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Thermodynamics
Paper 8: Physical Chemistry
1. Thermodynamics - I 15 hrs.
Definition: of Thermodynamic Terms: System, Surrounding types of system,
intensive and extensive properties. Thermodynamic Process, Concept of heat and
work. Work done in reversible and irreversible process, concept of maximum work
(Wmax), Numerical Problems. First law of Thermodynamics: Statement, Definition
of Internal energy and Enthalpy. Heat capacity, heat capacities at constant volume
pressure and their relationship. Calculation of W, q, dU and dH for the expansion of
ideal gases under isothermal and adiabatic conditions for reversible process,
Numerical problems, Hess’s law of heat Summation and its application.
2. Thermodynamic - II 20 hrs.
Second Law of Thermodynamics: Need for the law, different statement of the law
Carnot Cycle and its efficiency, Numerical Problems. Carnot Theorem. Concept of
Entropy: Definition, Physical significance, Entropy as a State Function, Entropy
change in Physical change, Entropy as criteria of Spontaneity & Equilibrium
Entropy Change in Ideal Gases. Gibbs and Helmholtz Functions: Gibbs Function
(G) and Helmoltz Function (A) as Thermodynamic Quantities. A and G as criteria
for Thermodynamic Equilibrium and Spontaneity, their Advantage over Entropy
change. Variation of A with P, V and T.
3. Chemical Equilibrium 10 hrs.
Equilibrium Constant and Free Energy. Thermodynamic Derivation of Law of Mass
Action. Le Chatelier’s Principle. Reaction Isotherm and Reaction Isochore.
Clapeyron Equation, Clausius-Clapeyron Equation and its Application.
Thermodynamics
Thermodynamics is the study of energy.
OR
Thermodynamics is the study of the movement of heat
from one body to another and the relations between heat
and other forms of energy.
OR
Thermodynamics is the study of the connection between
heat and work and the conversion of one into the other.
OR
Chemical thermodynamics is the study of energy
relationships in chemistry.
Why is the study of Thermodynamics important?
The study of
thermodynamics is
important because-
many machines and
modern devices change
heat into work, such as an
automobile engine or turn
work into heat or cooling,
such as with a refrigerator.
How the conversion of
energy from one from to
another, change in state,
change in matter etc occurs.
Different forms of energies
 Kinetic Energy : Having motion
 Potential Energy : Have ‘potential’
 Chemical Energy : Free energy of chemical sub.
 Thermal Energy : Function of temperature
 Nuclear Energy : Unstable nuclei (Binding energy)
 Molecular Energy: Translational, rotational,
vibrational and electronic energy.
• Kinetic Energy:
– Consider a baseball flying through the air. The ball is said to have
"kinetic energy" by virtue of the fact that its in motion relative to the
ground.
• Potential Energy:
– Consider a book sitting on a table. The book is said to have "potential
energy" because if it is nudged off, gravity will accelerate the book,
giving the book kinetic energy. Therefore, it has “potential.”
• Thermal or Heat Energy:
– Consider a hot cup of coffee. The coffee is said to possess "thermal
energy", or "heat energy," which is really the collective, microscopic,
kinetic, and potential energy of the molecules in the coffee. This energy
is function of temperature.
Chemical Energy:
Consider the ability of your body to do work. The glucose (blood sugar) in your
body is said to have "chemical energy" because the glucose releases energy
when chemically reacted (combusted) with oxygen.
Electrical Energy
All matter is made up of atoms, and atoms are made up of smaller
particles, called protons, neutrons, and electrons. Electrons orbit
around the center, or nucleus, of atoms, just like the moon orbits the
earth. The nucleus is made up of neutrons and protons.
Material, like metals, have certain electrons that are only loosely
attached to their atoms. They can easily be made to move from one
atom to another if an electric field is applied to them. When those
electrons move among the atoms of matter, a current of electricity is
created.
Electrochemical Energy:
Consider the energy stored in a battery. Like the example above
involving blood sugar, the battery also stores energy in a chemical
way. But electricity is also involved, so we say that the battery stores
energy "electro-chemically". Another electron chemical device is a
"fuel-cell".
Sound Energy:
Sound waves are compression waves associated with the potential and kinetic
energy of air molecules. When an object moves quickly, for example the head of
drum, it compresses the air nearby, giving that air potential energy. That air then
expands, transforming the potential energy into kinetic energy (moving air). The
moving air then pushes on and compresses other air, and so on down the chain.
Electromagnetic Energy (light): E = hν Nuclear Energy: E = 931.1 Δm
1) The most obvious and trivial way in which energy is transported is when an
object that possesses energy simply moves from one place to another. For
example, a baseball flying through the air is a simple form of
energy transport.
2) Kinetic energy can also be transferred from one object to another when objects
collide. This is also pretty trivial, except that we also know that the total energy,
including any heat or other forms of energy generated during the collision, is
conserved in this process, regardless of the relative sizes, shapes, and materials of
the objects.
3) There are three important ways that heat energy can be transported or
transferred, called conduction, convection, and radiation.
The first two refer to transfer of the thermal energy, whereas the last is really a
conversion of energy to a different form, (photons of light) and the subsequent
travel (transport) of those photons.
11/26/2021 8
How is energy transported from place to place and transferred between
objects?
4) The "diffusion" of thermal energy (heat) through a substance, which occurs
because hotter molecules (those that are vibrating, rotating, or traveling faster),
interact with colder molecules, and in the process transfer some of their energy.
Metals are excellent conductors of heat energy, whereas things like wood or
plastics are not good conductors of heat. Those that are not so good conductors
are called insulators.
5) The transfer of heat energy by the movement of a substance, such as a heated
gas or liquid from one place to another.
For example, hot air rising to the ceiling is an example of convection (in this case
called a convection current).
6) In the context of heat transfer, however, the term “emits radiation" refers
just to light (electro-magnetic waves), and in particular, to the surprising fact
that all objects, even those that are in equilibrium (at equal temperature) with
their surroundings, continuously emit, or radiate electromagnetic waves (that is,
light waves) into their surroundings.
The source of this radiation is the thermal energy of the materials, that is, the
movement of the object's molecules.
11/26/2021 9
Open
Close
Isolated
Energy &
Matter
Surrounding
SOME TERM USED IN THERMODYNAMICS
“The part of universe that under study in thermodynamics is called as system, it is
real or imaginary.”
“The Remaining part of the universe which is in contact or interacts with the system
is called as surrounding.”
For the thermodynamic study the interaction between system and surrounding is very
important. According to the interaction (exchange of matter and energy), systems are
classified in as:
1) Open system:- A system which can exchange both matter and energy with it’s
surrounding is called as open system. e.g.- Heating of water in open beaker.
2) Closed system:- A system which exchange only energy but not matter with it’s
surrounding is called as closed system. e.g.- Heating of water in closed/sealed
container.
3) Isolated system:- A system which neither exchange energy nor the matter with it’s
surrounding is called as isolated system. e.g.- Store of hot water in thermos flask.
The systems were further classified according to the phases as- homogeneous and
heterogeneous system.
Phase of the system is defined as - a homogeneous, physically distinct and
mechanically separable portion of a system.
• According to the nature of the changes occurs in the system, it may be reversible and
irreversible, system can be classified as reversible and irreversible system.
• “Chemical changes in which driving and opposing forces are differ by large amount in
which driving force large than opposing force, if opposing force increased by small amount,
reaction does not reverse is called as irreversible process.”
• “A reaction proceeds in only one direction is called as irreversible reaction while the
reaction proceeds in both directions is called as reversible reaction”.
“A chemical reaction in which driving and opposing forces
are differ by infinitely small amount and proceeds till
attends equilibrium state, if opposing force is increased by
small amount, reaction become reverse is called as
reversible reaction.
Substitution reaction
The properties of the system are classified into two groups as-
a) Extensive properties:- An extensive properties of the system is any property
whose magnitude is depends on the amount of present in the system.
e.g.- pressure, volume, entropy, free energy, etc.
b) Intensive properties:- Intensive properties of a system is any property whose
amount is independent on the total amount but depends on the concentration of the
substance or substances in a system.
e.g.- density, refractive index, chemical potential, viscosity, surface tension, etc.
Properties of system
System
Properties depends
on amount
Properties
independs on
amount
Extensive
properties
Intensive
properties
According
to process
occurred in
system
Isothermal
process
Isobaric
process
Isochronic
process
Cyclic
process
Adiabatic
process
• According to the process occurring in the system or according to the fundamental parameter change in a
system, it can be classified as-
• 1) Isothermal process:- A process occurring at constant temperature is called as isothermal process and
a system is called as isothermal system.
Δ T = 0
• 2) Isobaric process:- A process occurring at constant pressure is called as isothermal process and a
system is called as isobaric system.
Δ P = 0
• 3) Isochronic process:- A process occurring at constant volume is called as isothermal process and a
system is called as isochronic system.
Δ V = 0
• 4) Cyclic process:- A process in which a change in any state function such as internal energy is equal to
zero or a process in system having initial and final state are same is called as cyclic process and system is
called as cyclic system.
Change in state function = 0
• 5) Adiabatic process:- A process in which system does not exchange heat with it’s surrounding is called
as adiabatic process and a system is adiabatic system.
q = 0.
• All fundamental parameters or properties (P, V, T, H, E, G, A, density, surface tension, K, etc) are
classified into two types as-
• a) State function:- A parameter or properties of the system is depends on only initial and final state of
the system but not on the path through which change occurs is called as state function.
• e.g.- free energy, entropy, pressure, volume, etc.
• b) Path function:- A parameter or properties of the system is depends on the path through which change
occurs but not on the initial and final state of the system is called as path function.
The First law of Thermodynamics
-energy cannot be created or destroyed only converted from
one form to another.
-Energy can be changed from one form to another, but it
cannot be created or destroyed.
-The total amount of energy and matter in the Universe
remains constant, merely changing from one form to another.
-The First Law of Thermodynamics (Conservation) states that
energy is always conserved, it cannot be created or destroyed.
The First law of Thermodynamics
The supplied heat is converted to work and change in
internal energy, work and internal energy are forms of
energy transfer and hence energy is conserved.
change in
total internal energy
heat added
to system
State Function Process Functions
E = q + Won
system
done
work
:
positive
system
added
heat
:
positive
by
to
W
q
W
q
E 


The First law of Thermodynamics
• statement of energy conservation for
a thermodynamic system
• internal energy E is a state variable
• W, q process dependent
Sign conventions: (a) Heat:
 q is positive - heat is flow from surrounding to the system i.e. system absorb heat
from surrounding.
 q is negative - heat is flow from system to the surrounding i.e. system lost heat to its
surrounding.
(b) Work done:
 System does work on its surrounding, W is positive.
 Surrounding does work on its System, W is negative.
(c) Internal energy:
 Both heat (q) and Work done (W) is positive, internal energy of the system increase
i.e. ΔE increases.
 Internal energy of the system decreases, ΔE is negative.
.
11/26/2021 20
The First Law of Thermodynamics
by
dW
dq
dE 

int
What this means: The internal energy (dE) of a system tends to
increase if energy is added via heat (dq) and decrease via work (dW)
done by the system.
on
dW
dq
dE 

int
. . . and increase via work (W) done on the system.
on
by dW
dW 

Work done in reversible and irreversible process:
Consider a cylinder having cross-sectional area ‘A’ fitted
with weightless and frictionless piston. Let pressure of
gas on the piston is ‘p’ (p = f/A) and the force acting on
the piston is ‘f’.
For reversible isothermal expansion, piston moves in upward direction through a
distance dl so that the volume change is dV and the pressure (p – dp).
dw = (p – dp) dV
= pdV – dp.dV 1
But for reversible expansion, change in pressure and volume is very small, so that
dp.dV ≈ 0 (neglected)
dw = p.dV 2
For one mole of ideal gas - pV = RT or p = RT/V 3
From equation (2) and (3), dw = RT. (dV/V)
Integrate this equation with the limits V1 to V2 at constant temperature, we have: 𝑑𝑤 = 𝑅𝑇
𝑉1
𝑉2
𝑑𝑉
𝑉
𝑊 = 𝑅𝑇 ln 𝑉2 − ln 𝑉1
𝑊 = 2.303 𝑅𝑇 log
𝑉2
𝑉1
For n mole of ideal gas: 𝑊 = 2.303 𝑛𝑅𝑇 log
𝑉2
𝑉1
At constant temperature, according to the Boyle’s law: p.V = Constant. 𝑃1𝑉1 = 𝑃2𝑉2 or
𝑉2
𝑉1
=
𝑃1
𝑃2
Where, P1, V1 is the initial pressure and volume and P2, V2 is the final pressure and
volume of the system respectively.
𝑊 = 2.303 𝑛𝑅𝑇 log
𝑃1
𝑃2
The workdone is maximum, when opposing pressure ‘P’ differs only infinitesimally small (dp) in
magnitude from the internal pressure of the gas itself which is the condition for the reversibility of the
reaction.
𝑊
𝑚𝑎𝑥 = 2.303 𝑅𝑇 log
𝑉2
𝑉1
= 2.303 𝑛𝑅𝑇 log
𝑃1
𝑃2
In reversible expansion
In reversible compression 𝑊
𝑚𝑎𝑥 = 2.303 𝑅𝑇 log
𝑉1
𝑉2
= 2.303 𝑛𝑅𝑇 log
𝑃2
𝑃1
For irreversible process:
The gas was expand irreversibly from volume V1 toV2 at constant pressure P
and temperature T, then from equation (1)-
dw = p . dV
On integration between the limits V1 and V2, 𝑊 = 𝑝 𝑉1
𝑉2 𝑑𝑉
𝑉
= PΔV
Enthalpy
The thermal changes at constant pressure can be expressed by using new thermodynamic function called
as enthalpy (H).
“It is sum of internal energy (E) of the system and pressure-volume product (PV).”
Since E and PV are the state functions of the system, H is also state function. Therefore
change in enthalpy can be determined as:
ΔH = H2 - H1
H2 and H1 are enthalpies of the final and initial state of the system: H1 = E1 + P1V1 H2 = E2 + P2V2
and
ΔH = H2 - H1
= (E2 + P2V2) – (E1 + P1V1)
= (E2- E1) + (P2V2- P1V1)
= ΔE + ΔPV
At constant pressure, ΔH = ΔE + PΔV
For an ideal gas: PV = nRT
At constant P and T, PΔV = Δn RT
(Δn is change in number of moles during the reaction)
ΔH = ΔE + Δn RT
According to the first law of thermodynamics: q = W + ΔE
And workdone by irreversible process at constant P is: W = PΔV, therefore – q = ΔE + PΔV
Therefore, change in enthalpy - ΔH = q
Heat capacity
A small amount of heat is added to a system, causes rise in
temperature. Suppose a small amount of heat (dq) required to rising
the temperature of the system by one degree is called as heat capacity.
Heat capacity =
𝑑𝑞
dT
Heat capacity does not explain amount of the system therefore new term is
used containing amount of matter is called as molar heat capacity. “The
amount of heat (dq) required to rise the temperature of the one mole of the gas
through one degree is called as molar heat capacity.” It is denoted by latter C.
C =
𝑑𝑞
dT
But according to the first law of thermodynamics: dq = dE + PdV C =
(dE + PdV)
dT
The molar heat capacity at constant volume is called as (Cv). dV = 0 𝐶𝑣 =
𝑑𝐸
dT 𝑉
At constant volume, the amount of heat (dq) required to rise the temperature of the one mole
of the gas through one degree is called as molar heat capacity at constant volume (Cv).
At constant pressure, the heat absorbs ΔH = q. the molar heat capacity at
constant pressure is called as Cp.
𝐶𝑝 =
𝑑𝑞
dT
=
𝑑𝐻
dT 𝑃
At constant pressure, the amount of heat (dq) required to rise the temperature of the
one mole of the gas through one degree is called as molar heat capacity at constant
pressure (Cp).
Difference between two heat capacities is - Cp - Cv =
𝑑𝐻
dT 𝑃
-
𝑑𝐸
dT 𝑉
But according to the definition of enthalpy, H = E + PV. Differentiate
it with respect to temperature at constant pressure, then-
𝑑𝐻
dT 𝑃
=
𝑑𝐸
dT 𝑃
+ P
𝑑𝑉
dT 𝑃
But the internal energy E is the function of any two of the three
variables; P, V and T. If V and T are two independent variables, then;
E = f(V,T)
dE =
𝑑𝐸
dT 𝑉
dT +
𝑑𝐸
dV 𝑇
dV
Dividing both side by dT and apply the condition of constant pressure
as:
𝑑𝐸
dT 𝑃
=
𝑑𝐸
dT 𝑉
+
𝑑𝐸
dV 𝑇
𝑑𝑉
dT 𝑃
Cp - Cv =
𝑑𝐸
dT 𝑉
+
𝑑𝐸
dV 𝑇
𝑑𝑉
dT 𝑃
+ P
𝑑𝑉
dT 𝑃
-
𝑑𝐸
dT 𝑉
=
𝑑𝐸
dV 𝑇
𝑑𝑉
dT 𝑃
+ P
𝑑𝑉
dT 𝑃
For an ideal gas PV = nRT, differentiate it with respect to T at
constant V gives;
𝑑𝑃
dT 𝑉
=
𝑛𝑅
𝑉
But,
𝑑𝐸
dV 𝑇
= 𝑇
𝑑𝑃
dT 𝑉
− 𝑃
𝑑𝐸
dV 𝑇
= 𝑇
𝑛𝑅
𝑉
– P = P – P = 0.
The internal energy of the ideal gas is independs on the volume
The internal energy of the ideal gas is depends on the volume, but it
depends on temperature. If we differentiate ideal gas equation with
respect to temperature at constant pressure, then-
𝑑𝑉
dT 𝑃
=
𝑛𝑅
𝑃
But,
𝑑𝐻
dP 𝑇
= V – T
𝑑𝑉
dT 𝑃
𝑑𝐻
dP 𝑇
= V – T
𝑛𝑅
𝑃
= V – V = 0
Enthalpy of an ideal gas is independent on the pressure
Cp - Cv =
𝑑𝐸
dT 𝑃
+ P
𝑑𝑉
dT 𝑃
-
𝑑𝐸
dT 𝑉
= P
𝑑𝑉
dT 𝑃
From the above explanation, a conclusion is drawn that Cp and Cv
are functions of T only and is independent on volume and pressure.
But for ideal gas; PV = RT (for one mole) Therefore, P
𝑑𝑉
dT 𝑃
= R.
