.
D
(fast)
Rate = k[X]
At equilibrium, k1[A][B] = k2[X]
kk1
Rate 
[A][B]  k3 [A][B]
k2
89
Effect of Catalysts
on Rates of
Reactions
90
Working Principle of Catalysts and
their Effects on Reaction Rates
Catalysis  Catalytic action
Catalysts alter the rates of reaction,
1. but remain chemically unchanged at
the end of the reaction
2. by providing new, alternative reaction
pathways with different activation
energies.
91
Working Principle of Catalysts and
their Effects on Reaction Rates
Positive catalyst:
• Provides an alternative reaction pathway
with a lower activation energy
92
• Lower Ea
 Greater fraction of molecules with
K.E. greater than or equal to Ea
 Reaction proceeds faster
Ea’
93
Working Principle of Catalysts and
their Effects on Reaction Rates
Negative catalyst:
• Provides an alternative reaction pathway
with a higher activation energy
94
• Higher Ea
 Smaller fraction of molecules with
K.E. greater than or equal to Ea
 Reaction proceeds slower
Ea”
95
Working Principle of Catalysts and
their Effects on Reaction Rates
With catalysts, the
contour diagrams
and thus the energy
profiles are totally
different from
those without
96
Catalyst
Homogeneous
Catalyst
Reactants & catalyst
are in the same
phase
97
Heterogeneous
Catalyst
Reactants & catalyst
are NOT in the same
phase
Characteristics of Catalysts
1. For a given reversible reaction,
Reactants
k1
K-1
Products
catalysts affect the rates of forward
reaction and backward reaction to the
same extent.
98
Q.29
Without
catalyst
With
catalyst
Reactants
Reactants
Show that
99
k1
k-1
k1'
k
'
-1
Products
Products
k
k

k1
k1
'
1
'
1
E1'
RT
k Ae

e
E1
k1
A1e RT
'
1
'
1
1
(E1 E1' )
RT
=
k
Ae

e
E-1
k-1
A-1e RT
'
-1
'
-1
1
(E-1 E-'1 )
RT
E1  E  E1  E
P.E.
'
1
'
1
E-1
E1
E’1
100
E-'1
RT
E’-1
Reaction coordinate
E1'
RT
1
(E1 E1' )
k1' A1'e
RT


