5-Numerical of thermodynamics-20-01-2025.pptx

Engineering Chemistry
Module -1
1
A. Laws of thermodynamics - entropy change (selected
processes) – spontaneity of a chemical reaction and Gibbs free
energy - Carnot cycle
B. Kinetics - Order of reactions- Concept of activation energy and
energy barrier - Arrhenius equation- Homogeneous,
heterogeneous and Enzyme catalysis (Lock and Key
mechanism)
According to the revised syllabus for Fall Semester 2022-2023

2
1st
class
Thermodynamics: Basic Terminologies
0th
Law of Thermodynamics
1st
2nd
class
Enthalpy & Heat Capacity
Application of 1st
Law to the Expansion Work
Numerical from 1st
Law
3rd
class
2nd
Entropy and Carnot Cycle
4th
class
Free Energy
Numerical from 2nd
law of Thermodynamics
Spontaneity of a Chemical Reaction
3rd
5th
class
Chemical kinetics
Factors affecting reaction rates
Rate laws and rate constants
First order reaction
Half-life of first order reaction
6th
class
Second-order reactions
Zero-order reactions
Pseudo first order reaction
7th
class
Temperature and rate of reactions
Collision model and activation energy
Arrhenius equation
Catalysis
8th
class
Types of catalysis
Homogeneous and Heterogeneous catalysis
Enzyme catalysis

3
Part-A
Thermodynamics
- Laws of thermodynamics
- Entropy change (selected processes)
- Spontaneity of a chemical reaction and
- Gibbs free energy
- Carnot cycle

Thermodynamics: Basic
Terminologies
Thermodynamic Systems:
the quantity of matter or a
region in space upon which
attention is concentrated in the
analysis of a problem
Everything
external to
the system
Properties of a system:
# Intensive Properties:
# Extensive
Properties
4

Terminologies
# State function:
Depends on the initial state & final state;
independent of the path used to reach from.
Example: T (Temperature), P (Pressure), U
(Internal energy), H (Enthalpy) etc.
# Path function:
Depends on the path between the initial & final
state
Example: W (work done), q (heat transferred) etc.
Internal energy (U)
= Kinetic energy + Potential energy
∙ It’s a state function & an extensive property
of the system.
∙ Internal energy of a system changes when
energy is transferred into or outside the
system in the form of heat or work
5

Terminologies
• Adiabatic – no heat transferred
• Isothermal – constant
temperature
• Isobaric – constant pressure
• Isochoric – constant volume
Work done on/by the system
W (Work) = F (force) x w (distance moved in the direction of
force)
⮚ Gas is heated ⇒ it will expand and pushes the piston, thereby doing work on the piston.
The work done (dw) when the system expands by dV against a pressure Pex:
dw = −PexdV
🗹 This is an example of the system doing the work on the surrounding
State of a system
⮚ The state of thermodynamic variables
such as pressure, temperature, volume,
composition which describes the system
is called state of the system.
⮚ when one/more variables undergo
change, the system is said to have
undergone a change of state
6

Reversible Process in
Thermodynamics
# Example 1:
Thermal equilibrium of two systems with the same temp.
▪ If the temperature of either system is lowered infinitesimally,
then energy flows into the system with the lower temperature.
▪ If the temperature of either system at thermal equilibrium is
raised infinitesimally, then energy flows out of the hotter system
❑ A thermodynamic process is
reversible if the process can be
turned back such that both the
system and the surroundings
return to their original states,
with no other change anywhere
else in the universe.
❑ In reality, no such processes as
reversible processes can exist.
❑ A change can be reversed by
an infinitesimal modification of
a variable.
# Example 2:
Reversible expansion:
Suppose a gas is confined by a piston.
external pressure (Pex) = pressure (P) of the confined gas.
▪ Such a system is in mechanical equilibrium with its
surroundings because an infinitesimal change in the
external pressure in either direction causes changes in
volume in the opposite directions.
7

0th
Law of
Thermodynamics
❑ Two physical systems are in
thermal equilibrium if there is no
net flow of heat (thermal energy)
between them when they are
connected by a path permeable to
heat.
⮚ According to 0th
law:
If two systems are in thermal equilibrium with a third
system, then those two systems are in thermal
equilibrium with each other.
⮚ The 0th
law of thermodynamics defines thermal equilibrium
and forms a basis for the definition of temperature.
8

1st
Law of
Thermodynamics
▪ It’s the law of conservation of energy
⮚ For a system, if w = work done on a system, q = energy transferred as heat to a system & ΔU =
resulting change in internal energy
The signs of w and q:
✔ +ve if energy is transferred to the system as work/heat
✔ -ve if energy is lost from the system.
❑ The energy of an isolated system remains constant. Whenever a quantity of energy (some
form) disappears, an exactly equivalent quantity of energy (some other form) must make an
appearance.
⮚ Heat (q) and work (w) are equivalent ways of changing the internal energy of a system
→ Example:
o If a weight has been raised/lowered in the surroundings, transfer of energy happens by doing the
work.
o If ice melts in the surroundings, it indicates the transfer of energy as heat.
9