Cp - Cv = R
For n moles of ideal gas;
Cp - Cv = nR
Alternative derivation:
According to definition of enthalpy - H = E + PV 1
For one mole of ideal gas - PV = RT 2
From (2), equation (1) become - H = E + RT 3
Differentiate (3) completely - dH = dE + R dT 4
From the definition of Cp & Cv Cp dT = Cv dT + R dT
Devided both side by dT - Cp - Cv = R
For n moles of ideal gas; Cp - Cv = nR
Adiabatic process in ideal gases
In an adiabatic process, there is no exchange of heat between a system and it’s surrounding i.e. q = 0.
According to the first law of thermodynamic - ΔE = -W
Work is done by the system decreases internal energy and therefore temperature of the system.
For infinitesimally small increase in volume dV at constant pressure P, the
workdone is PdV. This workdone is due to expense of internal energy of the gas.
The decrease in internal energy is dE.
dE = -P dV
But according to the definition of Cv: dE = Cv dT for one mole of gas. Cv dT = -PdV
According to the ideal gas equation- PV = RT or P = RT/V 𝐶𝑣 𝑑𝑇 = −
𝑅𝑇
𝑉
𝑑𝑉
𝑑𝑉
𝑉
= −
𝐶𝑣
𝑅
𝑑𝑇
𝑇
Consider Cv is constant and integrate above equation with the limits V1 at T1 and V2 at T2, we have:
𝑉1
𝑉2 𝑑𝑉
𝑉
= −
𝑪𝒗
𝑹 𝑇1
𝑇2 𝑑𝑇
𝑇
ln
𝑉2
𝑉1
= −
𝐶𝑣
𝑅
ln
𝑇2
𝑇1
ln
𝑉2
𝑉1
= ln
𝑇2
𝑇1
−
𝐶𝑣
𝑅
ln
𝑉2
𝑉1
= ln
𝑇1
𝑇2
𝐶𝑣
𝑅
ln
𝑉2
𝑉1
= ln
𝑇1
𝑇2
𝐶𝑣
𝑅
Taking antilog of both sides, we have: 𝑉2
𝑉1
=
𝑇1
𝑇2
𝐶𝑣
𝑅
or 𝑉1𝑇1
𝐶𝑣
𝑅
= 𝑉2𝑇2
𝐶𝑣
𝑅
Where, Cv is constant.
According to the Boyle’s law: PV = constant or
V1P1 = V2P2 or
𝑉2
𝑉1
=
𝑃1
𝑃2
𝑃1
𝑃2
=
𝑇1
𝑇2
𝐶𝑣
𝑅
𝑃1
𝑇1
𝐶𝑣
𝑅
=
𝑃2
𝑇2
𝐶𝑣
𝑅
According to the ideal gas equation-
PV = RT (for one mole of ideal gas).
T =
𝑃 𝑉
𝑅 𝑉1
𝑃1𝑉1
𝑅
𝐶𝑣
𝑅
= 𝑉2
𝑃2𝑉2
𝑅
𝐶𝑣
𝑅
𝑉1
𝐶𝑣+𝑅
𝑅 𝑃1
𝐶𝑣
𝑅 = 𝑉2
𝐶𝑣+𝑅
𝑅 𝑃2
𝐶𝑣
𝑅
𝑉1
𝐶𝑃
𝑅 𝑃1
𝐶𝑣
𝑅 = 𝑉2
𝐶𝑃
𝑅 𝑃2
𝐶𝑣
𝑅
because Cp - Cv = R
Taking Rth power of both side, 𝑉1
𝐶𝑃 𝑃1
𝐶𝑣 = 𝑉2
𝐶𝑃 𝑃2
𝐶𝑣 = constant.
Taking Cv
th root of both side, 𝑉1
𝐶𝑃
𝐶𝑣 𝑃1 = 𝑉2
𝐶𝑃
𝐶𝑣 𝑃2 = constant.
Put (Cp/Cv) = γ i.e. ratio of molar heat capacities at constant pressure and constant volume respectively.
𝑃1𝑉
1
𝛾
= 𝑃2𝑉
2
𝛾
The workdone performed in the adiabatic reversible expansion of ideal gas as:
Differentiate PVγ = constant completely. 𝛾 𝑃 𝑉𝛾−1𝑑𝑉 + 𝑉𝛾𝑑𝑃 = 0
Dividing both side by Vγ-1: γ PdV + V.dP = 0
V.dP = -γ PdV
Differentiate the equation PV = nRT completely, we have- PdV + VdP = nRdT
(1-γ) PdV = nRdT or
𝑃 𝑑𝑉 =
𝑛 𝑅 𝑑𝑇
1 − 𝛾
The work done: dW = PdV =
𝑛 𝑅 𝑑𝑇
1−𝛾
Integrate equation with the limits of temperature T1 and T2 as:
T2
Wm = ∫ nRdT/(1-γ) = [nR/(1-γ)](T2- T1)
T1
If T2 > T1, γ > 1, then, Wm is negative, i.e. work is done on the gas/system. If T2 < T1, γ > 1,
then Wm is positive, i.e. work is done by the gas.
Hess’s law of heat Summation and its application
If a reaction takes place in several steps then its standard reaction
enthalpy is the sum of the standard enthalpies of the intermediate
reactions into which the overall reaction may be divided at the same
temperature.
enthalpy is a state
function
Enthalpy of an overall reaction A → B along one route is ∆rH
∆rH = ∆rH1 + ∆rH2 + ∆rH3 + .....
Example: Given the thermochemical equations:
2 WO2 (s) + O2 (g) → 2 WO3 (s) H = -506 kJ
2 W(s) + 3 O2 (g) → 2 WO3 (s) H = -1686 kJ
Calculate the enthalpy change for: 2 W(s) + 2 O2 (g) → 2WO2 (s).
Hess’s law of heat Summation and its application
Standard enthalpy of combustion (symbol: ∆cH*)
Combustion reactions are exothermic in nature
Important in industry, rocketry, and other walks of life
“the enthalpy change per mole (or per unit amount) of a substance, when it undergoes
combustion and all the reactants and products being in their standard states at the specified
temperature”.
Cooking gas in cylinders contains
mostly butane (C4H10) which completely
burns release 2658 kJ of heat.
C4H10 (g) + 13/2 O2 (g) → 4CO2 (g) + 5H2O(1);
∆cH* = - 2658 kJ mol-1.
Combustion of glucose gives out 2802.0
kJ/mol of heat
C6H12O6 (g) + 6O2 (g) → 6CO2 (g) + 6H2O(1);
∆cH* = - 2802.0 kJ mol-1
Enthalpies of Formation
An enthalpy of formation, ΔHf, is defined as the enthalpy change for the reaction in
which a compound is made from its constituent elements in their elemental forms.
Standard Enthalpy of Formation
The standard enthalpy of formation is the enthalpy change when one mole of a substance
in its standard state is formed from the most stable form of the elements in their standard
states.
Enthalpy of atomization (symbol: ∆αH*)
It is the enthalpy change on breaking one mole of bonds completely to obtain atoms in
the gas phase.
In case of diatomic molecules, like dihydrogen, the enthalpy of atomization is also the
bond dissociation enthalpy.
H2 (g) → 2H (g); ∆αH* = 435.0 kJ mol–1
Bond Enthalpy (symbol: ∆bondH*)
Energy is required to break a bond and energy is released when a bond is formed.
The enthalpy changes associated with chemical bonds, two different terms are used in
thermodynamics.
(i) Bond dissociation enthalpy
The bond dissociation enthalpy is the change in enthalpy when one mole of covalent
bonds of a gaseous covalent compound is broken to form products in the gas phase.
(ii) Mean bond enthalpy
Mean of energy required to break the similar bonds.
The net enthalpy change of a reaction is the amount of energy required to break all
the bonds in the reactant molecules minus the amount of energy required to break
all the bonds in the product molecules.
CH4 (g) → CH3 (g) + H (g); ∆C-HH* = 427 kJ mol-1
CH3 (g) → CH2 (g) + H (g); ∆C-HH* = 439 kJ mol-1
CH2 (g) → CH (g) + H (g); ∆C-HH* = 452 kJ mol-1
CH (g) → C (g) + H (g); ∆C-HH* = 347 kJ mol-1
Therefore,
CH4 (g) → C (g) + 4H (g); ∆αH* = 1665 kJ mol-1
∆C–HH* = ¼ (∆αH*) = ¼ (1665 kJ mol-1) = 416 kJ mol-1
Mean bond enthalpy
Enthalpy of Solution (symbol: ∆solH*)
The enthalpy of solution at infinite dilution is the enthalpy change observed on dissolving
the substance in an infinite amount of solvent when the interactions between the ions
(or solute molecules) are negligible.
The enthalpy of solution of AB(s),
∆solH*, in water
∆solH* = ∆latticeH* + ∆hydH*
Lattice Enthalpy
The lattice enthalpy of an ionic compound is the enthalpy change which occurs when
one mole of an ionic compound dissociates into its ions in gaseous state.
It is impossible to determine lattice energies directly by experiment hence we use an
indirect approach to determine it by constructing energy diagram called a Born-Haber
cycle.
Na+
(g) + Cl (g)
Na+
(g) + Cl-
(g)
NaCl(s)
Na (g) + Cl (g)
Na (g) + 1/2Cl2 (g)
Na (s) + 1/2Cl2 (g)
- 348.8 kJ
-787.0 kJ
-
4
1
1
.
3
k
J
+107.8 kJ
+121.3 kJ
+495.4 kJ
Lattice energy
Ionization energy
of Na (g)
D
i
r
e
c
t
p
a
t
h
Electron affinity
of Cl(g)
energy needed to form
gaseous Cl atoms
energy needed to form
gaseous Na atoms
Heat of formation of NaCl
A metal pellet with mass 100.0 g, originally at 88.4 0C, is dropped into 125 g of water
originally at 25.1 0C. The final temperature of both the pellet and the water is 31.3 0C.
Calculate the heat capacity C (in J/0C) and specific heat capacity Cs (in J/g0C) of the pellet.
The specific heat of water is 4.184 J/g.0C.
Two litres of an ideal gas at a pressure of 10 atm expands isothermally into a vacuum until
its total volume is 10 litres. How much heat is absorbed and how much work is done in the
expansion?
If water vapour is assumed to be a perfect gas, molar enthalpy change for vapourisation of 1
mol of water at 1bar and 100°C is 41 kJ mol–1. Calculate the internal energy change, when
(i) 1 mol of water is vaporised at 1 bar pressure and 100°C. (ii) 1 mol of water is converted
into ice.
1g of graphite is burnt in a bomb calorimeter in excess of oxygen at 298 K and 1 atmospheric
pressure according to the equation:
C (graphite) + O2 (g) → CO2 (g)
During the reaction, temperature rises from 298 K to 299 K. If the heat capacity of the bomb
calorimeter is 20.7kJ/K, what is the enthalpy change for the above reaction at 298 K and 1 atm?
Carnot Cycle
Carnot heat engine is an ideal engine operating under ideal conditions.
Convert a fraction of absorbed heat into work.
Operating between the temperatures T1 (source) and T2 (sink) (T1 > T2).
Absorb a quantity of heat at some
temperature T1,
Convert a part of heat into work, and
remaining part of heat is rejected at
temperature T2.
Carnot by considering a sequence of
operations called as Carnot cycle (series of
operations so conducted and at the end; the
system is back to its original state).
cylinder fitted with weightless and frictionless
piston, It containing a given substance (gas) is
called as working substance.
insolating jacket used to
perform adiabatic process
a source having
temperature T1 and
supplied heat to the
working substance
sink having temperature
T2 and act as heat
acceptor
A Carnot cycle operates in four different steps- two are isothermal and two are adiabatic process
A(P4
,V4
)
B(P1
,V1
)
C(P2
,V2
)
D(P3
,V3
)
q1
q2
T1
T2
Wm
P
V
Isothermal reversible expansion
Isothermal reversible compression
I.
I
II
III
IV
III.
II. Adiabatic reversible expansion
IV. Adiabatic reversible Compression
1. The pressure, volume and temperature of the working substance is P4, V4, and T1 at point A.
A substance absorbs heat from source q1 at the temperature T1 and undergoes isothermal
and reversible expansion to a point B having pressure P1 and volume V1. The change in
internal energy is ΔE1 and work done is W1.
According to the first law of thermodynamics: ΔE1 = q1 - W1. 1
And 𝑾𝟏 = 𝒏𝑹𝑻𝟏 𝒍𝒏
𝑽𝟏
𝑽𝟒
2) Expand gas adiabatically and reversibly from point B (P1, V1, T1) to a point C (P2, V2, T2).
For the adiabatic process, q = 0, the work done by the system is W2 at the expense of an
internal energy ΔE2. During the expansion temperature must be drop from T1 to T2.
4) Expand gas adiabatically and reversibly from point D (P3, V3, T2) to a point A (P4, V4, T1).
For the adiabatic process, q = 0, the work done by the system is W4 at the expense of an
internal energy ΔE4. During the expansion temperature must be drop from T2 to T1.
According to the first law of thermodynamics:
ΔE2 = - W2. 2
And, 𝑊2 = 𝑛 𝑇1
𝑇2
𝐶𝑣
𝑑𝑇
𝑇
ΔE4 = - W4. 4
And 𝑊4 = 𝑛 𝑇2
𝑇1
𝐶𝑣
𝑑𝑇
𝑇
According to the first law of thermodynamics:
3) Gas is compressed isothermally and reversibly at temperature T2 from a point C (P2, V2)
to D (P3, V3). During this step, work done on the gas (system) is W3, a quantity of heat
given to sink or surrounding is q2.
According to the first law of thermodynamics: ΔE3 = - W3 – q2 3
And 𝑊3 = 𝑛𝑅𝑇2 ln
𝑉3
𝑉2
For complete cycle, the total change in internal energy is-
ΔE = ΔE1 + ΔE2 + ΔE3 + ΔE4
= q1 - W1 – W2 – q2 – W3 – W4.
= (q1 – q2) – (W1 + W2 + W3 + W4)
= (q1 – q2) – Wm
Wm is the total work done by the gas in the one complete cycle.
The principle of reproducibility of the states for cycle or reversible process, ΔE = 0.
E is state function.
Wm = (q1 – q2) 5
= (W1 + W2 + W3 + W4)
= 𝑛𝑅𝑇1 ln
𝑉1
𝑉4
+ 𝑛 𝑇1
𝑇2
𝐶𝑣
𝑑𝑇
𝑇
+ 𝑛𝑅𝑇2 ln
𝑉3
𝑉2
+ 𝑛 𝑇2
𝑇1
𝐶𝑣
𝑑𝑇
𝑇
= 𝑛𝑅𝑇1 ln
𝑉1
𝑉4
+ 𝑛𝑅𝑇2 ln
𝑉3
𝑉2
6
But the point B(P1,V1) at T1 and point C(P2,V2) at T2 are lie on same adiabatic curve BC.
𝑉1𝑇1
𝐶𝑣
𝑅
= 𝑉2𝑇2
𝐶𝑣
𝑅
7
𝑉3𝑇2
𝐶𝑣
𝑅
= 𝑉4𝑇1
𝐶𝑣
𝑅
8
But the point D(P3,V3) at T2 and point A(P4,V4) at T1 are lie on same adiabatic curve AD.
Divide equation (7) by (8)- 𝑉1
𝑉4
=
𝑉2
𝑉3
. 9
From equation (9) and (6), we have-
Wm = = 𝑛𝑅𝑇1 ln
𝑉1
𝑉4
- 𝑛𝑅𝑇2 ln
𝑉1
𝑉4
Wm = (q1 – q2) = nR (T1 - T2) 𝒍𝒏
𝑽𝟏
𝑽𝟒
10
Efficiency of Carnot heat engine:
Ratio of the total work done by the working substance and total heat absorbed from the
source. It is denoted by the latter η or ε.
ε =
𝑊𝑚𝑎𝑥
𝑞1
11
=
𝑞1− 𝑞2
𝑞1
= 1 -
𝑞2
𝑞1
12
Also from equation (10), (1a) and (11),
ε =
nR (T1
− T2
) ln
𝑉1
𝑉4
𝑛𝑅𝑇1 ln
𝑉1
𝑉4
=
T1
− T2
T1
= 1 -
T2
T1
13
efficiency of Carnot heat engine depends on heat absorbed or given out to sink
ε depends on the temperature of source and sink but not depends on the nature and
amount of the working substance
Thermodynamic efficiency
 For reversible and cyclic process of the system, if it absorb heat q1 at T1, then it
undergoes a temperature change (T1 - T2) and the maximum work done is equal to the
heat absorb q1 at T1 multiplied by the ratio (T1 - T2)/ T1.
ε =
𝑊𝑚𝑎𝑥
𝑞1
=
T1
− T2
T1
𝑊
𝑚𝑎𝑥 =
T1
− T2
T1
𝑥 𝑞1
 The some part of heat rejected at temperature T2 such that T1>T2.
 The magnitude of the work done is given by the area of the P-V diagram of Carnot cycle.
 For 100% thermodynamic efficiency, T2 = 0 i.e. temperature of the sink is zero K. To
maintain such temperature of the sink or surrounding is practically impossible.
Carnot Theorem/Rule
Developed by Nicolas Léonard Sadi Carnot in the year 1824, with the principle that there
are limits on maximum efficiency for any given heat engine.
Carnot’s theorem states that:
 Heat engines working between two heat reservoirs are less efficient than the
Carnot heat engine (reversible engine) that is operating between the same
reservoirs.
 Efficiency of all reversible engines working between the same two temperatures
(same source and the sink) is same whatever the working substance.
 Maximum efficiency is given as:
𝜖𝑚𝑎𝑥 = 𝜖𝐶𝑎𝑟𝑛𝑜𝑡 =
𝑇𝐶
𝑇𝐻
𝑇𝐶 is absolute temperature of the cold reservoir
𝑇𝐻 is absolute temperature of the hot reservoir
Consider two heat engines R (reversible
engine) and I (irreversible engine)
working between the same source (T1)
and the same sink (T2), where T1>T2.
Reversible engine R takes heat Q1 from the source (T1), perform work W and rejects some heat
Q2 = Q1 –W to the same sink (T2).
The efficiency of the reversible engine R is: 𝜖𝑅 =
𝑊
𝑄1
Suppose irreversible engine I takes heat 𝑄1
′
from the source (T1), perform work W and rejects
some heat 𝑄2
′
= 𝑄1
′
− 𝑊 to the same sink (T2).
𝜖𝐼 =
𝑊
𝑄1
′
The efficiency of the irreversible engine I is:
Suppose, the irreversible engine I is more efficient than the reversible engine R.
𝜖𝐼 > 𝜖𝑅 ⟹
𝑊
𝑄1
′ >
𝑊
𝑄1
⇒ 𝑄1 > 𝑄1
′
⇒ 𝑄1 − 𝑄1
′
𝑖𝑠 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑞𝑢𝑎𝑛𝑡𝑖𝑡𝑦
Where, W work done is same.
These two engines are coupled together so that the
engine R works in opposite direction i.e. the engine
R works as a refrigerator.