e
E1
k1
A1e RT
Given : T = 298 K, R = 8.31 J K1 mol1
k
Calculate
if (E1'  E1 )  50 kJ mol1
k1
'
1
k1'
e
k1
101
1
(E1 E1' )
RT
50103 J mol1
e
(8.31 J K 1 mol1 )(298 K)
= 5.9108
Characteristics of Catalysts
2. Catalysts are chemically unchanged at the
end of reactions, but may undergo physical
changes.
E.g. Lumps of MnO2 used in the decomposition
of H2O2 become
powdered at the end
of the reaction.
102
Characteristics of Catalysts
3. Only small quantity is sufficient to
catalyze a reaction because catalysts can be
regenerated.
However, if the catalysts are involved in the
rate equation, higher concentrations may
affect the rate more.
103
Characteristics of Catalysts
4. The effect of heterogeneous catalysts
depends on the surface area available for the
catalytic action.
Surface area of solid catalyst 
 number of reaction sites 
 catalytic activity
E.g. Finely divided Fe powder is used as the
catalyst in Haber process.
104
Characteristics of Catalysts
5. Catalytic actions are specific especially in
biological systems.
E.g. Enzymatic actions are highly specific.
105
Characteristics of Catalysts
6. The efficiency of a catalyst is often
enhanced by adding promoters. Promoters
have no catalytic actions on their own.
E.g. Fe2O3, KOH, Al2O3 in Haber process
106
Characteristics of Catalysts
7. The efficiency of a catalyst can be lowered
by adding poisons or inhibitors. Catalyst
poisons are specific in action.
E.g. Arsenic impurities may poison Pt but
not V2O5 in Contact process
107
Characteristics of Catalysts
8. Transition metals or compounds/ions
containing transition metals show marked
catalytic activities.
E.g. Pt, Ni, Fe, V2O5, MnO2, Mn2+ Fe3+
The catalytic actions are due to the presence
of low-lying partially filled d-orbitals.
108
Heterogeneous Catalysis – Adsorption
Occur on the surface of the catalyst.
1. Reactants are adsorbed on the surface,
forming new bonds with the catalyst while
weakening bonds in reactants
2. Products, once formed, are desorbed from
the surface,
109
Examples of heterogeneous catalysis
MnO2(s)
2H2O2(aq)  2H2O(l) + O2(g)
Fe(s)
3H2(g) + N2(g)  2NH3(g)
Al2O3/SiO2(s)
C8H18(g)  C4H10(g) + C4H8(g)
Ni(s)
CH2=CH2(g) + H2(g)  CH3–CH3(g)
110
Ni(s)
CH2=CH2(g) + H2(g)  CH3–CH3(g)
CH2=CH2 H–H
slow
H H2C
CH2 H
fast
fast
111
Ni(s)
CH2=CH2(g) + H2(g)  CH3–CH3(g)
CH2=CH2 H–H
slow
H H2C
CH2 H
fast
fast
112
Ni(s)
CH2=CH2(g) + H2(g)  CH3–CH3(g)
CH2=CH2 H–H
slow
H H2C
CH2 H
fast
fast
113
Q.30
Ea1  Ea2  Ea3
CH2=CH2 + H2
CH3-CH3
114
Homogeneous Catalysis – Intermediate Formation
Homogeneous catalysts participate in certain
stages of reactions and are regenerated at the
end or later stages of reactions.
intermediate
Stage 1: A + catalyst  A ─ catalyst
Stage 2: A ─ catalyst + B  A ─ B + catalyst
Overall reaction: A + B  A ─ B
115
Homogeneous Catalysis – Intermediate Formation
• Acid-catalyzed esterification of
ethanoic acid and methanol
CH3COOH(l) + CH3OH(l)
H+
116
CH3COOCH3(l) + H2O(l)
Homogeneous Catalysis – Intermediate Formation
117
Q.31
nucleophilic attack
O
H3C
C
CH3
O
+
OH
O
r.d.s.
H3C
H
C
OH
Rate = k[CH3COOH][CH3OH]
Uncatalyzed esterification
118
CH3
O
H
Q.31
O
H3C
CH3
C
O
r.d.s.
O
+
OH
H3C
H
119
H3C
C
O
O
H
-H+
O
O
+ H2O
C
OH
Uncatalyzed esterification
H3C
CH3
CH3
H+
C
OH
O
CH3
Acid-catalyzed esterification
Protonation at carbonyl O rather than
hydroxyl O since the former is more electron
sufficient due to polarization of pi electron
cloud
120
Acid-catalyzed esterification
121
Acid-catalyzed esterification
122
Acid-catalyzed esterification
123
Acid-catalyzed esterification
124
Acid-catalyzed esterification
Most probable
resonance structure
Carbonyl C becomes more electron-deficient
 More easily attacked by nucleophile
125
Acid-catalyzed esterification
126
Acid-catalyzed esterification
r.d.s.