Enthalpy & Heat
Capacity
⮚ If the change of a system is brought
about at constant pressure, there will be
change in volume.
❑ The quantity (U + PV) is called the
enthalpy (H) of the system
ΔU = ΔH – PΔV
❑ Heat capacity (C) of a system b/n any two
temperatures – the quantity of heat (q) required to
raise the temperature of the system from the lower
temperature (T1) to the higher temperature (T2) divided
by the temperature difference.
☻ If mass of the system is 1 g, the heat capacity is
called the specific heat of the system.
☻ For 1 mol of substance, the heat capacity is
termed as ‘molar heat capacity’
⮚ Molar heat capacity varies with temperature
❑ Molar heat capacity at constant volume
@ constant volume w = 0 & ΔU = q
ΔH = ΔU + PΔV
or
, 10

Heat
Capacity
❑ Molar heat capacity at constant pressure (Cp)
@ constant pressure,
there is change in
volume & some work is
done
ΔU = q - w
As quantity (U+PV) is the
enthalpy (H) of the system
❑ Relationship between Cp & Cv
⇒
▪ For ideal gas, PV = RT (for 1 mole)
Δ(PV) = R ΔT
Heat capacity ratio,
11

Application of 1st
Law to the Expansion
Work
⮚ Isothermal Process (constant temperature)
▪ In an isothermal process,
the temperature stays
constant, so the pressure
and volume are
inversely proportional to
one another.
⇒ For an ideal gas,
The internal energy (U) ∝ Temperature (T)
if T = fixed, ΔU = 0 (according to 1st
law, which deals with ΔU)
▪ If the system does work, the energy
comes from heat flowing into the
system from the surrounding
▪ If work is done on the system, heat
flows out of the system to the
surrounding.
# Magnitude of w depends on whether the
expansion is reversible or irreversible.
12
work done
Pressure-volume diagram

Application of 1st
Law to the Expansion
Work
⮚ Reversible isothermal expansion:
▪ Work done by the n moles of gas can be
evaluated as:
⮚ The work done by a perfect gas when it expands
reversibly and isothermally is equal to the area
under the isotherm p = nRT/V.
⮚ The work done during the irreversible
expansion against the same final pressure is
equal to the rectangular area shown slightly
darker. Note that the reversible work done is
greater than the irreversible work done.
(as PV = nRT)
in an isothermal expansion of a perfect gas)
 Irreversible Isothermal expansion
13

14
Application of 1st
 Adiabatic process
 no heat is added/ removed from a
system.
 1st
law of thermodynamics:
ΔU = w
 Example: A gas expanding so quickly that no heat
can be transferred. Due to the expansion work,
temperature drops. This is exactly what happens
with a carbon dioxide fire extinguisher, with the
gas coming out at high pressure and cooling as it
expands at atmos. pressure
(as no heat is allowed to enter/leave the system, q = 0)
® Expansion: w = ― ve, = ― ve;
So, T of the system falls
ÞWork is done by the system at the expense of
its internal energy
 w = = (for 1 mole of gas)
 (for 1 mole of gas)
 Reversible adiabatic expansion
 Relation between T, V and P
(𝑇𝑖
𝑇 𝑓
)=
(𝑉 𝑓
𝑉 𝑖
)
𝛾 −1
Þ ln
Þ ln
Þ
[𝑎𝑠 ,𝛾=
𝐶𝑝
𝐶𝑣
]
ln ― = R]
Þ ln
 Irreversible adiabatic expansion
 Free expansion (Pex = 0) :
 Expansion against a constant pressure:

15
Application of 1st
 Isobaric process (constant pressure)
 the pressure is kept constant.
 The work done by the system in an isobaric
process is simply the pressure multiplied by the
change in volume
 Example of an isobaric system: A gas, being
slowly heated or cooled, confined by a piston in a
cylinder.
 Isochoric process (constant volume)
 the volume is kept constant
 The work done is zero in an isochoric
process
 . Example of an isochoric system: A gas in a
box with fixed walls

2nd
Law of
Thermodynamics
▪ Why we need for the 2nd law of thermodynamics?
→ The 1st
law of thermodynamics does not tell us
anything about the direction of change. The
direction of spontaneous change of a process is
defined by the 2nd
law of thermodynamics
▪ Heat does not flow spontaneously from a cool
body to a hotter body.
▪ The entropy (S) of an isolated system
increases in the course of a spontaneous
change.
❑ 2nd
law of thermodynamics
# Note: when considering applications of the 2nd
law – it
is a statement about the total entropy of the overall
isolated system (the ‘universe’), not just about the
entropy of the system of interest.
ΔStot > 0
Where, Stot = S + Ssur
S = the entropy of the system of interest, &
Ssur = the entropy of the surroundings
⮚ The 1st
law uses the internal energy to
identify permissible changes
⮚ The 2nd
law uses the entropy to identify
which of these permissible changes are
spontaneous.
→ A spontaneous process points towards
the direction in which the total entropy
increases.
⮚ Entropy (S) is a state function.
→ The thermodynamic definition of entropy
concentrates on the change in entropy
(dS) that occurs as the result of a
physical or chemical process.
→ dqrev is the energy transferred as heat
reversibly to the system at the absolute
temperature T.
⮚ Thermodynamic definition of entropy
16