The engine I absorbs heat 𝑄1
′
from the source (T1),
perform work W and rejects some heat 𝑄2
′
= 𝑄1
′
−
𝑊 to the same sink (T2).
The engine R takes heat Q2 from the same sink (T2), perform work W on the engine R and
Q1 = Q2 + W heat is rejected to the source (T1).
Work done on the gas by the engine R is actually received from the work done by the gas
on the engine I.
Heat lost by the sink is: 𝑄2 − 𝑄2
′
= 𝑄1 − 𝑊 − 𝑄1
′
− 𝑊 ⇒ 𝑄1 − 𝑄1
′
𝑖𝑠 𝑎 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑞𝑢𝑎𝑛𝑡𝑖𝑡𝑦
i.e. external work done on the system is zero. both the engines R and I work as a self-acting machine
The coupled engines transfers heat continuously from a body at lower temperature to the
body at higher temperature. contrary to the second law of thermodynamics.
i.e. our assumption 𝜖𝐼 > 𝜖𝑅 is wrong.
i.e. no engine will be more efficient than a perfectly reversible engine working between the
same temperatures (same source and same sink)
Second law of thermodynamic
The first law of thermodynamic state that, when one form of energy is converted into
another, the total energy of the system is conserved.
It does not indicate any other restriction on this process and many of the process has a natural
direction. This direction can be explained by using a second law of thermodynamics.
e.g. Suppose gas in expand into a vacuum but although it does not violate first law of
thermodynamics; the reverse does not occur, etc.
Second law of thermodynamic tell us that whether the reaction take places in forward or in
backward direction or direction of flow energy.
The quantity tells us that whether a chemical change or physical process is spontaneous
or not in an isolated system in terms of a new thermodynamic function called as
entropy.
Entropy is a state function. Thermodynamic does not explain the rate of reaction/change to
approach the equilibrium state, but it explains at equilibrium state.
e.g. i) Gas is expand into vaccum, ii) Heat is always transfer from hot end to cold end of the rod, iii)
In presence of catalyst, molecular hydrogen is react with molecular oxygen to form water.
Different Statements of Second Law of Thermodynamics:
Definition by Clausius:
There is no thermodynamic transformation whose sole effect is to deliver heat from
a reservoir of lower temperature to a reservoir of higher temperature
Heat does not flow upwards i.e. lower to higher temperature spontaneously.
In an isolated system, the entropy never decreases.
Definition by Kelvin:
There is no thermodynamic transformation whose sole effect is to extract heat from
a reservoir and convert it entirely to work.
No heat engine is constructed having unit efficiency.
Entropy (S)
Second law of thermodynamic can be expressed mathematically by a new
thermodynamic functions S called as entropy of the system. Entropy of the system is a
state function
ΔS = S2 - S1
Where, S2 and S1 are entropies of final and initial states of the system respectively.
dqrev/T is the differential of the state function. Clausius named this state function is entropy dS.
For infinitesimal change: 𝑑𝑆 =
𝑑𝑞𝑟𝑒𝑣
𝑇
For finite change in state: ∆𝑆 =
𝑑𝑞𝑟𝑒𝑣
𝑇
=
𝑞𝑟𝑒𝑣
𝑇
Where dqrev is the infinitesimal quantity of heat absorbed under reversible conditions at temperature
T. Also qrev is the heat absorbed at any temperature T for any isothermal reversible process.
𝑞𝑟𝑒𝑣 is the positive (heat is absorbed)-
ΔS is positive (entropy of the system
increases)
𝑞𝑟𝑒𝑣 is the negative (heat is evolved)-
ΔS is negative (entropy of the system
decreases)
Endothermic Exothermic
Derivation of entropy from Carnot cycle
The heat q1 is taken by the working
substance from source at temperature T1
and heat q2 is given out by the working
substance at lower temperature T2.
Efficiency of Carnot engine ε =
𝑾𝒎𝒂𝒙
𝒒𝟏
𝑞1− 𝑞2
𝑞1
=
T1
− T2
T1
1 -
𝑞2
𝑞1
= 1 -
T2
T1
𝑞2
𝑞1
=
T2
T1
or
𝑞2
T2
=
𝑞1
T1
q1 has positive value and q2 has a negative value. − 𝑞2
T2
=
𝑞1
T1
or 𝑞1
T1
+
𝑞2
T2
= 0
In general:
𝑞
𝑇
= 0
For a complete reversible cycle represented by the close curve ABA
For reversible cycle ABA:
𝑞
𝑇
= 0
Since the cycle is performed in two steps; via
from A to B and then back from B to A, we have:
𝑑𝑞
𝑇
=
𝐴
𝐵
𝑑𝑞
𝑇
+
𝐵
𝐴
𝑑𝑞
𝑇
= 0
𝑑𝑞
𝑇
=
𝐴
𝐵
𝑑𝑞
𝑇
= −
𝐵
𝐴
𝑑𝑞
𝑇
Both these integrate shows that
𝒅𝒒
𝑻
is independent on the path but it depends on the state of
the system. This function
𝑑𝑞
𝑇
is called as entropy and is state function.
Second law of thermodynamic
i) Natural process are irreversible and spontaneous, it is accompanied by increase
in entropy of the universe. Universe means the system and its surrounding.
ΔSuniv. = ΔSsystem + ΔSsarr. > 0.
But for reversible process, entropy of the system remains constant.
ΔSuniv. = 0.
ii) Heat can not be converted to work without leaving permanent change either in
systems involved or their surrounding.
iii) It is impossible to take heat from a hot reservoir and convert it completely into
work by cyclic process without transforming a part of heat to a cooler reservoir i.e.
it impossible to construct an engine having unit efficiency.
iv) It is impossible for a cyclic process to transfer heat from a body at a lower
temperature to a body of higher temperature in absence of any agency.
Here are 5 different ways to state the second law of thermodynamic
1. For any spontaneous process the entropy of the universe increases.
2. It is impossible for heat to spontaneously flow from a cold object to a hot object.
3. No steam engine can be more efficient than a reversible steam engine (Carnot
theorem).
4. It is impossible to build a perfectly efficient steam engine.
5. It is impossible to build a perpetual motion machine of the second kind.
Entropy change in Physical transformations
Change in entropy occur not only variations of temperature, pressure and volume of the
system but also physical transformation such as change of state from solid to liquid
(melting), liquid to vapour (boiling) and solid to vapour (sublimation)
At constant temperature, two phases are in
equilibrium during phase transformation by
an absorbing or evolving heat.
∆𝑆 =
𝑞𝑟
𝑇
=
∆ 𝐻
𝑇
For melting
For one mole of substance, it absorb a heat is
called as latent/molar heat of fusion at
constant temperature T is called as melting
point. The change in entropy is:
∆𝑆𝑓=
∆ 𝐻𝑓
𝑇𝑓
For boiling
It absorb a heat is called as latent/molar heat of
vaporization at constant temperature T is called
as boiling point. The change in entropy is:
∆𝑆𝑣=
∆ 𝐻𝑣
𝑇𝑏
One mole of solid is change to another allotropic
form at constant allotropic transition
temperature. The change in entropy is:
For
allotropic
transition
∆𝑆𝑡=
∆ 𝐻𝑡
𝑇𝑡
• According to the physical significance of entropy-
In Carnot cycle, only some part of heat is
converted into work and remaining part of
heat is given to sink.
The part of heat which is not available to perform work is a
measure of entropy change accompanying the process. Hence the
fraction of energy unavailable to perform the work is T.ΔS.
S = q/T = ∆H/T
Fraction available to perform work = (ΔH) - T.ΔS.
Physical significance of entropy
(b)
i.e. melting at constant temperature Tf, the heat absorbed by the solid particles to go to
liquid state is ΔHf.
The addition of heat to the system introduces more randomness or disorder in the system
e.g. when solid melts, the ordered arrangement of the constituent particles is destroyed
and particles acquired greater freedom of movement.
Change in entropy is: ∆𝑆𝑓=
∆ 𝐻𝑓
𝑇𝑓
ΔHf is positive and hence ∆𝑆𝑓 is also positive i.e. Ssolid < Sliquid
Similarly for boiling of liquid, the constituent particles of gas have more degree of
freedom and more disorder system is formed.
entropy is taken as a measure of disorder of the system
(c) Entropy of the system is taken as a measure of mixtupness of the system
When organized from of energy such as electric, mechanical or chemical is converted
into heat, the randomness of the system increases as the constituent particles of the
system acquired greater degree of freedom of movement.
It increases the disorder in the system, increase its mixtupness.
(d) Entropy of the system is also used to measured thermodynamic probability of the system.
S = k lnW (put by Plank in 1912)
Where, S is entropy, W is thermodynamic probability, k is Boltzmann constant.
Entropy as criteria of Spontaneity & Equilibrium
(a) For irreversible process:
Consider a system in contact with a reservoir at constant temperature T, assume that in the
system an infinitesimal irreversible process (spontaneous) occurs and work done is of P-V
type.
The quantity of heat dq is exchanged between the system and reservoir. Therefore,
entropy change of system is dS and that of reservoir is – dq/T.
dS > dq/T
TdS > dq
dq – TdS < 0 (1)
According to the first law of thermodynamics - dq = dE + PdV (2)
From equation (1) and (2)-
dE + PdV – TdS < 0 (3)
Condition of entropy change:
Consider the system having constant volume and internal energy i.e. dE = dV = 0,
therefore equation (3) becomes:
-(TdS)E,V < 0
(TdS)E,V > 0 or (dS)E,V > 0 (4)
And for finite change - (ΔS)E,V > 0
Reversible change: change < , > replaced by = sign in above.
Entropy Change in Ideal Gases
The change in entropy of an isolated system is zero or greater than zero.
𝑑𝑆𝑖 = 𝑑𝑆𝑠𝑢𝑏. + 𝑑𝑆𝑟𝑒𝑠𝑒𝑟. ≥ 0
dSi is change in entropy of isolated system
dSsub is change in entropy of working substance
dSreser is change in entropy of heat reservoir
The heat is supplied by the heat reservoir to working substance at
constant temperature T.
𝑑𝑆𝑟𝑒𝑠𝑒𝑟. =
𝑑𝑞
𝑇
From above equation: 𝑑𝑆𝑠𝑢𝑏. +
𝑑𝑞
𝑇
≥ 0
According to the first law of thermodynamics dq = dE + dw
But if system does work is only P-V type, then dq = dE + pdV
And if the system does the work other than P-V type then: dq = dE + pdV + dw’
Put dq in above equation: dSsub – (dE + pdV + dw’)/T ≥ 0
Multiplying both side by T, TdSsub – (dE + pdV + dw’) ≥ 0
The inequality is applied to irreversible processes,
where only P-V work is done and dW’= 0 and p = P,
TdSsub – (dE + PdV) > 0
(for irreversible process)
For reversible processes, dw’ = dW’ and p = P, TdSsub – (dE + PdV + dW’) = 0
But when dW’ = 0, TdSsub – (dE + PdV) = 0
For reversible isothermal process dE = 0, TdSsub – PdV = 0
dSsub =
𝑷𝑑𝑉
𝑇
For n mole of ideal gas: PV = nRT P =
𝑛𝑅𝑇
𝑉
or
Therefore, dS is dSsub =
𝑛𝑅
𝑉
𝑑𝑉
Integrate with the limits of entropy S1 and S2 at the volume V1 and V2:
𝑆1
𝑆2
𝑑𝑆 = 𝑛𝑅
𝑉1
𝑉2 𝑑𝑉
𝑉
𝑆2 − 𝑆1 = 𝑛𝑅 ln
𝑉2
𝑉1
ΔS = 𝑛𝑅 ln
𝑉2
𝑉1
Expansion process ΔS = 𝟐. 𝟑𝟎𝟑 𝑛𝑅 log
𝑉2
𝑉1
Entropy change for the reversible isothermal compression process is: ΔS = 2.303 𝑛𝑅 log
𝑉1
𝑉2
*** Free Energy and Equilibrium ***
•
How much quantity of heat is used to performed work done is
determined by mew thermodynamic state function is called as free
energy. The heat energy is supplied in the form of internal energy or
enthalpy.
[Heat energy (E or H)] = (Heat available to perform work done) +
(Heat unavailable to perform work done)
If heat energy is supplied in the form of internal energy (E), the part of internal energy
available to perform work is called Helmholtz free energy. It is denoted by the latter A.
The internal energy unavailable to perform the work done is TS, S-is entropy at
temperature T.
[Heat energy (E)] = (E available to perform work done) +
(E unavailable to perform work done)
E = A + TS
A = E - TS (1)
If heat energy is supplied in the form of enthalpy (H), the part of
enthalpy available to perform work is called Gibbs free energy.
It is denoted by the latter G or F.
The enthalpy unavailable to perform the work done is TS, S-is
entropy at temperature T.
[Heat energy (H)] = (H available to perform work done) +
(H unavailable to perform work done)
H = G + TS
G = H - TS (2)
Helmholtz free energy
Helmholtz free energy of any system A is defined as -
A = E - TS
Where, A, E, S and T are state functions.
The change in A during any process is given as-
ΔA = A2 - A1
= (E2 – T2S2) – (E1 – T1S1)
= (E2 - E1) – (T2S2 – T1S1)
= ΔE – ΔTS (3)
Equation (3) is the general definition of ΔA. For isothermal process T2 = T1, then
equation (3) becomes-
ΔA = ΔE – TΔS (4)
Eq. (4) explain physical interpretation (significance) of ΔA, under isothermal conditions
– TΔS = qrev (according to the definition of change in entropy).
ΔA = ΔE – qrev. (5)
According to the first law of thermodynamics for reversible and isothermal process-
qrev = ΔE + Wmax or -Wmax = ΔE - qrev (a)
From equation (5) and (a)-
ΔA = -Wmax (6)
The equation (6) indicates that, at constant temperature, the maximum work done by
the system is due to expense of (decrease in) Helmholtz free energy.
Therefore, A is called as Work function of a system.
Relationship between G and A
Consider equation (3) is- ΔG = ΔH – TΔS
But H is also state function- ΔH = ΔE + PΔV (at constant P)
Therefore,
ΔG = ΔE + PΔV – TΔS
= (ΔE – TΔS) + PΔV (ΔA = ΔE – TΔS)
ΔG = ΔA + PΔV (4)
In general- G = A + PV (5)
The equation (4) and (5) shows the relationship between G and A.
A and G as criteria for Thermodynamic Equilibrium and Spontaneity
Condition of Helmoholtz free energy change
If system is not isolated system then its entropy can be changed. If volume of the system
remains constant for irreversible process i.e. dV = 0 then,
dE – TdS < 0
At constant V and T, for infinite small change:
(∂E – T∂S)T,V < 0
∂(E – TS)T,V < 0
Or
According to the definition of A, A = E – TS
∂(A)T,V < 0
then
And for finite change: (ΔA)T,V < 0
Thus, for any irreversible or spontaneous process in a system at constant volume and
temperature, the Helmoholtz free energy of the system decreases therefore stable state is
obtained at the lowest Helmoholtz free energy.
Helmoholtz free energy is state function, therefore at equilibrium state (ΔA)T,V = 0
dE + PdV – TdS < 0
Gibbs free energy is state function, therefore at equilibrium state (ΔG)T,P = 0
Condition of Gibbs free energy change
If system is not isolated system then its entropy can be changed. If pressure and temperature
of the system remains constant for irreversible process i.e. dP = dT = 0
dE + PdV – TdS < 0
At constant P and T, for infinite small change: (∂E + P∂V – T∂S)T,P < 0
Or ∂(E + PV– TS)T,P < 0
Or ∂(H – TS)T,P < 0
According to the definition of G, G = H – TS
∂(G)T,P < 0
And for finite change: (ΔG)T,P < 0
Thus, for any irreversible or spontaneous process in a system at constant pressure and
temperature, the Gibbs free energy of the system decreases therefore stable state is
obtained at the lowest Gibbs free energy.
Variation of A with T and V:
Differentiate it completely as -
dA = dE – TdS - SdT (7)
According to the definition of entropy - dqrev = TdS
Therefore equation (7) becomes-
dA = dE - dqrev - SdT (8)
According to the first law of thermodynamics- dqrev = dE + PdV (for reversible
process)
Or -PdV = dE - dqrev
Therefore equation (8) becomes -
dA = - PdV - SdT (9)
Equation (9) indicates that A is most conveniently expressed in terms of a T and V as
the independent variables, we have-
A = f (T,V)
Differentiate it completely-
dA = (δA/δT)v dT + (δA/δV)T dV (10)
Comparing equation (9) and (10) we have-
(δA/δV)T = -P (11)
(δA/δT)v = -S (12)
• An alternative equation for the variation of A with T can be obtained
as follows-
• Differentiate A/T with respect to T at constant V gives -
• [δ(A/T)/δT]v = [T(δA/δT)v – A]/T2
• = [T(-S) – A]/T2
• = - [A + TS]/T2
• = - E/T2 (13)
• Equation (9) shows that the dependence of A for a pure substance
on both T and V, these can be shown by (12) and (13) for T and (11)
for V.
Variation of G with P and T
According to the definition of G [i.e. from equation (1)]-
G = H - TS
Differentiate it completely as-
dG = dH - TdS - SdT (7)
According to the definition of entropy and first law of thermodynamics-
dqr = TdS = dE + PdV (for reversible process) (8)
According to the definition of enthalpy- H = E + PV
Differentiate it completely- dH = dE + PdV + VdP (9)
Therefore equation (8) and (9) dH = VdP + TdS (10)
From equation (7) and (10)-
dG = VdP + TdS -TdS -SdT
dG = VdP - SdT (11)
Equation (11) indicates that G is function of a T and P as the independent variables, we have-
G = f (T,P)
G = f (T,P)
Differentiate it completely-
dG = (δG/δT)P dT + (δG/δP)T dP (12)
Comparing equation (11) and (12) we have-
(δG/δP)T = V
(δG/δT)P = -S (13)
An alternative equation for the variation of G with T can be obtained as follows-
Differentiate G/T with respect to T at constant P gives-
[δ(G/T)/δT]P = [T(δG/δT)P – G]/T2
= [T(-S) – G]/T2
= - [G + TS]/T2
= -H/T2 (14)
Equation (11) shows that the dependence of G for a pure substance on both T and P,
these can be shown by (14) and (13) for T.
Advantage of G/A over Entropy change
The first law of thermodynamics /energy conservation, but one can't comment on the nature
of a process (spontaneous or non-spontaneous). Hence knowing only the quantity of energy
change is never sufficient.
The second law is defined in the terms of both entropy and Gibbs free energy.