Rate = k[RCOOH][H+]
127
Acid-catalyzed esterification
r.d.s.
step 2
128
Acid-catalyzed esterification
r.d.s.
step 3
step 4
129
step 5
step 2
step 6
Acid-catalyzed esterification
r.d.s.
step 3
step 4
130
step 2
step 6
Acid-catalyzed esterification
r.d.s.
step 3
step 4
131
step 5
step 2
Acid-catalyzed esterification
r.d.s.
step 3
step 2
H+ is
regenerated
step 4
132
step 5
step 6
Acid-catalyzed esterification
r.d.s.
step 3
step 4
133
step 5
step 2
For simplicity,
steps 3 to 6
are combined
step 6
H3C
+
H
C
H3C
O
+
IH2C
C
I2
H3C
Rate = k[CH3COCH3][H+]
134
O
+
HI
H3C
H3C
C
H+
O
C
fast
OH
H3C
H3C
H
OH
H2C
C
H2C
OH
OH
H2C
+
C
H
I2
CH3
135
slow
CH3
H3C
H
H+
+
C
I
I
C
C
H
CH3
I
I
C
C
H
CH3
fast
OH
IH2C
C
O
H
H3C
O
+
HI
fast
C
O
H3C
k1
H3C
+
H
C
k2
H3C
fast
OH
H3C
H
k3
H2C
C
OH
OH
H2C
H+
CH3
H3C
At equilibrium,
H3C
k1[CH3COCH3][H+] = k2[
Rate = k3[
H3C
C
OH
]
H3C
H3C
136
+
C
C
OH
k3k1
] 
[CH3COCH3 ][H ]
k2
= k[CH3COCH3][H+]
slow
Homogeneous Catalysis Using Transition
Metal Ions
Principle : - Transition metals exhibit
variable oxidation states
137
2I(aq) + S2O82(aq)
I2(aq) + 2SO42(aq)
The reaction is slow because
colliding particles carry like charges
138
2I(aq) + S2O82(aq)
Fe3+(aq)
I2(aq) + 2SO42(aq)
Mechanism of catalyzed reaction : 2I(aq) + 2Fe3+(aq)  I2(aq) + 2Fe2+(aq)
2Fe2+(aq) + S2O82(aq)  2Fe3+(aq) + 2SO42(aq)
Both steps are fast because
colliding particles carry opposite charges.
139
2I(aq) + S2O82(aq)
Fe3+(aq)
I2(aq) + 2SO42(aq)
Mechanism of catalyzed reaction : 2I(aq) + 2Fe3+(aq)  I2(aq) + 2Fe2+(aq)
2Fe2+(aq) + S2O82(aq)  2Fe3+(aq) + 2SO42(aq)
The mechanism is made possible by the
variable oxidation states of Fe
140
Q.31
2I(aq) + S2O82(aq)
Fe2+(aq)
I2(aq) + 2SO42(aq)
Mechanism of catalyzed reaction : 2Fe2+(aq) + S2O82(aq)  2Fe3+(aq) + 2SO42(aq)
2I(aq) + 2Fe3+(aq)  I2(aq) + 2Fe2+(aq)
141
2Ce4+(aq) + Tl+(aq)
Mn2+(aq)
2Ce3+(aq) + Tl3+(aq)
Mechanism of catalyzed reaction : Ce4+(aq) + Mn2+(aq)  Ce3+(aq) + Mn3+(aq)
Ce4+(aq) + Mn3+(aq)  Ce3+(aq) + Mn4+(aq)
Mn4+(aq) + Tl+(aq)  Mn2+(aq) + Tl3+(aq)
The mechanism is made possible by the
variable oxidation states of Fe
142
Applications of Catalysts
Industrial Catalysts
1. Iron is used in the Haber process
Fe
N2(g) + 3H2(g)
2NH3(g)
2. Platinum or vanadium(V) oxide is used in
the Contact process
2SO2(g) + O2(g)
143
Pt or V2O5
2SO3(g)
Applications of Catalysts
3. Nickel, platinium or palladium is used in the
hydrogenation of unsaturated oils to make
margarine
144
Applications of Catalysts
4. Nickel and nickel(II) oxide are used in
the production of town gas in Hong Kong.
Ni or
NiO
C5H12(g) + 5H2O(g)  5CO(g) + 11H2(g)
2CO(g) + 2H2(g)  CO2(g) + CH4(g)
145
Applications of Catalysts
Catalytic Converters in Car Exhaust Systems
146
Applications of Catalysts
Rh
2CO(g) + 2NO(g)  2CO2(g) + N2(g)
CxHy(g) + ( x + y/4) O2(g)
Pt
 xCO2(g) + y/2 H2O(g)
Pt
2CO(g) + O (g)  2CO (g)
2
147
2
Applications of Catalysts
Enzymes in the Production of Alcoholic Drinks
C6H12O6(aq)
enzyme
 2C2H5OH(aq) + 2CO2(g)
Fermentation
148
The END
149
15.1 Activation Energy and Arrhenius Equation (SB p.51)
For the following reaction:
C6H5N2 +Cl–(aq) + H2O(l)
 C6H5OH(aq) + N2(g) + H+(aq) + Cl–(aq)
the rate constants of the reaction at different
temperatures were measured and recorded in the
following table:
150
15.1 Activation Energy and Arrhenius Equation (SB p.51)
Temperature (K)
Rate constant
(10-5 s-1)
278.0
0.15
298.1
4.10
308.2
20.00
323.0
140.00
Determine the activation energy graphically.
(Given: R = 8.314 J K–1 mol–1)
151
Answer
15.1 Activation Energy and Arrhenius Equation (SB p.52)
152
ln k
1/T (k-1)
-13.41
3.597  10-3
-10.10
3.355  10-3
-8.52
3.245  10-3
-6.57
3.096  10-3
15.1 Activation Energy and Arrhenius Equation (SB p.52)
A graph of ln k against 1 gives a straight line with slope E a .