Entrop
y
▪ Process that leads to an increase in entropy (ΔS > 0)
☻ Notice the increasing disorder in above processes
⮚ Entropy change for the system of an
isothermal expansion of a perfect gas
⇒
17

Entrop
y
⮚ Total Entropy change in irreversible
(spontaneous) process of a perfect gas
✔ example: Isothermal expansion of an ideal gas
at constant temperature into vacuum
⮚ Isothermal Reversible expansion
⮚ Clausius inequality
⇒ All spontaneous process occurring in
Nature are irreversible and entropy of the
universe is increasing continuously. 18

Entropy
 The 1st
law & the 2nd
law of thermodynamics were summed up by German Physicist Rudolf Clausius as below:
The energy of the universe remains constant; the entropy of the universe tends towards a maximum
 For an ideal gas (1 mole) with variable T & V
 For an ideal gas with variable P & T
 For an ideal gas in an isothermal process
=
 For an ideal gas in an isobaric process
 For an ideal gas in an isochoric process
 Its a device which transforms heat into work
 This happens in a cyclic process
 Heat engines require a hot reservoir to supply energy
(QH) and a cold reservoir to take in the excess energy (QC)
– QH is defined as positive, QC is defined as negative
 A Carnot cycle (named after the French engineer Sadi
Carnot) consists of four reversible stages in which a gas
(the working substance) is either expanded/compressed in
various ways
 To demonstrate the maximum convertibility of heat into
work
 The system consists of 1 mole of an ideal gas which is
subjected to four strokes
 Entropy change during different processes  Heat engine
 Carnot Cycle
19

20
 Four stages of Carnot Cycle:
A. 1st
stroke:
Curve AB: AB: Isothermal expansion at Th
Work done by the gas
B. 2nd
stroke:
Curve BC (B → C): Adiabatic expansion,
C. 3rd
stroke:
Curve CD (C → D): Isothermal compression at TC,
Work done on the gas.
D. 4th
stroke:
Curve DA (D → A): Adiabatic compression
Work done on the gas
• The thermal efficiency () of a heat engine is
= =
=
 The “engine” statement of the 2nd
Law:
– it is impossible for any system to have an
efficiency of 100% ( = 1)
 Another statement of the 2nd
Law:
– It is impossible for any process to have as its sole result
the transfer of heat from a cooler object to a warmer
object

Carnot Cycle
 Four stages of Carnot Cycle: A. 1st
stroke:
Curve AB: AB: Isothermal expansion at Th
𝒒𝒉=−𝒘𝟏 =𝑹𝑻 𝒉 𝒍𝒏
𝑽 𝑩
𝑽 𝑨
Þ The gas is placed in thermal contact with QH (at Th) and
undergoes reversible isothermal expansion from A to B.
Þ The entropy change is qh/Th ( qh = the energy supplied to the
system as heat from the hot source)
B. 2nd
stroke:
Curve BC (B → C): Adiabatic expansion,
− 𝒘𝟐=− 𝑪𝒗 (𝑻 𝒉 − 𝑻𝒄)
Þ Contact with QH is broken & the gas undergoes
reversible adiabatic expansion from B to C.
Þ No energy leaves the system as heat, ΔS = 0
Þ The expansion is carried on until the temperature of
the gas falls from Th to Tc (the temperature of Qc)
21

Carnot Cycle
C. 3rd
stroke:
Curve CD (C → D): Isothermal compression at TC,
Work done on the gas.
−𝑞𝑐=𝑤3=𝑅𝑇 𝑐 ln
𝑉 𝐷
𝑉 𝐶
 The gas is placed in contact with the cold sink (Qc) and
undergoes a reversible isothermal compression from C to D at Tc.
 Energy is released as heat to the cold sink; the entropy change of
the system = qc/Tc, where qc is negative.
D. 4th
stroke:
Curve DA (D → A): Adiabatic compression
Work done on the gas
𝑤4=𝐶𝑣 (𝑇 h − 𝑇𝑐 )
 Contact with Qc is broken and the gas undergoes
reversible adiabatic compression from D to A such
that the final temperature is Th.
 No energy enters the system as heat, so the change
in entropy is zero.
22

Carnot Cycle
 The area enclosed by the four curves represents the net
work done by the engine in one cycle
 The total change in entropy around the cycle is the sum
of the changes in each of these four steps:
∮𝑑𝑆=¿
𝑞h
𝑇h
+
𝑞𝑐
𝑇𝑐
¿
 For an ideal gas, and
 The “engine” statement of the 2nd
Law:
– it is impossible for any system to have an
efficiency of 100% ( = 1)
Efficiency of a heat engine
• The thermal efficiency of a heat engine is
= = =
• Another statement of the 2nd
Law:
– It is impossible for any process to have as its sole
result the transfer of heat from a cooler object to
a warmer object
[Kelvin’s statement]
[Clausius’s statement]
23

Application of the Carnot Cycle
 Energy efficiency of the
Carnot cycle is independent
of its working substance.
 Any cyclic process that
absorbs heat at one
temperature and rejects
heat at another
temperature and is
reversible has the energy
efficiency of a Carnot cycle.
® Refrigeration, Air conditioners & Heat pumps
● These appliances are heat engines operating
in reverse.
● By doing work, heat is extracted from the cold
reservoir & exhausted to the hot reservoir
® Thermal devices or thermal machines are one of
the applications of this cycle.
 The heat pumps to produce heating,
 the refrigerators to produce cooling,
 the steam turbines used in the ships,
 the combustion engines of the combustion vehicles
 the reaction turbines of the aircraft 24