For a process to be spontaneous: ∆ 𝑆 𝑡𝑜𝑡𝑎𝑙 > 0 𝑎𝑛𝑑 ∆𝐺 < 0
Hence both entropy and Gibbs free energy changes can help you determine the spontaneity of a process
Entropy change for a reversible process: ∆𝑆 =
𝑞𝑟𝑒𝑣
𝑇
Criteria for spontaneity is the total entropy: ∆ 𝑆 𝑡𝑜𝑡𝑎𝑙 > 0 and
∆ 𝑆 𝑡𝑜𝑡𝑎𝑙 = ∆ 𝑆 𝑠𝑦𝑠𝑡𝑒𝑚 + ∆ 𝑆 𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔
It is extremely difficult task to determine the entropy change in surrounding therefore converting
it to Gibbs free energy.
∆ 𝑆 𝑡𝑜𝑡𝑎𝑙 =
𝑞𝑠𝑦𝑠𝑡𝑒𝑚
𝑇
+
𝑞𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔
𝑇
& 𝑞𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔 = ∆𝐻𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔 = −∆𝐻𝑠𝑦𝑠𝑡𝑒𝑚
Hence, ∆ 𝑆 𝑡𝑜𝑡𝑎𝑙 = ∆ 𝑆 𝑠𝑦𝑠𝑡𝑒𝑚 −
∆𝐻𝑠𝑦𝑠𝑡𝑒𝑚
𝑇
Multiplying by T and rearranging, 𝑇 ∆ 𝑆 𝑡𝑜𝑡𝑎𝑙 = 𝑇 ∆ 𝑆 𝑠𝑦𝑠𝑡𝑒𝑚 − ∆𝐻𝑠𝑦𝑠𝑡𝑒𝑚
∆𝐺 = − 𝑇 ∆ 𝑆 𝑡𝑜𝑡𝑎𝑙 = − 𝑇 ∆ 𝑆 𝑠𝑦𝑠𝑡𝑒𝑚 + ∆𝐻𝑠𝑦𝑠𝑡𝑒𝑚
ΔS(total) must be greater than zero, ΔG must be less than zero.
Hence, Gibbs free energy reduces the tedious work of determining entropy change in
surrounding and it's value can determined by knowing the state of the system only.
Gibbs free energy is more advantageous than entropy change.
Chemical equilibrium
A + B ↔ C + D
Based on the extent to which the reactions proceed to reach the
state of chemical equilibrium:
(i) The reactions that proceed nearly to completion and only
negligible concentrations of the reactants are left. In some cases,
it may not be even possible to detect these experimentally.
(ii) The reactions in which only small amounts of products are
formed and most of the reactants remain unchanged at
equilibrium stage.
(iii) The reactions in which the concentrations of the reactants and
products are comparable, when the system is in equilibrium.
A + B ↔ C + D
There are number of important questions about the composition
of equilibrium mixtures:
What is the relationship
between the concentrations of
reactants and products in an
equilibrium mixture?
How can we determine
equilibrium
concentrations from
initial concentrations?
What factors can be exploited to alter the
composition of an equilibrium mixture?
Particular is important when choosing conditions for synthesis of
industrial chemicals such as H2, NH3, CaO etc.
Equilibrium constant
Law of mass action
The rate of a chemical reaction at any instant is proportional to the molar concentration of
the reacting substance at that instant
Consider the simplest reaction –
A → product.
If CA is the concentration of the reactant at time t,
then-
Rate =
 CA
The proportionality constant in the above rate equations is called rate
constant or velocity constant or specific rate constant
Reversible reaction
k
k
f
b
A B
Depending on the physical state of the reactants and products, two
physical constants for reversible reaction is defined as KP and KC.
At gaseous state, the equilibrium constant KP is in terms of partial
pressures of reactants and products
For other states or liquid/solution state, it is denoted by KC
which is in terms of partial concentrations.
aA + bB lL + mM
A & B are reactants (behaves ideally)
L & M are products (behaves ideally)
a & b and c & d are number of moles of reactants and products respectively.
general
reaction
are the molar concentrations of A, B, C and D respectively.
Characteristics of Kc:
•When concentrations of the reactants and products have attained constant value at equilibrium
state.
• Value of equilibrium constant is independent of initial concentrations of the reactants and
products.
• Equilibrium constant is temperature dependent having one unique value for a particular
reaction represented by a balanced equation at a given temperature.
• The equilibrium constant for the reverse reaction is equal to the inverse of the equilibrium
constant for the forward reaction.
KC
Because gas pressures are easily measured, equilibrium equations for gas-phase
reactions are often written using partial pressures rather than molar
concentrations. For the gaseous state, the partial pressure is also measure the
activity of the component, KP is -
are the molar concentrations of A, B, C and D respectively.
KP
aA + bB lL + mM
A & B are reactants (behaves ideally)
L & M are products (behaves ideally)
a & b and c & d are number of moles of reactants and products respectively.
Relationship between KP and KC
For n mole of ideal gas - pV = nRT
p = =
But n/V = C, therefore, p = C R T.
Equations for KP and KP are -
aA + bB lL + mM
A & B are reactants (behaves ideally)
L & M are products (behaves ideally)
a & b and c & d are number of moles of reactants and products respectively.
But (l + m) - (a + b) = ∆n, therefore,
K =
p K C (RT)
n
Le Chatelier’s Principle
Equilibrium constant, Kc is independent of initial concentrations.
But, if a system at equilibrium is subjected to a change in the concentration of one or more of
the reacting substances, change in temperature or pressure of the system or catalyst can
affect equilibrium can be explain by Le Chatelier’s principle.
A change in any of the factors that determine the equilibrium conditions of a system will
cause the system to change in such a manner so as to reduce or to counteract the effect of
the change.
Effect of Concentration Change
Effect of Pressure Change
Effect of Inert Gas Addition
Effect of Temperature Change
Effect of a Catalyst
Effect of Concentration Change
When the concentration of any of the reactants or products in a reaction at equilibrium
is changed, the composition of the equilibrium mixture changes so as to minimize the
effect of concentration changes
)
𝐻2 𝑔 + 𝐼2 𝑔 ⇋ 2 𝐻𝐼 (𝑔
𝐹𝑒3+
𝑎𝑞 + 𝑆𝐶𝑁−
𝑎𝑞 ⇋ 𝐹𝑒 𝑆𝐶𝑁 2+
𝑎𝑞
yellow colourless deep red 𝐾𝑐 =
𝐹𝑒 𝑆𝐶𝑁 2+
𝑎𝑞
𝐹𝑒3+ 𝑎𝑞 𝑆𝐶𝑁− 𝑎𝑞
The intensity of the red colour becomes constant on attaining equilibrium
Oxalic acid (H2C2O4), reacts with Fe3+ ions to form the stable complex ion [Fe(C2O4)3]3–, thus
decreasing the concentration of free Fe3+ (aq); the concentration of [Fe(SCN)]2+ decreases, the
intensity of red colour decreases.
Similar effect was found by the addition of aq. HgCl2 because Hg2+ reacts with SCN– ions to form
stable complex ion [Hg(SCN)4]2–
Effect of Pressure Change
Affect in case of a gaseous reaction where the total number of moles of gaseous reactants
and total number of moles of gaseous products are different.
The effect of pressure changes on solids and liquids can be ignored because the volume
(and concentration) of a solution/liquid is nearly independent of pressure
𝑪𝑶 𝒈 + 𝟑 𝑯𝟐 ⇋ 𝑪𝑯𝟒 𝒈 + 𝑯𝟐𝑶 𝒈 4 mol of gaseous reactants (CO + 3H2) become
2 mol of gaseous products (CH4 + H2O)
As, total pressure will be doubled (according to p.V = constant), partial pressure and
therefore, concentration of reactants and products have changed and the mixture is no
longer at equilibrium. The equilibrium now shifts in the forward direction.
Effect of Inert Gas Addition
If the volume is kept constant and an inert gas such as argon is added which does not take
part in the reaction, the equilibrium remains undisturbed because it does not change the
partial pressures or the molar concentrations of the substance involved in the reaction
Effect of Temperature Change
when a change in temperature occurs, the value of equilibrium constant, Kc is changed.
In general, the temperature dependence of the equilibrium constant depends on the sign of
∆H or ∆G for the reaction.
 The equilibrium constant for an exothermic reaction (negative ∆H) decreases as
the temperature increases.
 The equilibrium constant for an endothermic reaction (positive ∆H) increases as
the temperature increases.
)
𝟑 𝑯𝟐 𝒈 + 𝑵𝟐 𝒈 ⇋ 𝟐 𝑵𝑯𝟑 (𝒈 ∆H= – 92.38 kJ mol–1. exothermic process
raising the temperature shifts the equilibrium to left and decreases the equilibrium
concentration of ammonia.
The Reaction isotherm (Van’t Hoff’s Isotherm)
Van’t Hoff’s isotherm gives the net work done that can be obtained from gaseous reactants at
constant temperature, when both reactants and products are at suitable orbitary pressure.
Consider a general reaction-
If aA, aB, aC and aD are activities of reactants A, B and products C, D
respectively.
The free energies of each of these substances per mole at T is-
GA = Go
A + RT lnaA (1)
GB = Go
B + RT lnaB (2)
GC = Go
C + RT lnaC (3)
GD = Go
D + RT lnaD (4)
Where, Go
A, Go
B, Go
C and Go
D are the free energies at unit activity of the A, B,
C and D respectively or are the standard free energies of A, B, C and D
respectively.
aA + bB + ............ cC + dD + ..................
A quantitative relation of the free energy change for the chemical reaction has been developed
which is known as Reaction isotherm or Van’t Hoff’s isotherm.
Therefore the free energy change of the chemical reaction -
ΔG = ΣGp – ΣGR
= [cGC + dGD + …….] – [aGA +bGB + ……..]
= [cGoC + RT lnaC
c + dGoD + RT lnaD
d + …….] –
[aGoA + RT lnaA
a + bGo
B + RT lnaB
b + ……..]
= [(cGoC + dGoD + ……) – (aGoA + bGoB +……)] +
RT[(lnaC
c + lnaD
d + …….) – (lnaA
a + lnaB
b + ……..)]
ΔG = ΔGo
+ RT ln{(aC
c.aD
d…….)/(aA
a .aB
b……..)} (5)
The equation (5) is called as the Van’t Hoff’s isotherm or reaction isotherm. When activities of
all reactants and products are equal to one then ΔG = ΔGo
.
At equilibrium step of the reaction - ΔG = 0
0 = ΔGo + RT ln{(acC x adD…….)/(aaA x abB……..)}
At equilibrium, according to the law of mass action, Ka = (acC adD…….)/(aaAabB……..)
Therefore, 0 = ΔGo + RT ln Ka
Or ΔGo = - RT ln Ka (6)
Ka = e -ΔGo/RT (7)
Form equation (5) and (6)-
ΔG = RT ln{(acC adD…….)/(aaAabB……..)} - RT ln Ka
= RT ln Qa - RT ln Kaa where Qa = {(acC adD…….)/(aaAabB……..)}
= RT ln (Qa/Ka) (8)
For gases activities are proportional to the partial pressures of the components of reaction
mixture hence the activities of the equation (5) can be replaced by the partial pressures,
therefore-
ΔG = ΔGo + RT ln{(PC
c.PD
d…….)/(PA
a.PB
b……..)} (9)
= RT ln Qp - RT ln Kp (10)
Where Kp is the equilibrium constant and
Qp = (Pc
C.Pd
D…….)/(Pa
A.Pb
B……..)
At equilibrium state, ΔG = ΔGo and Qp = 1.
ΔGo = - RT ln Kp = -W or Kp = e -ΔGo/RT (11)
i.e. net work done of the reaction is equal to the decrease in free energy of the system and
it is calculated by using expression -
RT ln Kp or 2.303RT log Kp.
It will be observed that, ΔG is positive when Kp is less than unity.
If ΔG is negative when Kp is greater than unity and is zero when Kp = 0 or 1.
*** Van’t Hoff Isochore ***
The Van’t Hoff isochore is obtained by comparing the Van’t Hoff isotherm with Gibbs -
Helmeholtz equation.
The Gibbs-Helmeholtz equation is ΔG = ΔH + T[d(ΔG)/dT]p
Or -ΔH = T[d(ΔG)/dT]p - ΔG
Dividing both side by T2 gives:- -ΔH/T2 = {T[d(ΔG)/dT]p - ΔG}/T2
The right hand side of the above equation is obtained by differentiating ΔG/T w.r.t. T at constant P-
[d(ΔG/T)/dT]p = {T[d(ΔG)/dT]p - ΔG}/ T2
-ΔH/T2 = [d(ΔG/T)/dT]p (12)
• According to Van’t Hoff isotherm- ΔG = -RT ln Kp (11)
• Dividing both side by T and differentiating it w.r.t. T at constant P-
• [d(ΔG/T)/dT]p = -R d(lnKp)/dT (13)
• Comparing equation (13) and (12), we have-
• ΔH/T2 = R d(lnKp)/dT
• Or (ΔH/R).dT/T2 = d(lnKp) (14)
• The above equation is known as Van’t Hoff isochore. If ΔH remains same over the range of
temperature, we have on integration -
• lnKp = (ΔH/R)∫dT/T2 ∫dx/xn = 1/(-n+1).1/xn-1
• = -ΔH/RT + constant of integration.
• Appling the limits T1 and T2 at the equilibrium constants Kp1 and Kp2 respectively, we have-
• lnKp2 - lnKp1 = -(ΔH/R)[1/T2 - 1/T1]
• ln[Kp2/Kp1] = (ΔH/R)[1/T1 - 1/T2]
• log[Kp2/Kp1] = (ΔH/2.303R)[(T2 – T1)/T1T2] (15)
• By using this equation, equilibrium constant at any temperature or heat of reaction is calculated.
Vapour pressure of Liquid
Clapeyron equation
For any pure substance in a single phase such as liquid or gaseous, are in contact with
each other then any variation in free energy is given by the equation.
dG = VdP – SdT (1)
If these two phases are in equilibrium to each other, then dG = 0 at constant pressure and
temperature i.e. dP = dT = 0 i.e. these phases are in equilibrium with other when the
pressure and temperature are constant and uniform throughout the phase.
“A useful thermodynamic equation is applicable to a system consisting of two phases of
same substance in equilibrium is called as Clapeyron equation.”
Consider the transition of a pure substance from one phase into another phase e.g.-
1. Solid convert to liquid- at melting point of solid- melting or fusion.
2. Liquid convert to vapour- at boiling point of liquid- boiling or vaporization.
3. Solid convert to vapour- at sublimation point of solid- sublimation.
4. Allotropic transitions at the transition temperature of the two allotropic forms.
e.g. S(rhombic) is in equilibrium with S(monoclinic).
All above changes are represented by the reaction-
For which ΔG is given as- ΔG = G2 –G1 (2)
Where, G2 and G1 are the molar free energies of the substances in final and initial states
respectively. At equilibrium state ΔG = 0, at constant P and T i.e. G1 = G2. All such
transformations will be in equilibrium at constant pressure and at constant temperature,
when molar free energies of the substances are identical in both phases.
A1
A2
Clapeyron equation
ΔG = G2 –G1 = 0.
i.e. G1 = G2
Or dG1 = dG2 (3)
The variation of free energies for pure substances with respect to P and T in single phase is given by the equation-
dG1 = V1dP – S1dT (4)
dG2 = V2dP – S2dT (5)
Where, order to maintain equilibrium, the temperature is changed by the amount dT and that of pressure is dP.
From equation (3), (4) and (5)-
V2dP – S2dT = V1dP – S1dT
(V2 – V1) dP = (S2 – S1) dT
dP/dT = (S2 – S1)/(V2 – V1) = ΔS/ΔV (6)
Where, ΔS is change in entropy and ΔV is change in volume during transformation.
But- ΔG = ΔH – TΔS,
But at equilibrium, ΔG = 0.
Or ΔH = TΔS or ΔS = ΔH/T (7)
From equation (6) and (7)-
dP/dT = ΔH/T.ΔV (8)
Where, ΔH is the change in enthalpy for reversible transformation occurring at temperature T. this equation is called as
Clapeyron equation.
The Clapeyron equation gives relation between the temperature and pressure of the system containing two phases of a
pure substance which are in equilibrium with each others. This equation also shows that dP/dT is directly proportional to
the enthalpy of the transition and inversely proportional to temperature and the volume change accompanying during
the transition i.e. ΔH and ΔV are the functions of temperature and or pressure. The equation (8) can be modified as-
(P2 – P1)/(T2 – T1) = ΔH/T.ΔV (9)
Where T is taken as average of T2 and T1 or equation (8) can be integrated with the proper limits of T and P by assuming
that ΔH and ΔV are constant.
P2 T2
∫dP = ΔH/ΔV ∫ dT/T
P1 T1
(P2 – P1) = (ΔH/ΔV) ln T2/T1.
(P2 – P1) = (2.303ΔH/ΔV) ln T2/T1. (10)
Vapour pressure of Liquid
The variation of vapour pressure with temperature can be expressed thermodynamically
by means of Clausius-Clapeyron equation. Let us consider a system-
For the transition of liquid to vapour, P is the vapour pressure at given temperature T; ΔHv
is the heat of vaporization of liquid. The volume of given liquid is Vl and that of vapour is
Vg. Therefore, the Clepeyron equation can be written as-
dP/dT = [ΔHv /T(Vg – Vl)] (1)
But molar volume of the substance in vapour phase is very large as compaired to
volume of same liquid in liquid state.
(Vg – Vl) ≈ Vg. (18 ml water liquid = 22400 ml water vapour)
Therefore, the Clepeyron equation becomes-
dP/dT = ΔHv/TVg. (2)
If we assume that the vapour behaves ideally then Vg per mole of substance can be
calculated as- Vg = RT/P (3)
Substituting (3) in equation (2) as-
dP/dT = PΔHv/RT2. (4)
dP/P = (ΔHv/R). dT/T2 (5)
d(lnP) = (ΔHv/R). dT/T2 (6)
The equation (6) is known as the Clausius-Clapeyron equation. The heat of vaporization
is a function of temperature. Assume that ΔHv will be independent on the temperature
(constant over the range of temperature and pressure).
Liquid vapour
Vapour pressure of Liquid
Then equation (6) can be integrated-
(a) Integrated without limits of temperature and pressure:-
∫d(lnP) = (ΔHv/R) ∫dT/T2 + C (C is constant of integration)
(lnP) = (ΔHv/R) . (-1/T) + C
(logP) = (-ΔHv/2.303RT) + C’ (7)
The equation (7) shows that logarithm of the vapour pressure should be a function of the
reciprocal of the absolute temperature T. It is also a equation of straight line of the type, y = mx+c.
Plot a graph of logP against 1/T is a straight line having slope = m = (-ΔHv/2.303R) and Y-intercept c =
C’. From the slope of the graph, heat of vaporization of the various liquids can be calculated as-
m = (-ΔHv/2.303R) (8)
ΔHv = (-2.303R . m) = -4.576 . m cal/mol.