T
R
153
15.1 Activation Energy and Arrhenius Equation (SB p.52)
Back
y = -11.8 – (-7)
= -4.8
x = (3.48 – 3.13)  10-3
= 0.35 10-3 K-1
Slope =
 4 .8
0.35  10  3 K 1
= -13.7  103 K
  E a = -13.7  103 K
R
154
Ea = 13.7  103 K  8.314 J K-1 mol-1
= 113.9  103 J mol-1
= 113.9 kJ mol-1
 The activation energy of the reaction is 113.9 kJ mol-1.
15.1 Activation Energy and Arrhenius Equation (SB p.53)
Back
The rate constant for a reaction at 110°C is found to
be twice the value of that at 100°C. Calculate the
activation of the reaction.
(Given : R = 8.314 J K-1 mol-1)
Answer
k100 C
Ea
1
1

(

)
k110 C
8.314 273  100 273  100
Since k110 oC = 2 k100 oC,
ln
o
o
Ea
1
1
1
ln  
(

)
2
8.314 273  100 273  100
Ea = 82 327 J mol-1
=82.3 kJ mol-1
 The activation energy of the reaction is 82.3 kJ mol-1.
155
15.1 Activation Energy and Arrhenius Equation (SB p.53)
(a) The reaction
2A(g) + B(g)  C(g)
was studied at a number of temperatures, and the
following results were obtained:
Tempera
12
60
112
203
292
ture (oC)
Rate
2.34
13.2
52.5
316
1000
constant
(dm6
mol-2 s-1)
Determine the activation energy of the reaction graphically.
(Given: R = 8.314 J K–1 mol–1)
156
Answer
15.1 Activation Energy and Arrhenius Equation (SB p.53)
157
T (K)
1 / T (K-1)
ln k
285
3.51  10-3
0.85
333
3.00  10-3
2.58
385
2.60  10-3
3.96
476
2.10  10-3
5.76
565
1.77  10-3
6.91
15.1 Activation Energy and Arrhenius Equation (SB p.53)
A graph of ln k against 1 gives a straight line with slope  E a
T
R
158
.
15.1 Activation Energy and Arrhenius Equation (SB p.53)
Slope =
6.91  0.85
1.77  10 - 3  3.51 10 - 3
= -3.48  103
Ea
 3.48  10 3
R
Ea = 3.48  103  8.314
= 28.93 kJ mol-1
 The activation energy of the reactions is 28.93 kJ mol-1.