Temperature & Free Energy
 Suppose an engine works reversibly
between a hot source at a temperature
Th and a cold sink at a temperature Tc,
then it follows that
Tc = (1 − η)Th
 The Kelvin scale (which is a special case of
the thermodynamic temperature scale) is
currently defined by using water at its
triple point as the notional hot source and
defining that temperature as 273.16 K
exactly.
 Kelvin used this expression to define the
thermodynamic temperature scale in
terms of the efficiency of a heat engine in
which the hot source is at a known
temperature and the cold sink is the object
of interest.
 Free energy and the spontaneity
 Thermodynamic Temperature
 It is not always easy to know the entropy change of
both system and the surrounding
 We can express this in terms of G (the free energy
function) and thus can device a criterion of spontaneity
in terms of the state function of the system.
 As,
and is a criterion for the spontaneity/feasibility of a process.
 There are few assumptions:
® Assumption no 1:
The system is in thermal equilibrium with its surroundings at
a temperature T.
® Assumption no 2:
Heat transfer between the system and the surrounding
happens at constant volume
® Assumption no 3:
Heat transfer between the system and the surrounding
happens at constant pressure. 25

Free Energy
 Under the Assumption 1 & Assumption 2
Clausius inequality becomes: dS ≥ 0
dA=dU−TdS [A = Helmholtz free energy]
 TdS ≥ dU

 The criterion of spontaneity in terms of Gibbs Free
energy change (dG) and Helmholtz energy (dA)
 Under the Assumption 1 & Assumption 3
Clausius inequality becomes: dS ≥ 0
[constant P, no additional work
 TdS ≥ dH
dG=dH−TdS [G = Gibbs free energy]

A=U−TS
G=H−TS
&
dH > 0
but if such a reaction is to be spontaneous at constant
temperature and pressure, G must decrease.
 as dG = dH − TdS
it is possible for dG to be
negative provided that
the entropy of the system
increases so much that
TdS outweighs dH.
Þ Endothermic
reactions are
therefore driven by
the increase of
entropy of the system
[we get this by applying the 1st
law &
the  In an endothermic reaction:
26

27
Free Energy
 Commonly spontaneous because dH < 0 & then
dG < 0 provided TdS which is not so negative
that it outweighs the decrease in enthalpy.
 dG = 0
 Free energy change with
temperature and pressure:
 In an exothermic reactions
 At chemical equilibrium

Spontaneity of a Chemical Reaction
• A spontaneous reaction is a reaction that favors the formation of products at the conditions
under which the reaction is occurring. Spontaneous processes may be fast or slow, but they
occur without outside intervention.
Ex. i) Conversion of graphite to diamond is slow;
ii) A burning fire is relatively a fast reaction.
• "In any spontaneous process there is always an increase in the entropy of the universe“
• For a given change to be spontaneous, ΔSuniverse must be positive.
28
• The change in enthalpy, change in entropy and change in free energy of a reaction are the
driving forces behind all chemical reactions.
• Free energy is energy that is available to do work. The free energy change of a reaction is a
mathematical combination of the enthalpy change and the entropy change.
Gibbs Free Energy

Conditions for Spontaneity of a Chemical Reaction
(Changes in Enthalpy (ΔH), Entropy (ΔS), and Free Energy (ΔG))
29
ΔH◦ ΔS◦ ΔG◦
Negative Positive Always negative
Positive Positive
Negative at higher temperatures, positive at
lower temperatures
Negative Negative
Negative at lower temperatures, positive at
higher temperatures
Positive Negative Always positive
• A spontaneous reaction is one that releases free energy, and so the sign of ΔG must be negative.
Since both ΔH and ΔS can be either positive or negative, depending on the characteristics of the
particular reaction, there are four different possible combinations as shown in the table below.

3rd
Law of
Thermodynamics
The entropy of all perfect crystalline substances is zero at T = 0.
⮚ Third law of thermodynamics:
▪ At T = 0, all energy of thermal motion has
been quenched and in a perfect crystal all
the atoms/ions are in a regular, uniform
array.
▪ The localization of matter and the absence
of thermal motion suggest that such
materials also have zero entropy.
▪ Statistical or microscopic definition of entropy:
S = k ln W
where, S = the entropy,
k = Boltzmann constant,
W = the number of microstates
or the total number of ways a
molecular state can be
distributed over the energy
states for a specific value of total
energy.
→ When T = 0, W = 1
∴ S = k lnW
= 0
⇒ if the value zero is ascribed to the entropies of elements
in their perfect crystalline form at T = 0, then all perfect
crystalline compounds also have zero entropy at T = 0
30

Part-B
Chemical Kinetics
31
- Order of reactions
- Concept of activation energy and energy barrier
- Arrhenius equation
- Homogeneous catalysis
- Heterogeneous catalysis
- Enzyme catalysis (Lock and Key mechanism)