(b) Integration with limits:-
Integrate equation (6) with the limits of pressure P1 and P2 corresponding to the temperature T1 and
T2, then-
P2 T2
∫d(lnP) = (ΔHv/R) ∫dT/T2
P1 T1
P2 T2
[lnP] = (ΔHv/R) [-1/T]
P1 T1
ln(P2/P1) = (ΔHv/R) [(T2 – T1)/T1T2] (9)
log(P2/P1) = (ΔHv/2.303R) [(T2 – T1)/T1T2] (10)
The equation (10) can be used to calculated ΔHv from the values of the vapour pressure at any two
temperatures or when ΔHv is known, P at some desired temperature can be calculated from a single
available vapour pressure at a given temperature.
Thank you
First and second law thermodynamics (sy p 8)

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First and second law thermodynamics (sy p 8)

  • 2. Paper 8: Physical Chemistry 1. Thermodynamics - I 15 hrs. Definition: of Thermodynamic Terms: System, Surrounding types of system, intensive and extensive properties. Thermodynamic Process, Concept of heat and work. Work done in reversible and irreversible process, concept of maximum work (Wmax), Numerical Problems. First law of Thermodynamics: Statement, Definition of Internal energy and Enthalpy. Heat capacity, heat capacities at constant volume pressure and their relationship. Calculation of W, q, dU and dH for the expansion of ideal gases under isothermal and adiabatic conditions for reversible process, Numerical problems, Hess’s law of heat Summation and its application. 2. Thermodynamic - II 20 hrs. Second Law of Thermodynamics: Need for the law, different statement of the law Carnot Cycle and its efficiency, Numerical Problems. Carnot Theorem. Concept of Entropy: Definition, Physical significance, Entropy as a State Function, Entropy change in Physical change, Entropy as criteria of Spontaneity & Equilibrium Entropy Change in Ideal Gases. Gibbs and Helmholtz Functions: Gibbs Function (G) and Helmoltz Function (A) as Thermodynamic Quantities. A and G as criteria for Thermodynamic Equilibrium and Spontaneity, their Advantage over Entropy change. Variation of A with P, V and T. 3. Chemical Equilibrium 10 hrs. Equilibrium Constant and Free Energy. Thermodynamic Derivation of Law of Mass Action. Le Chatelier’s Principle. Reaction Isotherm and Reaction Isochore. Clapeyron Equation, Clausius-Clapeyron Equation and its Application.
  • 3. Thermodynamics Thermodynamics is the study of energy. OR Thermodynamics is the study of the movement of heat from one body to another and the relations between heat and other forms of energy. OR Thermodynamics is the study of the connection between heat and work and the conversion of one into the other. OR Chemical thermodynamics is the study of energy relationships in chemistry.
  • 4. Why is the study of Thermodynamics important? The study of thermodynamics is important because- many machines and modern devices change heat into work, such as an automobile engine or turn work into heat or cooling, such as with a refrigerator. How the conversion of energy from one from to another, change in state, change in matter etc occurs.
  • 5. Different forms of energies  Kinetic Energy : Having motion  Potential Energy : Have ‘potential’  Chemical Energy : Free energy of chemical sub.  Thermal Energy : Function of temperature  Nuclear Energy : Unstable nuclei (Binding energy)  Molecular Energy: Translational, rotational, vibrational and electronic energy.
  • 6. • Kinetic Energy: – Consider a baseball flying through the air. The ball is said to have "kinetic energy" by virtue of the fact that its in motion relative to the ground. • Potential Energy: – Consider a book sitting on a table. The book is said to have "potential energy" because if it is nudged off, gravity will accelerate the book, giving the book kinetic energy. Therefore, it has “potential.” • Thermal or Heat Energy: – Consider a hot cup of coffee. The coffee is said to possess "thermal energy", or "heat energy," which is really the collective, microscopic, kinetic, and potential energy of the molecules in the coffee. This energy is function of temperature. Chemical Energy: Consider the ability of your body to do work. The glucose (blood sugar) in your body is said to have "chemical energy" because the glucose releases energy when chemically reacted (combusted) with oxygen.
  • 7. Electrical Energy All matter is made up of atoms, and atoms are made up of smaller particles, called protons, neutrons, and electrons. Electrons orbit around the center, or nucleus, of atoms, just like the moon orbits the earth. The nucleus is made up of neutrons and protons. Material, like metals, have certain electrons that are only loosely attached to their atoms. They can easily be made to move from one atom to another if an electric field is applied to them. When those electrons move among the atoms of matter, a current of electricity is created. Electrochemical Energy: Consider the energy stored in a battery. Like the example above involving blood sugar, the battery also stores energy in a chemical way. But electricity is also involved, so we say that the battery stores energy "electro-chemically". Another electron chemical device is a "fuel-cell". Sound Energy: Sound waves are compression waves associated with the potential and kinetic energy of air molecules. When an object moves quickly, for example the head of drum, it compresses the air nearby, giving that air potential energy. That air then expands, transforming the potential energy into kinetic energy (moving air). The moving air then pushes on and compresses other air, and so on down the chain. Electromagnetic Energy (light): E = hν Nuclear Energy: E = 931.1 Δm
  • 8. 1) The most obvious and trivial way in which energy is transported is when an object that possesses energy simply moves from one place to another. For example, a baseball flying through the air is a simple form of energy transport. 2) Kinetic energy can also be transferred from one object to another when objects collide. This is also pretty trivial, except that we also know that the total energy, including any heat or other forms of energy generated during the collision, is conserved in this process, regardless of the relative sizes, shapes, and materials of the objects. 3) There are three important ways that heat energy can be transported or transferred, called conduction, convection, and radiation. The first two refer to transfer of the thermal energy, whereas the last is really a conversion of energy to a different form, (photons of light) and the subsequent travel (transport) of those photons. 11/26/2021 8 How is energy transported from place to place and transferred between objects?
  • 9. 4) The "diffusion" of thermal energy (heat) through a substance, which occurs because hotter molecules (those that are vibrating, rotating, or traveling faster), interact with colder molecules, and in the process transfer some of their energy. Metals are excellent conductors of heat energy, whereas things like wood or plastics are not good conductors of heat. Those that are not so good conductors are called insulators. 5) The transfer of heat energy by the movement of a substance, such as a heated gas or liquid from one place to another. For example, hot air rising to the ceiling is an example of convection (in this case called a convection current). 6) In the context of heat transfer, however, the term “emits radiation" refers just to light (electro-magnetic waves), and in particular, to the surprising fact that all objects, even those that are in equilibrium (at equal temperature) with their surroundings, continuously emit, or radiate electromagnetic waves (that is, light waves) into their surroundings. The source of this radiation is the thermal energy of the materials, that is, the movement of the object's molecules. 11/26/2021 9
  • 11. SOME TERM USED IN THERMODYNAMICS “The part of universe that under study in thermodynamics is called as system, it is real or imaginary.” “The Remaining part of the universe which is in contact or interacts with the system is called as surrounding.” For the thermodynamic study the interaction between system and surrounding is very important. According to the interaction (exchange of matter and energy), systems are classified in as: 1) Open system:- A system which can exchange both matter and energy with it’s surrounding is called as open system. e.g.- Heating of water in open beaker. 2) Closed system:- A system which exchange only energy but not matter with it’s surrounding is called as closed system. e.g.- Heating of water in closed/sealed container. 3) Isolated system:- A system which neither exchange energy nor the matter with it’s surrounding is called as isolated system. e.g.- Store of hot water in thermos flask. The systems were further classified according to the phases as- homogeneous and heterogeneous system. Phase of the system is defined as - a homogeneous, physically distinct and mechanically separable portion of a system.
  • 12. • According to the nature of the changes occurs in the system, it may be reversible and irreversible, system can be classified as reversible and irreversible system. • “Chemical changes in which driving and opposing forces are differ by large amount in which driving force large than opposing force, if opposing force increased by small amount, reaction does not reverse is called as irreversible process.” • “A reaction proceeds in only one direction is called as irreversible reaction while the reaction proceeds in both directions is called as reversible reaction”. “A chemical reaction in which driving and opposing forces are differ by infinitely small amount and proceeds till attends equilibrium state, if opposing force is increased by small amount, reaction become reverse is called as reversible reaction. Substitution reaction
  • 13. The properties of the system are classified into two groups as- a) Extensive properties:- An extensive properties of the system is any property whose magnitude is depends on the amount of present in the system. e.g.- pressure, volume, entropy, free energy, etc. b) Intensive properties:- Intensive properties of a system is any property whose amount is independent on the total amount but depends on the concentration of the substance or substances in a system. e.g.- density, refractive index, chemical potential, viscosity, surface tension, etc. Properties of system System Properties depends on amount Properties independs on amount Extensive properties Intensive properties
  • 15. • According to the process occurring in the system or according to the fundamental parameter change in a system, it can be classified as- • 1) Isothermal process:- A process occurring at constant temperature is called as isothermal process and a system is called as isothermal system. Δ T = 0 • 2) Isobaric process:- A process occurring at constant pressure is called as isothermal process and a system is called as isobaric system. Δ P = 0 • 3) Isochronic process:- A process occurring at constant volume is called as isothermal process and a system is called as isochronic system. Δ V = 0 • 4) Cyclic process:- A process in which a change in any state function such as internal energy is equal to zero or a process in system having initial and final state are same is called as cyclic process and system is called as cyclic system. Change in state function = 0 • 5) Adiabatic process:- A process in which system does not exchange heat with it’s surrounding is called as adiabatic process and a system is adiabatic system. q = 0. • All fundamental parameters or properties (P, V, T, H, E, G, A, density, surface tension, K, etc) are classified into two types as- • a) State function:- A parameter or properties of the system is depends on only initial and final state of the system but not on the path through which change occurs is called as state function. • e.g.- free energy, entropy, pressure, volume, etc. • b) Path function:- A parameter or properties of the system is depends on the path through which change occurs but not on the initial and final state of the system is called as path function.
  • 16. The First law of Thermodynamics -energy cannot be created or destroyed only converted from one form to another. -Energy can be changed from one form to another, but it cannot be created or destroyed. -The total amount of energy and matter in the Universe remains constant, merely changing from one form to another. -The First Law of Thermodynamics (Conservation) states that energy is always conserved, it cannot be created or destroyed.
  • 17. The First law of Thermodynamics The supplied heat is converted to work and change in internal energy, work and internal energy are forms of energy transfer and hence energy is conserved. change in total internal energy heat added to system State Function Process Functions E = q + Won
  • 18. system done work : positive system added heat : positive by to W q W q E    The First law of Thermodynamics • statement of energy conservation for a thermodynamic system • internal energy E is a state variable • W, q process dependent
  • 19. Sign conventions: (a) Heat:  q is positive - heat is flow from surrounding to the system i.e. system absorb heat from surrounding.  q is negative - heat is flow from system to the surrounding i.e. system lost heat to its surrounding. (b) Work done:  System does work on its surrounding, W is positive.  Surrounding does work on its System, W is negative. (c) Internal energy:  Both heat (q) and Work done (W) is positive, internal energy of the system increase i.e. ΔE increases.  Internal energy of the system decreases, ΔE is negative.
  • 20. . 11/26/2021 20 The First Law of Thermodynamics by dW dq dE   int What this means: The internal energy (dE) of a system tends to increase if energy is added via heat (dq) and decrease via work (dW) done by the system. on dW dq dE   int . . . and increase via work (W) done on the system. on by dW dW  
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  • 27. Work done in reversible and irreversible process: Consider a cylinder having cross-sectional area ‘A’ fitted with weightless and frictionless piston. Let pressure of gas on the piston is ‘p’ (p = f/A) and the force acting on the piston is ‘f’. For reversible isothermal expansion, piston moves in upward direction through a distance dl so that the volume change is dV and the pressure (p – dp). dw = (p – dp) dV = pdV – dp.dV 1 But for reversible expansion, change in pressure and volume is very small, so that dp.dV ≈ 0 (neglected) dw = p.dV 2 For one mole of ideal gas - pV = RT or p = RT/V 3 From equation (2) and (3), dw = RT. (dV/V) Integrate this equation with the limits V1 to V2 at constant temperature, we have: 𝑑𝑤 = 𝑅𝑇 𝑉1 𝑉2 𝑑𝑉 𝑉 𝑊 = 𝑅𝑇 ln 𝑉2 − ln 𝑉1 𝑊 = 2.303 𝑅𝑇 log 𝑉2 𝑉1 For n mole of ideal gas: 𝑊 = 2.303 𝑛𝑅𝑇 log 𝑉2 𝑉1 At constant temperature, according to the Boyle’s law: p.V = Constant. 𝑃1𝑉1 = 𝑃2𝑉2 or 𝑉2 𝑉1 = 𝑃1 𝑃2 Where, P1, V1 is the initial pressure and volume and P2, V2 is the final pressure and volume of the system respectively. 𝑊 = 2.303 𝑛𝑅𝑇 log 𝑃1 𝑃2
  • 28. The workdone is maximum, when opposing pressure ‘P’ differs only infinitesimally small (dp) in magnitude from the internal pressure of the gas itself which is the condition for the reversibility of the reaction. 𝑊 𝑚𝑎𝑥 = 2.303 𝑅𝑇 log 𝑉2 𝑉1 = 2.303 𝑛𝑅𝑇 log 𝑃1 𝑃2 In reversible expansion In reversible compression 𝑊 𝑚𝑎𝑥 = 2.303 𝑅𝑇 log 𝑉1 𝑉2 = 2.303 𝑛𝑅𝑇 log 𝑃2 𝑃1 For irreversible process: The gas was expand irreversibly from volume V1 toV2 at constant pressure P and temperature T, then from equation (1)- dw = p . dV On integration between the limits V1 and V2, 𝑊 = 𝑝 𝑉1 𝑉2 𝑑𝑉 𝑉 = PΔV
  • 29. Enthalpy The thermal changes at constant pressure can be expressed by using new thermodynamic function called as enthalpy (H). “It is sum of internal energy (E) of the system and pressure-volume product (PV).” Since E and PV are the state functions of the system, H is also state function. Therefore change in enthalpy can be determined as: ΔH = H2 - H1 H2 and H1 are enthalpies of the final and initial state of the system: H1 = E1 + P1V1 H2 = E2 + P2V2 and ΔH = H2 - H1 = (E2 + P2V2) – (E1 + P1V1) = (E2- E1) + (P2V2- P1V1) = ΔE + ΔPV At constant pressure, ΔH = ΔE + PΔV For an ideal gas: PV = nRT At constant P and T, PΔV = Δn RT (Δn is change in number of moles during the reaction) ΔH = ΔE + Δn RT According to the first law of thermodynamics: q = W + ΔE And workdone by irreversible process at constant P is: W = PΔV, therefore – q = ΔE + PΔV Therefore, change in enthalpy - ΔH = q
  • 30. Heat capacity A small amount of heat is added to a system, causes rise in temperature. Suppose a small amount of heat (dq) required to rising the temperature of the system by one degree is called as heat capacity. Heat capacity = 𝑑𝑞 dT Heat capacity does not explain amount of the system therefore new term is used containing amount of matter is called as molar heat capacity. “The amount of heat (dq) required to rise the temperature of the one mole of the gas through one degree is called as molar heat capacity.” It is denoted by latter C. C = 𝑑𝑞 dT But according to the first law of thermodynamics: dq = dE + PdV C = (dE + PdV) dT The molar heat capacity at constant volume is called as (Cv). dV = 0 𝐶𝑣 = 𝑑𝐸 dT 𝑉 At constant volume, the amount of heat (dq) required to rise the temperature of the one mole of the gas through one degree is called as molar heat capacity at constant volume (Cv). At constant pressure, the heat absorbs ΔH = q. the molar heat capacity at constant pressure is called as Cp. 𝐶𝑝 = 𝑑𝑞 dT = 𝑑𝐻 dT 𝑃 At constant pressure, the amount of heat (dq) required to rise the temperature of the one mole of the gas through one degree is called as molar heat capacity at constant pressure (Cp).