159
15.1 Activation Energy and Arrhenius Equation (SB p.53)
(b) Determine the activation energy of the following
reaction using the data provided only.
A + B  C
Temperature (K)
Rate constant (mol dm-3 s-1)
350
0.096
400
0.400
(Given: R = 8.31 J K–1 mol–1)
160
Answer
15.1 Activation Energy and Arrhenius Equation (SB p.53)
(b)
ln
E
0.096
1
1
 a (

)
0.400
8.31 350 400
Ea = 33 206 J mol-1
= 33.2 kJ mol-1
Back
161
15.2 Interpretation of Rates of Gaseous Reactions at Molecular Level
(SB p.58)
(a) Explain why not all collisions between reactant
molecules lead to the formation of products.
162
Answer
(a) For a reaction to occur, colliding molecules must have
kinetic
energy equal to or greater than the activation energy to
break the bonds in the reactants, so that new bonds can
form in the products. Moreover, the collision must be in the
right geometrical orientation, and the atoms to be transferred
or shared do not come into direct contact with each other, so
that the atoms can rearrange to form products. Products
cannot be formed if the kinetic energy of the reactant
molecules cannot overcome the activation energy, or the
collision orientation is not appropriate.
15.2 Interpretation of Rates of Gaseous Reactions at Molecular Level
(SB p.58)
(b) Describe the effect of temperature on the
distribution of molecular speeds in a gaseous
system.
Answer
(b) An increase in temperature will lead to an increase in the
most probable speed of the molecules. The peak of the
curve of Maxwell-Boltzmann distribution of molecular speeds
shifts to the right and the curve becomes flattened. This
indicates that the distribution of molecular speed becomes
wider and the number of molecules having the most
probable speed decreases.
163
15.2 Interpretation of Rates of Gaseous Reactions at Molecular Level
(SB p.58)
Back
(c) Explain why the rates of chemical reactions
increase with temperature.
Answer
(c) As temperature rises, the proportion of fast-moving
molecules increases. The kinetic energy of the molecules
also increases. A greater fraction of molecules can overcome
the activation energy required for a reaction to occur.
Therefore, the number of effective collisions increases and
hence the rates of chemical reactions increase.
164
15.3 Energy Profile (SB p.60)
Back
Draw an energy profile of a typical single-stage
endothermic reaction.
Answer
165
15.3 Energy Profile (SB p.61)
The energy profile of a multi-stage reaction is
shown below:
166
15.3 Energy Profile (SB p.62)
Back
(a) Which stage is the rate determining step?
Explain your answer.
(b) Is the reaction exothermic or endothermic?
Explain your answer.
Answer
(a) Stage 2 is the rate determining step.
It is because stage 2 has the greatest amount of activation
energy.
(b) The reaction is exothermic.
It is because the potential energy of the products is lower
than that of the reactants.
167
15.3 Energy Profile (SB p.62)
Referring to the energy profiles below, answer the
questions that follow.
A
168
B
15.3 Energy Profile (SB p.62)
Referring to the energy profiles below, answer the
questions that follow.
C
169
D
15.3 Energy Profile (SB p.62)
(a) Which reaction(s) is/are exothermic?
(b) Which reaction is the fastest?
(c) Which reaction has the greatest amount of
activation energy?
Answer
(a) A, B and C
(b) B
(c) D
Back
170
15.4 Effect of Catalysts on Rates of Reactions (SB p.69)
(a) Explain what a negative homogeneous catalyst is.
Answer
(a) A negative homogeneous catalyst is a catalyst that
slows down a reaction. It exists in the same phase as
the reactants and products in the reaction, and
involves in the formation of an intermediate in the
reaction.
171
15.4 Effect of Catalysts on Rates of Reactions (SB p.69)
(b) Explain what a positive heterogeneous catalyst is.
Answer
(b) A positive heterogeneous catalyst is a catalyst that
speeds up a reaction but it is not in the same phase
as the reactant and products. It provides an active
surface for the reactant particles to adsorb in a
reaction.
172
15.4 Effect of Catalysts on Rates of Reactions (SB p.69)
Back
(c) Give three applications of catalysts.
173
Answer
(c) Iron used in the Haber process;
Platinum or vanadium(V) oxide used in the Contact process;
Nickel, platinum or palladium used in the hydrogenation of
unsaturated oils to make margarine;
Nickel and nickel(II) oxide used in the production of town gas;
Platinum (or palladium) and rhodium used in catalytic
converters;
Enzymes used in fermentation of glucose to produce ethanol;
Enzymes used in the manufacture of biological washing
powders.
(any 3)