• The chemistry that deals with the reaction rates is known as chemical
kinetics.
• It plays an important role in the production of chemicals on an industrial scale
and the decay of radioactive isotopes used in medicine.
• Chemical kinetics is also useful in providing information about how reactions
occur—the order in which chemical bonds are broken and formed during the
course of a reaction.
• Experimental information on the rate of a given reaction provides important
evidence that helps us formulate a reaction mechanism, which is a step-by-
step, molecular-level view of the pathway from reactants to products.
It is to be contrasted with thermodynamics, which deals with the direction in
which a process occurs but in itself tells nothing about its rate
Factors That Affect Reaction Rates
1. Physical state of the reactants- homogeneous, involving either all gases or all
liquids, or as heterogeneous, in which reactants are in different phases.
2. Reactant concentrations.
3. Reaction temperature.
4. The presence of a catalyst.

Reaction Rates
On a molecular level, reaction rates depend on the
frequency of collisions between molecules. The greater the
frequency of collisions, the higher the reaction rate.
 The change in concentration of reactants or products per
unit time.
 Here, the instantaneous rate of disappearance of one of
the reactants (A or B) at a given time, t (at constant
volume) is −d[R]/dt.
 Similarly, the rate of formation of one of the products is
d[P]/dt. (Note the change in the sign)
 The negative sign indicates that the concentration is
decreasing with time.

Concentration and the rates of reactions
• The rate of this reaction can be expressed either as the rate of disappearance of reactant
‘A’ and ‘B’ or as the rate of appearance of product ‘C’.
• The rate of reaction will be:
• Average rate of appearance /disappearance of A, B or C = change in concentration of
A, B or C
change in time
Consider a general reaction, A + B → C
     
dt
C
d
dt
B
d
dt
A
d




Now, consider another general reaction: aA + bB → cC + dD
The rate of reaction will be:
where a, b, c & d are stoichiometric coefficients
       
dt
D
d
d
dt
C
d
c
dt
B
d
b
dt
A
d
a
1
1
1
1






35
Write rate expressions for the following reactions:
Sample Questions:
1. NO2 (g) + CO (g) → NO (g) + CO2 (g)
2. 2HI (g) → H2 (g) + I2 (g)
Solution:
       
dt
CO
d
dt
NO
d
dt
CO
d
dt
NO
d 2
2





1.
2.      
dt
I
d
dt
H
d
dt
HI
d 2
2
2
1



The rates of reactions

36
The rate law is the relationship between the rate and the concentration, which
are related by a proportionality constant ‘k’, known as rate constant.
For the general reaction:
aA + bB → cC + dD
the rate law generally has the form
rate = k [A]m
[B]n
where exponents ‘m’ and ‘n’ are order of reaction in ‘A’ and ‘B’, respectively and
‘k’ is the rate constant.
This above equation is called the rate law of the reaction.
Rate laws and rate constants
Exponents m and n are typically small whole numbers, whose values are not necessarily
equal to the coefficients a and b from the balanced equation.

37
Important points about rate laws and rate constant:
 Rate law is a result of experimental observation. You CANNOT look
at the stoichiometry of the reaction and predict the rate law (unless
the reaction is an elementary reaction).
 The rate law is not limited to reactants. It can have a product term,
For example: rate = k[A]m
[B]n
[C]c
 The rate constant is independent of the concentrations but depends
on the temperature.
 The units for k vary. Determine units for k by considering units for
rate and for concentration.
Rate laws and rate constants…

38
Ex. For the reaction: A + B → C
rate = k [A]m
[B]n
• where, m is the order of reaction with respect to A, n is the order of reaction with respect to
B.
• The overall reaction order is the sum of the exponents in the rate law = m + n
 m = 0 (Zero order k [A]0
)
 m = 1 (First order k [A]1
)
 m = 2 (Second order k [A]2
)
Order of a reaction
Examples:
H2 + Cl2 → 2HCl. Rate = k [H2]0
[Cl2]0
(Zero order)
SO2Cl2(g) → SO2(g) + Cl2(g) Rate = k [SO2Cl2]1
(First order)
2NO2 → 2 NO + O2 Rate = k [NO2]2
(Second order)
2NO(g) + 2H2(g) → 2N2 (g) + 2H2O (g) Rate = k [NO]2
[H2]1
(Third order)
CH3COOC2H5 + H2O  CH3COOH + C2H5OH Rate = k [CH3COOC2H5]1
[H2O]0
(pseudo-first-order)
• It is the sum of the exponents of the concentrations in the rate law equation
• It can be integers, fractions, negative or positive.
• It can be determined only experimentally
• It may not be equal to the number of molecules of reactants

Molecularity of a reaction
• It is the number of molecules or ions that participate in the rate determining step
• It is a theoretical concept and it can be determined from rate determining step
(slowest step).
• It is always an integer between 1 to 3, as it is not possible to colliding of four or
more molecules simultaneously.
• For elementary reactions (single step reactions):
– It is the sum of stoichiometry coefficients of the reactants
Ex. H2 + I2  2HI
Molecularity = 1 + 1 = 2
• For complex reactions:
– The sum of the number of reactant in the rate determining steps gives the
molecularity