  • 31. Difference between two heat capacities is - Cp - Cv = 𝑑𝐻 dT 𝑃 - 𝑑𝐸 dT 𝑉 But according to the definition of enthalpy, H = E + PV. Differentiate it with respect to temperature at constant pressure, then- 𝑑𝐻 dT 𝑃 = 𝑑𝐸 dT 𝑃 + P 𝑑𝑉 dT 𝑃 But the internal energy E is the function of any two of the three variables; P, V and T. If V and T are two independent variables, then; E = f(V,T) dE = 𝑑𝐸 dT 𝑉 dT + 𝑑𝐸 dV 𝑇 dV Dividing both side by dT and apply the condition of constant pressure as: 𝑑𝐸 dT 𝑃 = 𝑑𝐸 dT 𝑉 + 𝑑𝐸 dV 𝑇 𝑑𝑉 dT 𝑃 Cp - Cv = 𝑑𝐸 dT 𝑉 + 𝑑𝐸 dV 𝑇 𝑑𝑉 dT 𝑃 + P 𝑑𝑉 dT 𝑃 - 𝑑𝐸 dT 𝑉 = 𝑑𝐸 dV 𝑇 𝑑𝑉 dT 𝑃 + P 𝑑𝑉 dT 𝑃 For an ideal gas PV = nRT, differentiate it with respect to T at constant V gives; 𝑑𝑃 dT 𝑉 = 𝑛𝑅 𝑉 But, 𝑑𝐸 dV 𝑇 = 𝑇 𝑑𝑃 dT 𝑉 − 𝑃 𝑑𝐸 dV 𝑇 = 𝑇 𝑛𝑅 𝑉 – P = P – P = 0. The internal energy of the ideal gas is independs on the volume
  • 32. The internal energy of the ideal gas is depends on the volume, but it depends on temperature. If we differentiate ideal gas equation with respect to temperature at constant pressure, then- 𝑑𝑉 dT 𝑃 = 𝑛𝑅 𝑃 But, 𝑑𝐻 dP 𝑇 = V – T 𝑑𝑉 dT 𝑃 𝑑𝐻 dP 𝑇 = V – T 𝑛𝑅 𝑃 = V – V = 0 Enthalpy of an ideal gas is independent on the pressure Cp - Cv = 𝑑𝐸 dT 𝑃 + P 𝑑𝑉 dT 𝑃 - 𝑑𝐸 dT 𝑉 = P 𝑑𝑉 dT 𝑃 From the above explanation, a conclusion is drawn that Cp and Cv are functions of T only and is independent on volume and pressure. But for ideal gas; PV = RT (for one mole) Therefore, P 𝑑𝑉 dT 𝑃 = R. Cp - Cv = R For n moles of ideal gas; Cp - Cv = nR
  • 33. Alternative derivation: According to definition of enthalpy - H = E + PV 1 For one mole of ideal gas - PV = RT 2 From (2), equation (1) become - H = E + RT 3 Differentiate (3) completely - dH = dE + R dT 4 From the definition of Cp & Cv Cp dT = Cv dT + R dT Devided both side by dT - Cp - Cv = R For n moles of ideal gas; Cp - Cv = nR
  • 34. Adiabatic process in ideal gases In an adiabatic process, there is no exchange of heat between a system and it’s surrounding i.e. q = 0. According to the first law of thermodynamic - ΔE = -W Work is done by the system decreases internal energy and therefore temperature of the system. For infinitesimally small increase in volume dV at constant pressure P, the workdone is PdV. This workdone is due to expense of internal energy of the gas. The decrease in internal energy is dE. dE = -P dV But according to the definition of Cv: dE = Cv dT for one mole of gas. Cv dT = -PdV According to the ideal gas equation- PV = RT or P = RT/V 𝐶𝑣 𝑑𝑇 = − 𝑅𝑇 𝑉 𝑑𝑉 𝑑𝑉 𝑉 = − 𝐶𝑣 𝑅 𝑑𝑇 𝑇 Consider Cv is constant and integrate above equation with the limits V1 at T1 and V2 at T2, we have: 𝑉1 𝑉2 𝑑𝑉 𝑉 = − 𝑪𝒗 𝑹 𝑇1 𝑇2 𝑑𝑇 𝑇 ln 𝑉2 𝑉1 = − 𝐶𝑣 𝑅 ln 𝑇2 𝑇1 ln 𝑉2 𝑉1 = ln 𝑇2 𝑇1 − 𝐶𝑣 𝑅 ln 𝑉2 𝑉1 = ln 𝑇1 𝑇2 𝐶𝑣 𝑅
  • 35. ln 𝑉2 𝑉1 = ln 𝑇1 𝑇2 𝐶𝑣 𝑅 Taking antilog of both sides, we have: 𝑉2 𝑉1 = 𝑇1 𝑇2 𝐶𝑣 𝑅 or 𝑉1𝑇1 𝐶𝑣 𝑅 = 𝑉2𝑇2 𝐶𝑣 𝑅 Where, Cv is constant. According to the Boyle’s law: PV = constant or V1P1 = V2P2 or 𝑉2 𝑉1 = 𝑃1 𝑃2 𝑃1 𝑃2 = 𝑇1 𝑇2 𝐶𝑣 𝑅 𝑃1 𝑇1 𝐶𝑣 𝑅 = 𝑃2 𝑇2 𝐶𝑣 𝑅 According to the ideal gas equation- PV = RT (for one mole of ideal gas). T = 𝑃 𝑉 𝑅 𝑉1 𝑃1𝑉1 𝑅 𝐶𝑣 𝑅 = 𝑉2 𝑃2𝑉2 𝑅 𝐶𝑣 𝑅 𝑉1 𝐶𝑣+𝑅 𝑅 𝑃1 𝐶𝑣 𝑅 = 𝑉2 𝐶𝑣+𝑅 𝑅 𝑃2 𝐶𝑣 𝑅 𝑉1 𝐶𝑃 𝑅 𝑃1 𝐶𝑣 𝑅 = 𝑉2 𝐶𝑃 𝑅 𝑃2 𝐶𝑣 𝑅 because Cp - Cv = R Taking Rth power of both side, 𝑉1 𝐶𝑃 𝑃1 𝐶𝑣 = 𝑉2 𝐶𝑃 𝑃2 𝐶𝑣 = constant. Taking Cv th root of both side, 𝑉1 𝐶𝑃 𝐶𝑣 𝑃1 = 𝑉2 𝐶𝑃 𝐶𝑣 𝑃2 = constant. Put (Cp/Cv) = γ i.e. ratio of molar heat capacities at constant pressure and constant volume respectively. 𝑃1𝑉 1 𝛾 = 𝑃2𝑉 2 𝛾
  • 36. The workdone performed in the adiabatic reversible expansion of ideal gas as: Differentiate PVγ = constant completely. 𝛾 𝑃 𝑉𝛾−1𝑑𝑉 + 𝑉𝛾𝑑𝑃 = 0 Dividing both side by Vγ-1: γ PdV + V.dP = 0 V.dP = -γ PdV Differentiate the equation PV = nRT completely, we have- PdV + VdP = nRdT (1-γ) PdV = nRdT or 𝑃 𝑑𝑉 = 𝑛 𝑅 𝑑𝑇 1 − 𝛾 The work done: dW = PdV = 𝑛 𝑅 𝑑𝑇 1−𝛾 Integrate equation with the limits of temperature T1 and T2 as: T2 Wm = ∫ nRdT/(1-γ) = [nR/(1-γ)](T2- T1) T1 If T2 > T1, γ > 1, then, Wm is negative, i.e. work is done on the gas/system. If T2 < T1, γ > 1, then Wm is positive, i.e. work is done by the gas.
  • 37. Hess’s law of heat Summation and its application If a reaction takes place in several steps then its standard reaction enthalpy is the sum of the standard enthalpies of the intermediate reactions into which the overall reaction may be divided at the same temperature. enthalpy is a state function Enthalpy of an overall reaction A → B along one route is ∆rH ∆rH = ∆rH1 + ∆rH2 + ∆rH3 + .....
  • 38. Example: Given the thermochemical equations: 2 WO2 (s) + O2 (g) → 2 WO3 (s) H = -506 kJ 2 W(s) + 3 O2 (g) → 2 WO3 (s) H = -1686 kJ Calculate the enthalpy change for: 2 W(s) + 2 O2 (g) → 2WO2 (s).
  • 39. Hess’s law of heat Summation and its application Standard enthalpy of combustion (symbol: ∆cH*) Combustion reactions are exothermic in nature Important in industry, rocketry, and other walks of life “the enthalpy change per mole (or per unit amount) of a substance, when it undergoes combustion and all the reactants and products being in their standard states at the specified temperature”. Cooking gas in cylinders contains mostly butane (C4H10) which completely burns release 2658 kJ of heat. C4H10 (g) + 13/2 O2 (g) → 4CO2 (g) + 5H2O(1); ∆cH* = - 2658 kJ mol-1. Combustion of glucose gives out 2802.0 kJ/mol of heat C6H12O6 (g) + 6O2 (g) → 6CO2 (g) + 6H2O(1); ∆cH* = - 2802.0 kJ mol-1
  • 40. Enthalpies of Formation An enthalpy of formation, ΔHf, is defined as the enthalpy change for the reaction in which a compound is made from its constituent elements in their elemental forms. Standard Enthalpy of Formation The standard enthalpy of formation is the enthalpy change when one mole of a substance in its standard state is formed from the most stable form of the elements in their standard states. Enthalpy of atomization (symbol: ∆αH*) It is the enthalpy change on breaking one mole of bonds completely to obtain atoms in the gas phase. In case of diatomic molecules, like dihydrogen, the enthalpy of atomization is also the bond dissociation enthalpy. H2 (g) → 2H (g); ∆αH* = 435.0 kJ mol–1
  • 41. Bond Enthalpy (symbol: ∆bondH*) Energy is required to break a bond and energy is released when a bond is formed. The enthalpy changes associated with chemical bonds, two different terms are used in thermodynamics. (i) Bond dissociation enthalpy The bond dissociation enthalpy is the change in enthalpy when one mole of covalent bonds of a gaseous covalent compound is broken to form products in the gas phase. (ii) Mean bond enthalpy Mean of energy required to break the similar bonds. The net enthalpy change of a reaction is the amount of energy required to break all the bonds in the reactant molecules minus the amount of energy required to break all the bonds in the product molecules. CH4 (g) → CH3 (g) + H (g); ∆C-HH* = 427 kJ mol-1 CH3 (g) → CH2 (g) + H (g); ∆C-HH* = 439 kJ mol-1 CH2 (g) → CH (g) + H (g); ∆C-HH* = 452 kJ mol-1 CH (g) → C (g) + H (g); ∆C-HH* = 347 kJ mol-1 Therefore, CH4 (g) → C (g) + 4H (g); ∆αH* = 1665 kJ mol-1 ∆C–HH* = ¼ (∆αH*) = ¼ (1665 kJ mol-1) = 416 kJ mol-1 Mean bond enthalpy
  • 42. Enthalpy of Solution (symbol: ∆solH*) The enthalpy of solution at infinite dilution is the enthalpy change observed on dissolving the substance in an infinite amount of solvent when the interactions between the ions (or solute molecules) are negligible. The enthalpy of solution of AB(s), ∆solH*, in water ∆solH* = ∆latticeH* + ∆hydH* Lattice Enthalpy The lattice enthalpy of an ionic compound is the enthalpy change which occurs when one mole of an ionic compound dissociates into its ions in gaseous state. It is impossible to determine lattice energies directly by experiment hence we use an indirect approach to determine it by constructing energy diagram called a Born-Haber cycle.
  • 43. Na+ (g) + Cl (g) Na+ (g) + Cl- (g) NaCl(s) Na (g) + Cl (g) Na (g) + 1/2Cl2 (g) Na (s) + 1/2Cl2 (g) - 348.8 kJ -787.0 kJ - 4 1 1 . 3 k J +107.8 kJ +121.3 kJ +495.4 kJ Lattice energy Ionization energy of Na (g) D i r e c t p a t h Electron affinity of Cl(g) energy needed to form gaseous Cl atoms energy needed to form gaseous Na atoms Heat of formation of NaCl
  • 44. A metal pellet with mass 100.0 g, originally at 88.4 0C, is dropped into 125 g of water originally at 25.1 0C. The final temperature of both the pellet and the water is 31.3 0C. Calculate the heat capacity C (in J/0C) and specific heat capacity Cs (in J/g0C) of the pellet. The specific heat of water is 4.184 J/g.0C. Two litres of an ideal gas at a pressure of 10 atm expands isothermally into a vacuum until its total volume is 10 litres. How much heat is absorbed and how much work is done in the expansion? If water vapour is assumed to be a perfect gas, molar enthalpy change for vapourisation of 1 mol of water at 1bar and 100°C is 41 kJ mol–1. Calculate the internal energy change, when (i) 1 mol of water is vaporised at 1 bar pressure and 100°C. (ii) 1 mol of water is converted into ice.
  • 45. 1g of graphite is burnt in a bomb calorimeter in excess of oxygen at 298 K and 1 atmospheric pressure according to the equation: C (graphite) + O2 (g) → CO2 (g) During the reaction, temperature rises from 298 K to 299 K. If the heat capacity of the bomb calorimeter is 20.7kJ/K, what is the enthalpy change for the above reaction at 298 K and 1 atm?
  • 46. Carnot Cycle Carnot heat engine is an ideal engine operating under ideal conditions. Convert a fraction of absorbed heat into work. Operating between the temperatures T1 (source) and T2 (sink) (T1 > T2). Absorb a quantity of heat at some temperature T1, Convert a part of heat into work, and remaining part of heat is rejected at temperature T2. Carnot by considering a sequence of operations called as Carnot cycle (series of operations so conducted and at the end; the system is back to its original state). cylinder fitted with weightless and frictionless piston, It containing a given substance (gas) is called as working substance. insolating jacket used to perform adiabatic process a source having temperature T1 and supplied heat to the working substance sink having temperature T2 and act as heat acceptor
  • 47.
  • 48.
  • 49. A Carnot cycle operates in four different steps- two are isothermal and two are adiabatic process A(P4 ,V4 ) B(P1 ,V1 ) C(P2 ,V2 ) D(P3 ,V3 ) q1 q2 T1 T2 Wm P V Isothermal reversible expansion Isothermal reversible compression I. I II III IV III. II. Adiabatic reversible expansion IV. Adiabatic reversible Compression 1. The pressure, volume and temperature of the working substance is P4, V4, and T1 at point A. A substance absorbs heat from source q1 at the temperature T1 and undergoes isothermal and reversible expansion to a point B having pressure P1 and volume V1. The change in internal energy is ΔE1 and work done is W1. According to the first law of thermodynamics: ΔE1 = q1 - W1. 1 And 𝑾𝟏 = 𝒏𝑹𝑻𝟏 𝒍𝒏 𝑽𝟏 𝑽𝟒
  • 50. 2) Expand gas adiabatically and reversibly from point B (P1, V1, T1) to a point C (P2, V2, T2). For the adiabatic process, q = 0, the work done by the system is W2 at the expense of an internal energy ΔE2. During the expansion temperature must be drop from T1 to T2. 4) Expand gas adiabatically and reversibly from point D (P3, V3, T2) to a point A (P4, V4, T1). For the adiabatic process, q = 0, the work done by the system is W4 at the expense of an internal energy ΔE4. During the expansion temperature must be drop from T2 to T1. According to the first law of thermodynamics: ΔE2 = - W2. 2 And, 𝑊2 = 𝑛 𝑇1 𝑇2 𝐶𝑣 𝑑𝑇 𝑇 ΔE4 = - W4. 4 And 𝑊4 = 𝑛 𝑇2 𝑇1 𝐶𝑣 𝑑𝑇 𝑇 According to the first law of thermodynamics: 3) Gas is compressed isothermally and reversibly at temperature T2 from a point C (P2, V2) to D (P3, V3). During this step, work done on the gas (system) is W3, a quantity of heat given to sink or surrounding is q2. According to the first law of thermodynamics: ΔE3 = - W3 – q2 3 And 𝑊3 = 𝑛𝑅𝑇2 ln 𝑉3 𝑉2
  • 51. For complete cycle, the total change in internal energy is- ΔE = ΔE1 + ΔE2 + ΔE3 + ΔE4 = q1 - W1 – W2 – q2 – W3 – W4. = (q1 – q2) – (W1 + W2 + W3 + W4) = (q1 – q2) – Wm Wm is the total work done by the gas in the one complete cycle. The principle of reproducibility of the states for cycle or reversible process, ΔE = 0. E is state function. Wm = (q1 – q2) 5 = (W1 + W2 + W3 + W4) = 𝑛𝑅𝑇1 ln 𝑉1 𝑉4 + 𝑛 𝑇1 𝑇2 𝐶𝑣 𝑑𝑇 𝑇 + 𝑛𝑅𝑇2 ln 𝑉3 𝑉2 + 𝑛 𝑇2 𝑇1 𝐶𝑣 𝑑𝑇 𝑇 = 𝑛𝑅𝑇1 ln 𝑉1 𝑉4 + 𝑛𝑅𝑇2 ln 𝑉3 𝑉2 6
  • 52. But the point B(P1,V1) at T1 and point C(P2,V2) at T2 are lie on same adiabatic curve BC. 𝑉1𝑇1 𝐶𝑣 𝑅 = 𝑉2𝑇2 𝐶𝑣 𝑅 7 𝑉3𝑇2 𝐶𝑣 𝑅 = 𝑉4𝑇1 𝐶𝑣 𝑅 8 But the point D(P3,V3) at T2 and point A(P4,V4) at T1 are lie on same adiabatic curve AD. Divide equation (7) by (8)- 𝑉1 𝑉4 = 𝑉2 𝑉3 . 9 From equation (9) and (6), we have- Wm = = 𝑛𝑅𝑇1 ln 𝑉1 𝑉4 - 𝑛𝑅𝑇2 ln 𝑉1 𝑉4 Wm = (q1 – q2) = nR (T1 - T2) 𝒍𝒏 𝑽𝟏 𝑽𝟒 10
  • 53. Efficiency of Carnot heat engine: Ratio of the total work done by the working substance and total heat absorbed from the source. It is denoted by the latter η or ε. ε = 𝑊𝑚𝑎𝑥 𝑞1 11 = 𝑞1− 𝑞2 𝑞1 = 1 - 𝑞2 𝑞1 12 Also from equation (10), (1a) and (11), ε = nR (T1 − T2 ) ln 𝑉1 𝑉4 𝑛𝑅𝑇1 ln 𝑉1 𝑉4 = T1 − T2 T1 = 1 - T2 T1 13 efficiency of Carnot heat engine depends on heat absorbed or given out to sink ε depends on the temperature of source and sink but not depends on the nature and amount of the working substance
  • 54. Thermodynamic efficiency  For reversible and cyclic process of the system, if it absorb heat q1 at T1, then it undergoes a temperature change (T1 - T2) and the maximum work done is equal to the heat absorb q1 at T1 multiplied by the ratio (T1 - T2)/ T1. ε = 𝑊𝑚𝑎𝑥 𝑞1 = T1 − T2 T1 𝑊 𝑚𝑎𝑥 = T1 − T2 T1 𝑥 𝑞1  The some part of heat rejected at temperature T2 such that T1>T2.  The magnitude of the work done is given by the area of the P-V diagram of Carnot cycle.  For 100% thermodynamic efficiency, T2 = 0 i.e. temperature of the sink is zero K. To maintain such temperature of the sink or surrounding is practically impossible.
  • 55. Carnot Theorem/Rule Developed by Nicolas Léonard Sadi Carnot in the year 1824, with the principle that there are limits on maximum efficiency for any given heat engine. Carnot’s theorem states that:  Heat engines working between two heat reservoirs are less efficient than the Carnot heat engine (reversible engine) that is operating between the same reservoirs.  Efficiency of all reversible engines working between the same two temperatures (same source and the sink) is same whatever the working substance.  Maximum efficiency is given as: 𝜖𝑚𝑎𝑥 = 𝜖𝐶𝑎𝑟𝑛𝑜𝑡 = 𝑇𝐶 𝑇𝐻 𝑇𝐶 is absolute temperature of the cold reservoir 𝑇𝐻 is absolute temperature of the hot reservoir
  • 56. Consider two heat engines R (reversible engine) and I (irreversible engine) working between the same source (T1) and the same sink (T2), where T1>T2. Reversible engine R takes heat Q1 from the source (T1), perform work W and rejects some heat Q2 = Q1 –W to the same sink (T2). The efficiency of the reversible engine R is: 𝜖𝑅 = 𝑊 𝑄1 Suppose irreversible engine I takes heat 𝑄1 ′ from the source (T1), perform work W and rejects some heat 𝑄2 ′ = 𝑄1 ′ − 𝑊 to the same sink (T2). 𝜖𝐼 = 𝑊 𝑄1 ′ The efficiency of the irreversible engine I is: Suppose, the irreversible engine I is more efficient than the reversible engine R. 𝜖𝐼 > 𝜖𝑅 ⟹ 𝑊 𝑄1 ′ > 𝑊 𝑄1 ⇒ 𝑄1 > 𝑄1 ′ ⇒ 𝑄1 − 𝑄1 ′ 𝑖𝑠 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑞𝑢𝑎𝑛𝑡𝑖𝑡𝑦 Where, W work done is same.