A first-order reaction is one whose rate depends on the concentration of a single reactant raised to
the first power.
If a reaction of the type A  products is first order, the rate law is:
First-Order Reactions
Rate =
Separate concentration and time terms
Integrating over the limits [A]0
to [A]t
and 0 to
t,
    kt
A
A t 

 0
ln
ln
 
  
 

t
t
A
A
dt
k
A
A
d
0
]
[
0
]
[
 
 
kdt
A
A
d


 
  
 
 dt
k
A
A
d
   
A
k
dt
A
d


Straight line equation (y = mx + c)
y = mx + c
If we plot ln [A]t versus time, then we
will get a straight line having negative
slope (-k).
Rate constants can be determined
from experiment by plotting data in this
manner. -k
ln[A]0

The half-life of a reaction (t1/2)
• It is the time required for the concentration of a reactant to reach half its initial value, [A]t1/2 = ½
[A]0.
• Half-life is a convenient way to describe how fast a reaction occurs, especially if it is a first-
order process.
• A fast reaction has a short half-life.
• We can determine the half-life of a first-order reaction by substituting
[A]t1/2 = ½ [A]0 for [A]t and t1/2 for t in Equation :
ln ½ [A]0 = -k t1/2
[A] 0
ln ½ = -k t1/2
t1/2 = - ln ½ = 0.693
k k
    kt
A
A t 

 0
ln
ln
Half-Life of First Order Reaction

Ex. The conversion of methyl isonitrile (CH3NC) to its isomer
acetonitrile (CH3CN) at 199 °C.
ln[CH3NC]t = -kt + ln[CH3NC]0
Figure (a) shows how the pressure of this gas varies with time.
Figure (b) shows that a plot of the natural logarithm of the pressure versus time is a straight line.
The slope of this line is -5.1 X 10-5
s-1
.k = 5.1 X 10-5
s-1

A second-order reaction is one for which the rate depends either on a reactant concentration raised
to the second power or on the concentrations of two reactants each raised to the first power.
For Ex: in reactions A  products or A + B  products that are second order with respect to only
one reactant, A:
Second-Order Reactions
   2
A
k
dt
A
d


 
 
kdt
A
A
d


2
   
kt
A
A t


0
1
1
Rate =
With the use of calculus, this differential rate
law can be used to derive the integrated rate
law for second-order reactions:
• This equation, has four variables, k, t, [A]0, and [A]t,
and any one of these can be calculated knowing the
other three.
• This Equation also has the form of a straight line (y =
mx + c).
• If the reaction is second order, a plot of ‘1/ [A]t’
versus ‘t’ yields a straight line with slope k and y-
intercept 1/ [A]0.
• One way to distinguish between first and second-
order rate laws is to graph both ln[A]t and 1/ [A]t
against t.
• If the ln[A]t plot is linear, the reaction is first order;
• if the 1/ [A]t plot is linear, the reaction is second order.

[A]t = and, t = t1/2
Now, substituting these values in the integral form of the rate equation of second order
reactions, we get:
– = k t1/2
Therefore, the required equation for the half life of second order reactions can be written as
follows.
t1/2 =
This equation for the half life implies that the half life is inversely proportional to the
concentration of the reactants.
Half-Life of Second-Order Reactions
The half-life of a chemical reaction is the time taken for half of the initial amount of reactant to
undergo the reaction.
Therefore, while attempting to calculate the half life of a reaction, the following substitutions must
be made:
   
kt
A
A t


0
1
1

• We have seen that in a first-order reaction the concentration of a reactant ‘A’ decreases
nonlinearly, as shown by the red curve in Figure.
Zero-Order Reactions
• As [A] declines, the rate at which it disappears declines in
proportion.
• A zero-order reaction is one in which the rate of
disappearance of A is independent of [A].
• The rate law for a zero-order reaction is
   0
A
k
dt
A
d


    kt
A
A t 

 0
Rate =
The integrated rate law for a zero-order reaction is Figure. Comparison of first-order and zeroorder
reactions for the disappearance of reactant A with time
where [A]t is the concentration of A at time t and [A]0
is the initial concentration.
This is the equation for a straight line with vertical
intercept [A]0 and slope -kt, as shown in the blue
curve in Figure.
Ex. The most common type of zero-order reaction occurs
when a gas undergoes decomposition on the surface of
a solid.
2N2O Pt(hot) 2N2+O2
Photochemical reaction:
H2 (g) + Cl2 (g) hv 2HCl(g)

From the integral form, we have the following equation
[A] = [A0] – k t
Replacing t with half-life t1/2 we get:
[A0] = [A0] – k t1/2
Therefore, t1/2 can be written as:
k t1/2 = [A0] t1/2 = [A0]
Half-Life of a Zero Order Reaction
The timescale in which there is a 50% reduction in the initial population is referred to as half-
life. Half-life is denoted by the symbol ‘t1/2’.

A pseudo first-order reaction can be defined as a second-order or bimolecular reaction
that is made to behave like a first-order reaction.
⮚ This reaction occurs when one reacting material is present in great excess or is maintained at
a constant concentration compared with the other substance.
A + B → C
So, if component B is in large excess and the concentration of B is very high as compared to that
of A, the reaction is considered to be a pseudo-first-order reaction with respect to A.
If component A is in large excess and the concentration of A is very high as compared to that of B,
the reaction is considered to be pseudo-first order with respect to B.
CH3COOC2H5 + H2O → CH3COOH + C2H5OH
For example:
Rate = k [CH3COOC2H5]
**The concentration of water is very high and thus does not change much during the
course of the reaction.
Pseudo First Order Reaction
47

Temperature and Rate
• The rates of most chemical reactions increase as the
temperature rises.
• The faster rate at higher temperature is due to an increase
in the rate constant with increasing temperature.
• For example, CH3NC  CH3CN
The rate constant and, hence, the rate of the reaction increase
rapidly with temperature, approximately doubling for each 10
°C rise. Temperature dependence of the rate constant
for methyl isonitrile conversion to acetonitrile
The Collision Model
• The greater the number of collisions per second, the greater the reaction rate.
• As reactant concentration increases, therefore, the number of collisions increases, leading
to an increase in reaction rate.
• According to the kinetic-molecular theory of gases, increasing the temperature increases
molecular speeds. As molecules move faster, they collide more forcefully (with more
energy) and more frequently, both of which increase the reaction rate.

The Orientation Factor:
• In most reactions, collisions between molecules result in a chemical reaction only if the
molecules are oriented in a certain way when they collide.
• The relative orientations of the molecules during collision determine whether the atoms are
suitably positioned to form new bonds.
For example,
Cl + NOCl  NO + Cl2
• Molecular collisions may or may not lead to a chemical reaction between Cl and NOCl.

Activation Energy:
• In 1888 the Swedish chemist Svante Arrhenius suggested that molecules must possess a
certain minimum amount of energy to react.
• According to the collision model, this energy comes from the kinetic energies of the colliding
molecules.
• Upon collision, the kinetic energy of the molecules can be used to stretch, bend, and
ultimately break bonds, leading to chemical reactions.
• The minimum energy required to initiate a chemical reaction is called the activation
energy, Ea, and its value varies from reaction to reaction.
Energy is needed to overcome a barrier between initial and final states.

For example, • The difference between the energy of the starting molecule and the highest energy
along the reaction pathway is the activation energy, Ea.
• The molecule having the arrangement of atoms shown at the top of the barrier is
called either the activated complex or the transition state.
• The rate constant depends on the magnitude of Ea
• Generally, the lower the value of Ea is, the larger the rate constant and the faster the
reaction.
Energy profile for conversion of methyl isonitrile
(H3CNC) to its isomer acetonitrile (H3CCN).
The effect of temperature on the distribution of kinetic energies of
molecules in a sample.
• Figure shows the distribution of kinetic energies for two temperatures, comparing
them with the minimum energy needed for reaction, Ea.
• At the higher temperature a much greater fraction of the molecules have kinetic
energy greater than Ea, which leads to a greater rate of reaction.

Arrhenius equation (non exponential form)
where, k is rate constant, Ea is the activation
energy,
A is the pre-exponential (frequency) factor is
constant, and RT is the average kinetic energy i.e.,
R is the gas constant (8.314 J/mol-K), and T is the
absolute temperature.
❖ A graph of lnk versus 1/T is a straight line
when the reaction follows the behavior
described by the Arrhenius equation.
❖ The higher the activation energy, the
stronger the temperature dependence of the
rate constant (i.e., the steeper the slope).
❖ If a reaction has zero activation energy, its
rate is independent of temperature.
• value of from intercept at infinite
T (i.e.; 1/T=0)
• value of Ea from the slope.
53
Arrhenius noted that for most reactions the increase in rate with increasing temperature is nonlinear.
The Arrhenius equation gives the dependence of the rate constant of a chemical reaction on
the absolute temperature as k = A .
ln k = ln
ln k = 
Arrhenius equation

Consider a series of reactions having these energy profiles:
Rank the forward rate constants from smallest to largest assuming all three reactions have
nearly the same value for the frequency factor A.
SOLUTION
• The lower the activation energy, the larger the rate constant and the faster the reaction.
• The value of ΔE does not affect the value of the rate constant.
• Hence, the order of the rate constants is 2 < 3 < 1.

Catalysis
• A catalyst is a substance that changes the speed of a chemical reaction without undergoing a
permanent chemical change itself.
• Most reactions in the body, the atmosphere, and the oceans occur with the help of catalysts.
• Much industrial chemical research is devoted to the search for more effective catalysts for
reactions of commercial importance.
• The phenomenon of alteration of the rate of a reaction by a catalyst is known catalysis.
• The science and technology of catalysis is of great significance as it affects our daily life.
• The catalytic processes contribute greater than 30-40% of global GDP
• Four major sectors of the world economy involve catalytic processes.
 Petroleum and Energy Production,
 Chemicals and Polymer Production,
 Food industry and
 Pollution control.