  • 57. These two engines are coupled together so that the engine R works in opposite direction i.e. the engine R works as a refrigerator. The engine I absorbs heat 𝑄1 ′ from the source (T1), perform work W and rejects some heat 𝑄2 ′ = 𝑄1 ′ − 𝑊 to the same sink (T2). The engine R takes heat Q2 from the same sink (T2), perform work W on the engine R and Q1 = Q2 + W heat is rejected to the source (T1). Work done on the gas by the engine R is actually received from the work done by the gas on the engine I. Heat lost by the sink is: 𝑄2 − 𝑄2 ′ = 𝑄1 − 𝑊 − 𝑄1 ′ − 𝑊 ⇒ 𝑄1 − 𝑄1 ′ 𝑖𝑠 𝑎 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑞𝑢𝑎𝑛𝑡𝑖𝑡𝑦 i.e. external work done on the system is zero. both the engines R and I work as a self-acting machine The coupled engines transfers heat continuously from a body at lower temperature to the body at higher temperature. contrary to the second law of thermodynamics. i.e. our assumption 𝜖𝐼 > 𝜖𝑅 is wrong. i.e. no engine will be more efficient than a perfectly reversible engine working between the same temperatures (same source and same sink)
  • 58.
  • 59.
  • 60. Second law of thermodynamic The first law of thermodynamic state that, when one form of energy is converted into another, the total energy of the system is conserved. It does not indicate any other restriction on this process and many of the process has a natural direction. This direction can be explained by using a second law of thermodynamics. e.g. Suppose gas in expand into a vacuum but although it does not violate first law of thermodynamics; the reverse does not occur, etc. Second law of thermodynamic tell us that whether the reaction take places in forward or in backward direction or direction of flow energy. The quantity tells us that whether a chemical change or physical process is spontaneous or not in an isolated system in terms of a new thermodynamic function called as entropy. Entropy is a state function. Thermodynamic does not explain the rate of reaction/change to approach the equilibrium state, but it explains at equilibrium state. e.g. i) Gas is expand into vaccum, ii) Heat is always transfer from hot end to cold end of the rod, iii) In presence of catalyst, molecular hydrogen is react with molecular oxygen to form water.
  • 61. Different Statements of Second Law of Thermodynamics: Definition by Clausius: There is no thermodynamic transformation whose sole effect is to deliver heat from a reservoir of lower temperature to a reservoir of higher temperature Heat does not flow upwards i.e. lower to higher temperature spontaneously. In an isolated system, the entropy never decreases. Definition by Kelvin: There is no thermodynamic transformation whose sole effect is to extract heat from a reservoir and convert it entirely to work. No heat engine is constructed having unit efficiency.
  • 62.
  • 63. Entropy (S) Second law of thermodynamic can be expressed mathematically by a new thermodynamic functions S called as entropy of the system. Entropy of the system is a state function ΔS = S2 - S1 Where, S2 and S1 are entropies of final and initial states of the system respectively. dqrev/T is the differential of the state function. Clausius named this state function is entropy dS. For infinitesimal change: 𝑑𝑆 = 𝑑𝑞𝑟𝑒𝑣 𝑇 For finite change in state: ∆𝑆 = 𝑑𝑞𝑟𝑒𝑣 𝑇 = 𝑞𝑟𝑒𝑣 𝑇 Where dqrev is the infinitesimal quantity of heat absorbed under reversible conditions at temperature T. Also qrev is the heat absorbed at any temperature T for any isothermal reversible process. 𝑞𝑟𝑒𝑣 is the positive (heat is absorbed)- ΔS is positive (entropy of the system increases) 𝑞𝑟𝑒𝑣 is the negative (heat is evolved)- ΔS is negative (entropy of the system decreases) Endothermic Exothermic
  • 64. Derivation of entropy from Carnot cycle The heat q1 is taken by the working substance from source at temperature T1 and heat q2 is given out by the working substance at lower temperature T2. Efficiency of Carnot engine ε = 𝑾𝒎𝒂𝒙 𝒒𝟏 𝑞1− 𝑞2 𝑞1 = T1 − T2 T1 1 - 𝑞2 𝑞1 = 1 - T2 T1 𝑞2 𝑞1 = T2 T1 or 𝑞2 T2 = 𝑞1 T1 q1 has positive value and q2 has a negative value. − 𝑞2 T2 = 𝑞1 T1 or 𝑞1 T1 + 𝑞2 T2 = 0 In general: 𝑞 𝑇 = 0 For a complete reversible cycle represented by the close curve ABA For reversible cycle ABA: 𝑞 𝑇 = 0 Since the cycle is performed in two steps; via from A to B and then back from B to A, we have: 𝑑𝑞 𝑇 = 𝐴 𝐵 𝑑𝑞 𝑇 + 𝐵 𝐴 𝑑𝑞 𝑇 = 0 𝑑𝑞 𝑇 = 𝐴 𝐵 𝑑𝑞 𝑇 = − 𝐵 𝐴 𝑑𝑞 𝑇 Both these integrate shows that 𝒅𝒒 𝑻 is independent on the path but it depends on the state of the system. This function 𝑑𝑞 𝑇 is called as entropy and is state function.
  • 65. Second law of thermodynamic i) Natural process are irreversible and spontaneous, it is accompanied by increase in entropy of the universe. Universe means the system and its surrounding. ΔSuniv. = ΔSsystem + ΔSsarr. > 0. But for reversible process, entropy of the system remains constant. ΔSuniv. = 0. ii) Heat can not be converted to work without leaving permanent change either in systems involved or their surrounding. iii) It is impossible to take heat from a hot reservoir and convert it completely into work by cyclic process without transforming a part of heat to a cooler reservoir i.e. it impossible to construct an engine having unit efficiency. iv) It is impossible for a cyclic process to transfer heat from a body at a lower temperature to a body of higher temperature in absence of any agency.
  • 66. Here are 5 different ways to state the second law of thermodynamic 1. For any spontaneous process the entropy of the universe increases. 2. It is impossible for heat to spontaneously flow from a cold object to a hot object. 3. No steam engine can be more efficient than a reversible steam engine (Carnot theorem). 4. It is impossible to build a perfectly efficient steam engine. 5. It is impossible to build a perpetual motion machine of the second kind.
  • 67. Entropy change in Physical transformations Change in entropy occur not only variations of temperature, pressure and volume of the system but also physical transformation such as change of state from solid to liquid (melting), liquid to vapour (boiling) and solid to vapour (sublimation) At constant temperature, two phases are in equilibrium during phase transformation by an absorbing or evolving heat. ∆𝑆 = 𝑞𝑟 𝑇 = ∆ 𝐻 𝑇 For melting For one mole of substance, it absorb a heat is called as latent/molar heat of fusion at constant temperature T is called as melting point. The change in entropy is: ∆𝑆𝑓= ∆ 𝐻𝑓 𝑇𝑓 For boiling It absorb a heat is called as latent/molar heat of vaporization at constant temperature T is called as boiling point. The change in entropy is: ∆𝑆𝑣= ∆ 𝐻𝑣 𝑇𝑏 One mole of solid is change to another allotropic form at constant allotropic transition temperature. The change in entropy is: For allotropic transition ∆𝑆𝑡= ∆ 𝐻𝑡 𝑇𝑡
  • 68. • According to the physical significance of entropy- In Carnot cycle, only some part of heat is converted into work and remaining part of heat is given to sink. The part of heat which is not available to perform work is a measure of entropy change accompanying the process. Hence the fraction of energy unavailable to perform the work is T.ΔS. S = q/T = ∆H/T Fraction available to perform work = (ΔH) - T.ΔS.
  • 69. Physical significance of entropy (b) i.e. melting at constant temperature Tf, the heat absorbed by the solid particles to go to liquid state is ΔHf. The addition of heat to the system introduces more randomness or disorder in the system e.g. when solid melts, the ordered arrangement of the constituent particles is destroyed and particles acquired greater freedom of movement. Change in entropy is: ∆𝑆𝑓= ∆ 𝐻𝑓 𝑇𝑓 ΔHf is positive and hence ∆𝑆𝑓 is also positive i.e. Ssolid < Sliquid Similarly for boiling of liquid, the constituent particles of gas have more degree of freedom and more disorder system is formed. entropy is taken as a measure of disorder of the system
  • 70. (c) Entropy of the system is taken as a measure of mixtupness of the system When organized from of energy such as electric, mechanical or chemical is converted into heat, the randomness of the system increases as the constituent particles of the system acquired greater degree of freedom of movement. It increases the disorder in the system, increase its mixtupness. (d) Entropy of the system is also used to measured thermodynamic probability of the system. S = k lnW (put by Plank in 1912) Where, S is entropy, W is thermodynamic probability, k is Boltzmann constant.
  • 71. Entropy as criteria of Spontaneity & Equilibrium (a) For irreversible process: Consider a system in contact with a reservoir at constant temperature T, assume that in the system an infinitesimal irreversible process (spontaneous) occurs and work done is of P-V type. The quantity of heat dq is exchanged between the system and reservoir. Therefore, entropy change of system is dS and that of reservoir is – dq/T. dS > dq/T TdS > dq dq – TdS < 0 (1) According to the first law of thermodynamics - dq = dE + PdV (2) From equation (1) and (2)- dE + PdV – TdS < 0 (3) Condition of entropy change: Consider the system having constant volume and internal energy i.e. dE = dV = 0, therefore equation (3) becomes: -(TdS)E,V < 0 (TdS)E,V > 0 or (dS)E,V > 0 (4) And for finite change - (ΔS)E,V > 0 Reversible change: change < , > replaced by = sign in above.
  • 72. Entropy Change in Ideal Gases The change in entropy of an isolated system is zero or greater than zero. 𝑑𝑆𝑖 = 𝑑𝑆𝑠𝑢𝑏. + 𝑑𝑆𝑟𝑒𝑠𝑒𝑟. ≥ 0 dSi is change in entropy of isolated system dSsub is change in entropy of working substance dSreser is change in entropy of heat reservoir The heat is supplied by the heat reservoir to working substance at constant temperature T. 𝑑𝑆𝑟𝑒𝑠𝑒𝑟. = 𝑑𝑞 𝑇 From above equation: 𝑑𝑆𝑠𝑢𝑏. + 𝑑𝑞 𝑇 ≥ 0 According to the first law of thermodynamics dq = dE + dw But if system does work is only P-V type, then dq = dE + pdV And if the system does the work other than P-V type then: dq = dE + pdV + dw’ Put dq in above equation: dSsub – (dE + pdV + dw’)/T ≥ 0 Multiplying both side by T, TdSsub – (dE + pdV + dw’) ≥ 0 The inequality is applied to irreversible processes, where only P-V work is done and dW’= 0 and p = P, TdSsub – (dE + PdV) > 0 (for irreversible process)
  • 73. For reversible processes, dw’ = dW’ and p = P, TdSsub – (dE + PdV + dW’) = 0 But when dW’ = 0, TdSsub – (dE + PdV) = 0 For reversible isothermal process dE = 0, TdSsub – PdV = 0 dSsub = 𝑷𝑑𝑉 𝑇 For n mole of ideal gas: PV = nRT P = 𝑛𝑅𝑇 𝑉 or Therefore, dS is dSsub = 𝑛𝑅 𝑉 𝑑𝑉 Integrate with the limits of entropy S1 and S2 at the volume V1 and V2: 𝑆1 𝑆2 𝑑𝑆 = 𝑛𝑅 𝑉1 𝑉2 𝑑𝑉 𝑉 𝑆2 − 𝑆1 = 𝑛𝑅 ln 𝑉2 𝑉1 ΔS = 𝑛𝑅 ln 𝑉2 𝑉1 Expansion process ΔS = 𝟐. 𝟑𝟎𝟑 𝑛𝑅 log 𝑉2 𝑉1 Entropy change for the reversible isothermal compression process is: ΔS = 2.303 𝑛𝑅 log 𝑉1 𝑉2
  • 74. *** Free Energy and Equilibrium *** • How much quantity of heat is used to performed work done is determined by mew thermodynamic state function is called as free energy. The heat energy is supplied in the form of internal energy or enthalpy. [Heat energy (E or H)] = (Heat available to perform work done) + (Heat unavailable to perform work done) If heat energy is supplied in the form of internal energy (E), the part of internal energy available to perform work is called Helmholtz free energy. It is denoted by the latter A. The internal energy unavailable to perform the work done is TS, S-is entropy at temperature T. [Heat energy (E)] = (E available to perform work done) + (E unavailable to perform work done) E = A + TS A = E - TS (1)
  • 75. If heat energy is supplied in the form of enthalpy (H), the part of enthalpy available to perform work is called Gibbs free energy. It is denoted by the latter G or F. The enthalpy unavailable to perform the work done is TS, S-is entropy at temperature T. [Heat energy (H)] = (H available to perform work done) + (H unavailable to perform work done) H = G + TS G = H - TS (2)
  • 76. Helmholtz free energy Helmholtz free energy of any system A is defined as - A = E - TS Where, A, E, S and T are state functions. The change in A during any process is given as- ΔA = A2 - A1 = (E2 – T2S2) – (E1 – T1S1) = (E2 - E1) – (T2S2 – T1S1) = ΔE – ΔTS (3) Equation (3) is the general definition of ΔA. For isothermal process T2 = T1, then equation (3) becomes- ΔA = ΔE – TΔS (4) Eq. (4) explain physical interpretation (significance) of ΔA, under isothermal conditions – TΔS = qrev (according to the definition of change in entropy). ΔA = ΔE – qrev. (5) According to the first law of thermodynamics for reversible and isothermal process- qrev = ΔE + Wmax or -Wmax = ΔE - qrev (a) From equation (5) and (a)- ΔA = -Wmax (6) The equation (6) indicates that, at constant temperature, the maximum work done by the system is due to expense of (decrease in) Helmholtz free energy. Therefore, A is called as Work function of a system.
  • 77. Relationship between G and A Consider equation (3) is- ΔG = ΔH – TΔS But H is also state function- ΔH = ΔE + PΔV (at constant P) Therefore, ΔG = ΔE + PΔV – TΔS = (ΔE – TΔS) + PΔV (ΔA = ΔE – TΔS) ΔG = ΔA + PΔV (4) In general- G = A + PV (5) The equation (4) and (5) shows the relationship between G and A.
  • 78. A and G as criteria for Thermodynamic Equilibrium and Spontaneity Condition of Helmoholtz free energy change If system is not isolated system then its entropy can be changed. If volume of the system remains constant for irreversible process i.e. dV = 0 then, dE – TdS < 0 At constant V and T, for infinite small change: (∂E – T∂S)T,V < 0 ∂(E – TS)T,V < 0 Or According to the definition of A, A = E – TS ∂(A)T,V < 0 then And for finite change: (ΔA)T,V < 0 Thus, for any irreversible or spontaneous process in a system at constant volume and temperature, the Helmoholtz free energy of the system decreases therefore stable state is obtained at the lowest Helmoholtz free energy. Helmoholtz free energy is state function, therefore at equilibrium state (ΔA)T,V = 0 dE + PdV – TdS < 0
  • 79. Gibbs free energy is state function, therefore at equilibrium state (ΔG)T,P = 0 Condition of Gibbs free energy change If system is not isolated system then its entropy can be changed. If pressure and temperature of the system remains constant for irreversible process i.e. dP = dT = 0 dE + PdV – TdS < 0 At constant P and T, for infinite small change: (∂E + P∂V – T∂S)T,P < 0 Or ∂(E + PV– TS)T,P < 0 Or ∂(H – TS)T,P < 0 According to the definition of G, G = H – TS ∂(G)T,P < 0 And for finite change: (ΔG)T,P < 0 Thus, for any irreversible or spontaneous process in a system at constant pressure and temperature, the Gibbs free energy of the system decreases therefore stable state is obtained at the lowest Gibbs free energy.
  • 80. Variation of A with T and V: Differentiate it completely as - dA = dE – TdS - SdT (7) According to the definition of entropy - dqrev = TdS Therefore equation (7) becomes- dA = dE - dqrev - SdT (8) According to the first law of thermodynamics- dqrev = dE + PdV (for reversible process) Or -PdV = dE - dqrev Therefore equation (8) becomes - dA = - PdV - SdT (9) Equation (9) indicates that A is most conveniently expressed in terms of a T and V as the independent variables, we have- A = f (T,V) Differentiate it completely- dA = (δA/δT)v dT + (δA/δV)T dV (10) Comparing equation (9) and (10) we have- (δA/δV)T = -P (11) (δA/δT)v = -S (12)
  • 81. • An alternative equation for the variation of A with T can be obtained as follows- • Differentiate A/T with respect to T at constant V gives - • [δ(A/T)/δT]v = [T(δA/δT)v – A]/T2 • = [T(-S) – A]/T2 • = - [A + TS]/T2 • = - E/T2 (13) • Equation (9) shows that the dependence of A for a pure substance on both T and V, these can be shown by (12) and (13) for T and (11) for V.
  • 82. Variation of G with P and T According to the definition of G [i.e. from equation (1)]- G = H - TS Differentiate it completely as- dG = dH - TdS - SdT (7) According to the definition of entropy and first law of thermodynamics- dqr = TdS = dE + PdV (for reversible process) (8) According to the definition of enthalpy- H = E + PV Differentiate it completely- dH = dE + PdV + VdP (9) Therefore equation (8) and (9) dH = VdP + TdS (10) From equation (7) and (10)- dG = VdP + TdS -TdS -SdT dG = VdP - SdT (11) Equation (11) indicates that G is function of a T and P as the independent variables, we have- G = f (T,P)
  • 83. G = f (T,P) Differentiate it completely- dG = (δG/δT)P dT + (δG/δP)T dP (12) Comparing equation (11) and (12) we have- (δG/δP)T = V (δG/δT)P = -S (13) An alternative equation for the variation of G with T can be obtained as follows- Differentiate G/T with respect to T at constant P gives- [δ(G/T)/δT]P = [T(δG/δT)P – G]/T2 = [T(-S) – G]/T2 = - [G + TS]/T2 = -H/T2 (14) Equation (11) shows that the dependence of G for a pure substance on both T and P, these can be shown by (14) and (13) for T.