Catalytic reactions
• Catalysts work by providing alternative mechanism involving a
different transition state of lower energy.
• Thereby, the activation energy* of the catalytic reaction is lowered
compared to the uncatalyzed reaction .
Figure: Schematic diagram to illustrate the
effect of catalyst on activation energy of
reactions.
Types of catalysis
Following are the main types of catalysis:
1. Homogeneous catalysis
2. Heterogeneous catalysis
3. Enzyme catalysis

1. Homogeneous catalysis
In a reaction, if the catalyst is present in the same phase as the
reactants, it is called a homogeneous catalyst and the phenomenon is
homogeneous catalysis. Such catalysis can take place in gaseous
reaction or reactions in solution.
Figure: Schematic diagram to illustrate the homogeneous catalysis.
57
These chemicals help in attaining the equilibrium more quickly by
increasing the rates of both the forward and reverse reactions to an
extent.

(1) Oxidation of sulphur dioxide, SO2, by oxygen to sulphur trioxide, SO3, in
presence of nitric oxide, NO, in the Chamber Process for sulphuric acid
manufacture.
2SO2 (g) + O2 (g) → 2SO3 (g)
here, NO acts as a catalyst.
(2) The following reaction in the gas phase is catalyzed by traces of chlorine gas,
particularly in presence of light.
2N2O (g) → 2N2 (g) + O2 (g)
In presence of light chlorine forms chlorine radicals, which react with N2O forming
the intermediate radical ClO*.
The proposed mechanism is:
Step 1: N2O (g) + Cl* (g) → N2 (g) + ClO*(g)
Step 2: 2ClO*(g) → Cl2 (g) + O2 (g)
Examples of homogeneous catalysis in the gas phase
58

Examples of homogeneous catalysis in the solution phase
(3) Hydrolysis of ester in the presence of acid and alkali:
CH3COOC2H5 (l) + H2O (l) → CH3COOH (aq) + C2H5OH (aq)
(4) Hydrolysis of sucrose (cane sugar) into glucose and fructose in
presence of minerals acids acting as catalysts:
(cane sugar)
(glucose) (fructose
)
C12H22O11 (aq) + H2O (l) → C6H12O6 (aq) + C6H12O6
(aq)
59

60
The decomposition of aqueous hydrogen peroxide, H2O2(aq) , into water and
oxygen:
2 H2O2 (aq)  H2O(l) + O2 (g) 
In presence of catalyst:
2Br 
(aq) + H2O2 (aq) + 2H+
(aq)  Br2 (aq) + 2H2O(l)
Br2 (aq) + H2O2 (aq)  2Br 
(aq) + 2H+
(aq) + O2 (g) 

If the catalyst is present in a different phase than the reactants is called
heterogeneous catalyst and the phenomenon is known heterogeneous
catalysis.
In heterogeneous catalysis the reactions take place at the interface of two
phases. The catalyst is, often a solid and adsorbs a liquid or a gas. This type
of catalysis is of great importance in many industrial processes.
2. Heterogeneous catalysis
Figure: Schematic diagram to illustrate the heterogeneous catalysis.
61

(a) Manufacture of ammonia by the Haber process. Iron (Fe) acts as catalyst.
N2 (g) + 3H3 (g) → 2NH3 (g)
(b) Manufacture of sulphuric acid by the Contact process. Vanadium pentoxide (V2O5)
or platinum are catalysts for the production of SO3 (g) from SO2 (g) and O2 (g).
2SO2 (g) + O2 (g) → 2SO3 (g)
(c) Catalysts used in many reactions in the petroleum and polymer industries. There
are cases of heterogeneous catalysis where a reaction in the liquid phase is
catalysed by a substance in the solid state. An example is the decomposition of
H2O2 (aqueous) by MnO2 (s).
2H2O2 (aq) → 2 H2O (l) + O2 (g)
(d) Examples of reactions in which both the reactant and the catalyst are in the
solid phase. The decomposition of KClO3 is catalysed by solid MnO2.
2 KClO3 (s) → 2KCl (s) + 3O2 (g)
Examples of Heterogeneous catalyst
62

Enzyme catalysis
• Biological catalysts known as enzymes are necessary for many of the chemical reactions occur
in an extremely complex system of the human body.
Enzymes speed up reactions.
• Enzymes are protein-based molecules that can
process certain chemical reactions
• These reactions occur at a localized site, called
the active site, at the rate much faster than a
normal chemical reaction.
• Substrate is the molecule that fits into the active
site of the enzyme and undergoes transformation
to a product.
• Example of an enzyme catalysed reaction is catalase that converts
hydrogen peroxide to water and oxygen.
• It is an important enzyme protecting the cell from oxidative damage by
reactive oxygen species (ROS)
H2O2 → H2O + O2
Reaction catalysed by the
enzyme, catalase

Mechanism of enzyme-catalyzed reactions
Enzymes’ effect on the activation energy
❑ Enzyme-catalyzed reactions work in a lock
and key fashion.
❑ The substrate uniquely fits like a key into
the active site of the enzyme, forming a
lock-key complex.
❑ The substrate is converted into the product
by the enzyme at the active site.
❑ The product is then released from the
active site.
❑ Enzymes lower the activation energy for reactions. The
lower the activation energy, the faster the rate of the
reactions.
❑ For example: the enzyme catalase reduces the
activation energy for the decomposition of hydrogen
peroxide to 8 kJ mol-1
, corresponding to an acceleration
of the reaction by a factor of 1015
at 298 K.
❑ A generic equation for the complex formation is as
follows:
E + S ES P + E

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