  • 84. Advantage of G/A over Entropy change The first law of thermodynamics /energy conservation, but one can't comment on the nature of a process (spontaneous or non-spontaneous). Hence knowing only the quantity of energy change is never sufficient. The second law is defined in the terms of both entropy and Gibbs free energy. For a process to be spontaneous: ∆ 𝑆 𝑡𝑜𝑡𝑎𝑙 > 0 𝑎𝑛𝑑 ∆𝐺 < 0 Hence both entropy and Gibbs free energy changes can help you determine the spontaneity of a process Entropy change for a reversible process: ∆𝑆 = 𝑞𝑟𝑒𝑣 𝑇 Criteria for spontaneity is the total entropy: ∆ 𝑆 𝑡𝑜𝑡𝑎𝑙 > 0 and ∆ 𝑆 𝑡𝑜𝑡𝑎𝑙 = ∆ 𝑆 𝑠𝑦𝑠𝑡𝑒𝑚 + ∆ 𝑆 𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔 It is extremely difficult task to determine the entropy change in surrounding therefore converting it to Gibbs free energy. ∆ 𝑆 𝑡𝑜𝑡𝑎𝑙 = 𝑞𝑠𝑦𝑠𝑡𝑒𝑚 𝑇 + 𝑞𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔 𝑇 & 𝑞𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔 = ∆𝐻𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔 = −∆𝐻𝑠𝑦𝑠𝑡𝑒𝑚 Hence, ∆ 𝑆 𝑡𝑜𝑡𝑎𝑙 = ∆ 𝑆 𝑠𝑦𝑠𝑡𝑒𝑚 − ∆𝐻𝑠𝑦𝑠𝑡𝑒𝑚 𝑇
  • 85. Multiplying by T and rearranging, 𝑇 ∆ 𝑆 𝑡𝑜𝑡𝑎𝑙 = 𝑇 ∆ 𝑆 𝑠𝑦𝑠𝑡𝑒𝑚 − ∆𝐻𝑠𝑦𝑠𝑡𝑒𝑚 ∆𝐺 = − 𝑇 ∆ 𝑆 𝑡𝑜𝑡𝑎𝑙 = − 𝑇 ∆ 𝑆 𝑠𝑦𝑠𝑡𝑒𝑚 + ∆𝐻𝑠𝑦𝑠𝑡𝑒𝑚 ΔS(total) must be greater than zero, ΔG must be less than zero. Hence, Gibbs free energy reduces the tedious work of determining entropy change in surrounding and it's value can determined by knowing the state of the system only. Gibbs free energy is more advantageous than entropy change.
  • 86. Chemical equilibrium A + B ↔ C + D Based on the extent to which the reactions proceed to reach the state of chemical equilibrium: (i) The reactions that proceed nearly to completion and only negligible concentrations of the reactants are left. In some cases, it may not be even possible to detect these experimentally. (ii) The reactions in which only small amounts of products are formed and most of the reactants remain unchanged at equilibrium stage. (iii) The reactions in which the concentrations of the reactants and products are comparable, when the system is in equilibrium.
  • 87. A + B ↔ C + D
  • 88. There are number of important questions about the composition of equilibrium mixtures: What is the relationship between the concentrations of reactants and products in an equilibrium mixture? How can we determine equilibrium concentrations from initial concentrations? What factors can be exploited to alter the composition of an equilibrium mixture? Particular is important when choosing conditions for synthesis of industrial chemicals such as H2, NH3, CaO etc.
  • 89. Equilibrium constant Law of mass action The rate of a chemical reaction at any instant is proportional to the molar concentration of the reacting substance at that instant Consider the simplest reaction – A → product. If CA is the concentration of the reactant at time t, then- Rate =  CA The proportionality constant in the above rate equations is called rate constant or velocity constant or specific rate constant
  • 90. Reversible reaction k k f b A B Depending on the physical state of the reactants and products, two physical constants for reversible reaction is defined as KP and KC. At gaseous state, the equilibrium constant KP is in terms of partial pressures of reactants and products For other states or liquid/solution state, it is denoted by KC which is in terms of partial concentrations.
  • 91. aA + bB lL + mM A & B are reactants (behaves ideally) L & M are products (behaves ideally) a & b and c & d are number of moles of reactants and products respectively. general reaction are the molar concentrations of A, B, C and D respectively. Characteristics of Kc: •When concentrations of the reactants and products have attained constant value at equilibrium state. • Value of equilibrium constant is independent of initial concentrations of the reactants and products. • Equilibrium constant is temperature dependent having one unique value for a particular reaction represented by a balanced equation at a given temperature. • The equilibrium constant for the reverse reaction is equal to the inverse of the equilibrium constant for the forward reaction. KC
  • 92. Because gas pressures are easily measured, equilibrium equations for gas-phase reactions are often written using partial pressures rather than molar concentrations. For the gaseous state, the partial pressure is also measure the activity of the component, KP is - are the molar concentrations of A, B, C and D respectively. KP aA + bB lL + mM A & B are reactants (behaves ideally) L & M are products (behaves ideally) a & b and c & d are number of moles of reactants and products respectively.
  • 93. Relationship between KP and KC For n mole of ideal gas - pV = nRT p = = But n/V = C, therefore, p = C R T. Equations for KP and KP are - aA + bB lL + mM A & B are reactants (behaves ideally) L & M are products (behaves ideally) a & b and c & d are number of moles of reactants and products respectively. But (l + m) - (a + b) = ∆n, therefore, K = p K C (RT) n
  • 94. Le Chatelier’s Principle Equilibrium constant, Kc is independent of initial concentrations. But, if a system at equilibrium is subjected to a change in the concentration of one or more of the reacting substances, change in temperature or pressure of the system or catalyst can affect equilibrium can be explain by Le Chatelier’s principle. A change in any of the factors that determine the equilibrium conditions of a system will cause the system to change in such a manner so as to reduce or to counteract the effect of the change. Effect of Concentration Change Effect of Pressure Change Effect of Inert Gas Addition Effect of Temperature Change Effect of a Catalyst
  • 95. Effect of Concentration Change When the concentration of any of the reactants or products in a reaction at equilibrium is changed, the composition of the equilibrium mixture changes so as to minimize the effect of concentration changes ) 𝐻2 𝑔 + 𝐼2 𝑔 ⇋ 2 𝐻𝐼 (𝑔 𝐹𝑒3+ 𝑎𝑞 + 𝑆𝐶𝑁− 𝑎𝑞 ⇋ 𝐹𝑒 𝑆𝐶𝑁 2+ 𝑎𝑞 yellow colourless deep red 𝐾𝑐 = 𝐹𝑒 𝑆𝐶𝑁 2+ 𝑎𝑞 𝐹𝑒3+ 𝑎𝑞 𝑆𝐶𝑁− 𝑎𝑞 The intensity of the red colour becomes constant on attaining equilibrium Oxalic acid (H2C2O4), reacts with Fe3+ ions to form the stable complex ion [Fe(C2O4)3]3–, thus decreasing the concentration of free Fe3+ (aq); the concentration of [Fe(SCN)]2+ decreases, the intensity of red colour decreases. Similar effect was found by the addition of aq. HgCl2 because Hg2+ reacts with SCN– ions to form stable complex ion [Hg(SCN)4]2–
  • 96. Effect of Pressure Change Affect in case of a gaseous reaction where the total number of moles of gaseous reactants and total number of moles of gaseous products are different. The effect of pressure changes on solids and liquids can be ignored because the volume (and concentration) of a solution/liquid is nearly independent of pressure 𝑪𝑶 𝒈 + 𝟑 𝑯𝟐 ⇋ 𝑪𝑯𝟒 𝒈 + 𝑯𝟐𝑶 𝒈 4 mol of gaseous reactants (CO + 3H2) become 2 mol of gaseous products (CH4 + H2O) As, total pressure will be doubled (according to p.V = constant), partial pressure and therefore, concentration of reactants and products have changed and the mixture is no longer at equilibrium. The equilibrium now shifts in the forward direction. Effect of Inert Gas Addition If the volume is kept constant and an inert gas such as argon is added which does not take part in the reaction, the equilibrium remains undisturbed because it does not change the partial pressures or the molar concentrations of the substance involved in the reaction
  • 97. Effect of Temperature Change when a change in temperature occurs, the value of equilibrium constant, Kc is changed. In general, the temperature dependence of the equilibrium constant depends on the sign of ∆H or ∆G for the reaction.  The equilibrium constant for an exothermic reaction (negative ∆H) decreases as the temperature increases.  The equilibrium constant for an endothermic reaction (positive ∆H) increases as the temperature increases. ) 𝟑 𝑯𝟐 𝒈 + 𝑵𝟐 𝒈 ⇋ 𝟐 𝑵𝑯𝟑 (𝒈 ∆H= – 92.38 kJ mol–1. exothermic process raising the temperature shifts the equilibrium to left and decreases the equilibrium concentration of ammonia.
  • 98. The Reaction isotherm (Van’t Hoff’s Isotherm) Van’t Hoff’s isotherm gives the net work done that can be obtained from gaseous reactants at constant temperature, when both reactants and products are at suitable orbitary pressure. Consider a general reaction- If aA, aB, aC and aD are activities of reactants A, B and products C, D respectively. The free energies of each of these substances per mole at T is- GA = Go A + RT lnaA (1) GB = Go B + RT lnaB (2) GC = Go C + RT lnaC (3) GD = Go D + RT lnaD (4) Where, Go A, Go B, Go C and Go D are the free energies at unit activity of the A, B, C and D respectively or are the standard free energies of A, B, C and D respectively. aA + bB + ............ cC + dD + .................. A quantitative relation of the free energy change for the chemical reaction has been developed which is known as Reaction isotherm or Van’t Hoff’s isotherm.
  • 99. Therefore the free energy change of the chemical reaction - ΔG = ΣGp – ΣGR = [cGC + dGD + …….] – [aGA +bGB + ……..] = [cGoC + RT lnaC c + dGoD + RT lnaD d + …….] – [aGoA + RT lnaA a + bGo B + RT lnaB b + ……..] = [(cGoC + dGoD + ……) – (aGoA + bGoB +……)] + RT[(lnaC c + lnaD d + …….) – (lnaA a + lnaB b + ……..)] ΔG = ΔGo + RT ln{(aC c.aD d…….)/(aA a .aB b……..)} (5) The equation (5) is called as the Van’t Hoff’s isotherm or reaction isotherm. When activities of all reactants and products are equal to one then ΔG = ΔGo . At equilibrium step of the reaction - ΔG = 0 0 = ΔGo + RT ln{(acC x adD…….)/(aaA x abB……..)} At equilibrium, according to the law of mass action, Ka = (acC adD…….)/(aaAabB……..) Therefore, 0 = ΔGo + RT ln Ka Or ΔGo = - RT ln Ka (6) Ka = e -ΔGo/RT (7) Form equation (5) and (6)- ΔG = RT ln{(acC adD…….)/(aaAabB……..)} - RT ln Ka = RT ln Qa - RT ln Kaa where Qa = {(acC adD…….)/(aaAabB……..)} = RT ln (Qa/Ka) (8)
  • 100. For gases activities are proportional to the partial pressures of the components of reaction mixture hence the activities of the equation (5) can be replaced by the partial pressures, therefore- ΔG = ΔGo + RT ln{(PC c.PD d…….)/(PA a.PB b……..)} (9) = RT ln Qp - RT ln Kp (10) Where Kp is the equilibrium constant and Qp = (Pc C.Pd D…….)/(Pa A.Pb B……..) At equilibrium state, ΔG = ΔGo and Qp = 1. ΔGo = - RT ln Kp = -W or Kp = e -ΔGo/RT (11) i.e. net work done of the reaction is equal to the decrease in free energy of the system and it is calculated by using expression - RT ln Kp or 2.303RT log Kp. It will be observed that, ΔG is positive when Kp is less than unity. If ΔG is negative when Kp is greater than unity and is zero when Kp = 0 or 1.
  • 101. *** Van’t Hoff Isochore *** The Van’t Hoff isochore is obtained by comparing the Van’t Hoff isotherm with Gibbs - Helmeholtz equation. The Gibbs-Helmeholtz equation is ΔG = ΔH + T[d(ΔG)/dT]p Or -ΔH = T[d(ΔG)/dT]p - ΔG Dividing both side by T2 gives:- -ΔH/T2 = {T[d(ΔG)/dT]p - ΔG}/T2 The right hand side of the above equation is obtained by differentiating ΔG/T w.r.t. T at constant P- [d(ΔG/T)/dT]p = {T[d(ΔG)/dT]p - ΔG}/ T2 -ΔH/T2 = [d(ΔG/T)/dT]p (12) • According to Van’t Hoff isotherm- ΔG = -RT ln Kp (11) • Dividing both side by T and differentiating it w.r.t. T at constant P- • [d(ΔG/T)/dT]p = -R d(lnKp)/dT (13) • Comparing equation (13) and (12), we have- • ΔH/T2 = R d(lnKp)/dT • Or (ΔH/R).dT/T2 = d(lnKp) (14) • The above equation is known as Van’t Hoff isochore. If ΔH remains same over the range of temperature, we have on integration - • lnKp = (ΔH/R)∫dT/T2 ∫dx/xn = 1/(-n+1).1/xn-1 • = -ΔH/RT + constant of integration. • Appling the limits T1 and T2 at the equilibrium constants Kp1 and Kp2 respectively, we have- • lnKp2 - lnKp1 = -(ΔH/R)[1/T2 - 1/T1] • ln[Kp2/Kp1] = (ΔH/R)[1/T1 - 1/T2] • log[Kp2/Kp1] = (ΔH/2.303R)[(T2 – T1)/T1T2] (15) • By using this equation, equilibrium constant at any temperature or heat of reaction is calculated.
  • 102. Vapour pressure of Liquid Clapeyron equation For any pure substance in a single phase such as liquid or gaseous, are in contact with each other then any variation in free energy is given by the equation. dG = VdP – SdT (1) If these two phases are in equilibrium to each other, then dG = 0 at constant pressure and temperature i.e. dP = dT = 0 i.e. these phases are in equilibrium with other when the pressure and temperature are constant and uniform throughout the phase. “A useful thermodynamic equation is applicable to a system consisting of two phases of same substance in equilibrium is called as Clapeyron equation.” Consider the transition of a pure substance from one phase into another phase e.g.- 1. Solid convert to liquid- at melting point of solid- melting or fusion. 2. Liquid convert to vapour- at boiling point of liquid- boiling or vaporization. 3. Solid convert to vapour- at sublimation point of solid- sublimation. 4. Allotropic transitions at the transition temperature of the two allotropic forms. e.g. S(rhombic) is in equilibrium with S(monoclinic). All above changes are represented by the reaction- For which ΔG is given as- ΔG = G2 –G1 (2) Where, G2 and G1 are the molar free energies of the substances in final and initial states respectively. At equilibrium state ΔG = 0, at constant P and T i.e. G1 = G2. All such transformations will be in equilibrium at constant pressure and at constant temperature, when molar free energies of the substances are identical in both phases. A1 A2
  • 103. Clapeyron equation ΔG = G2 –G1 = 0. i.e. G1 = G2 Or dG1 = dG2 (3) The variation of free energies for pure substances with respect to P and T in single phase is given by the equation- dG1 = V1dP – S1dT (4) dG2 = V2dP – S2dT (5) Where, order to maintain equilibrium, the temperature is changed by the amount dT and that of pressure is dP. From equation (3), (4) and (5)- V2dP – S2dT = V1dP – S1dT (V2 – V1) dP = (S2 – S1) dT dP/dT = (S2 – S1)/(V2 – V1) = ΔS/ΔV (6) Where, ΔS is change in entropy and ΔV is change in volume during transformation. But- ΔG = ΔH – TΔS, But at equilibrium, ΔG = 0. Or ΔH = TΔS or ΔS = ΔH/T (7) From equation (6) and (7)- dP/dT = ΔH/T.ΔV (8) Where, ΔH is the change in enthalpy for reversible transformation occurring at temperature T. this equation is called as Clapeyron equation. The Clapeyron equation gives relation between the temperature and pressure of the system containing two phases of a pure substance which are in equilibrium with each others. This equation also shows that dP/dT is directly proportional to the enthalpy of the transition and inversely proportional to temperature and the volume change accompanying during the transition i.e. ΔH and ΔV are the functions of temperature and or pressure. The equation (8) can be modified as- (P2 – P1)/(T2 – T1) = ΔH/T.ΔV (9) Where T is taken as average of T2 and T1 or equation (8) can be integrated with the proper limits of T and P by assuming that ΔH and ΔV are constant. P2 T2 ∫dP = ΔH/ΔV ∫ dT/T P1 T1 (P2 – P1) = (ΔH/ΔV) ln T2/T1. (P2 – P1) = (2.303ΔH/ΔV) ln T2/T1. (10)
  • 104. Vapour pressure of Liquid The variation of vapour pressure with temperature can be expressed thermodynamically by means of Clausius-Clapeyron equation. Let us consider a system- For the transition of liquid to vapour, P is the vapour pressure at given temperature T; ΔHv is the heat of vaporization of liquid. The volume of given liquid is Vl and that of vapour is Vg. Therefore, the Clepeyron equation can be written as- dP/dT = [ΔHv /T(Vg – Vl)] (1) But molar volume of the substance in vapour phase is very large as compaired to volume of same liquid in liquid state. (Vg – Vl) ≈ Vg. (18 ml water liquid = 22400 ml water vapour) Therefore, the Clepeyron equation becomes- dP/dT = ΔHv/TVg. (2) If we assume that the vapour behaves ideally then Vg per mole of substance can be calculated as- Vg = RT/P (3) Substituting (3) in equation (2) as- dP/dT = PΔHv/RT2. (4) dP/P = (ΔHv/R). dT/T2 (5) d(lnP) = (ΔHv/R). dT/T2 (6) The equation (6) is known as the Clausius-Clapeyron equation. The heat of vaporization is a function of temperature. Assume that ΔHv will be independent on the temperature (constant over the range of temperature and pressure). Liquid vapour
  • 105. Vapour pressure of Liquid Then equation (6) can be integrated- (a) Integrated without limits of temperature and pressure:- ∫d(lnP) = (ΔHv/R) ∫dT/T2 + C (C is constant of integration) (lnP) = (ΔHv/R) . (-1/T) + C (logP) = (-ΔHv/2.303RT) + C’ (7) The equation (7) shows that logarithm of the vapour pressure should be a function of the reciprocal of the absolute temperature T. It is also a equation of straight line of the type, y = mx+c. Plot a graph of logP against 1/T is a straight line having slope = m = (-ΔHv/2.303R) and Y-intercept c = C’. From the slope of the graph, heat of vaporization of the various liquids can be calculated as- m = (-ΔHv/2.303R) (8) ΔHv = (-2.303R . m) = -4.576 . m cal/mol. (b) Integration with limits:- Integrate equation (6) with the limits of pressure P1 and P2 corresponding to the temperature T1 and T2, then- P2 T2 ∫d(lnP) = (ΔHv/R) ∫dT/T2 P1 T1 P2 T2 [lnP] = (ΔHv/R) [-1/T] P1 T1 ln(P2/P1) = (ΔHv/R) [(T2 – T1)/T1T2] (9) log(P2/P1) = (ΔHv/2.303R) [(T2 – T1)/T1T2] (10) The equation (10) can be used to calculated ΔHv from the values of the vapour pressure at any two temperatures or when ΔHv is known, P at some desired temperature can be calculated from a single available vapour pressure at a given temperature.