CHEM 815 Classical Chemical Thermodynamics:
Course content: First, Second and Third laws of Thermodynamics, the Thermodynamic
interpretation of enthalpy, entropy and Gibb’s energy. Applications of thermodynamics to
chemical systems at equilibrium (3 Credits units).
In the study of chemistry the following fundamental questions need to be answered.
(i)
Why does a reaction occur? That is, what is the driving force of a reaction?
(ii)
How far a reaction can occur? That is, what is the extent (or progress) of the
reaction?
(iii) How fast a reaction can occur? That is, what is the rate of the reaction?
We get the answer of the first two questions by the study of thermodynamics, while third
question forms the domain of the study of chemical kinetics. In this unit we shall focus our
attention mainly on thermodynamics.
Thermodynamics
Thermodynamics is the study of heat and its transformations. (thermo; means heat and
dynamics means transformation. So, that branch of science which deals with the quantitative
relationship between heat and other forms of energies is called thermodynamics. A more
encompassing definition however states that thermodynamics is the study of restrictions on
the possible properties of matter or on permissible physical processes. Thermodynamics tries
to establish relationships between properties of matter and their changes when they undergo
certain processes. The principles of thermodynamics have been built around a few laws of
thermodynamics called zeroth law, first law, second law and third law. These laws find
applications in physics, chemistry, engineering, medicine, biotechnology, biochemistry,
geology and space sciences. The branch of chemistry dealing with the investigation of
energetics and feasibility of chemical reactions and physical changes is called chemical
thermodynamics. Its principles are simple, and its predictions are powerful and extensive.
The predictive power of chemical thermodynamics is based on the characteristics of
thermodynamic properties namely internal energy (U), enthalpy (H), entropy (S) and free
energy functions (A and G) and their variations with variables like temperature, pressure,
volume and amount. The changes in these properties depend only on the initial and final
states of the system, and are independent of the path followed for the system Therefore, these
thermodynamic properties are called state functions. This aspect will be discussed later in
this unit.
Aim of the study of chemical thermodynamics
The main aim of the study of chemical thermodynamics is to learn (i) transformation of
energy from one form into another form, (ii) utilization of various forms of energy and (iii)
changes in the properties of systems produced by chemical or physical effects. Therefore, this
branch of science is also called chemical energetics.
Various forms of energy involved in the study of chemical thermodynamics
In the study of chemical thermodynamics most frequently we deal with the interconversions
of four forms of energy namely, electrical energy, thermal energy, mechanical energy, and
chemical energy. The energy involved in the chemical processes is called chemical energy.
That is, it is the energy liberated or absorbed when chemical bonds are formed, broken or
rearranged.
Conversion of chemical energy into other forms
Depending upon the conditions under which the reaction proceeds, the chemical energy
released can be made to appear as thermal energy, mechanical energy or electrical energy.
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 If the reaction takes place in a closed vessel immersed in a water bath, the chemical
energy will appear as thermal energy (heat), which would warm the vessel, the
reaction mixture and the bath.
 If the reaction proceeds in a cylinder fitted with a movable piston, the chemical
energy released can be made to produce mechanical work (work of expansion) by
forcing the piston to move in the cylinder against an external pressure.
 If the reaction is allowed to take place in an appropriate electrochemical cell, the
chemical energy can be converted into electrical energy.
Some Important terms used in Thermodynamics
(i) System A thermodynamic system is any three-dimensional region of physical space on
which we wish to focus our attention. i.e. the part of the universe under study.
(ii) Surroundings The part of universe other than the system is known as surroundings.
(ill) Boundary The wall that separates the system from the surroundings is called boundary.
The type of barrier that is used for this purpose determines the boundary conditions of the
system. A diathermal boundary for example will allow the transfer of heat between the
system and the surrounding. An adiabatic boundary on the other hand would not allow heat
transfer between system and surrounding.
(iv) Thermodynamic equilibrium A system in which the macroscopic properties
(macroscopic properties of a system are those properties of the system which are measurable,
such as volume, elastic moduli, temperature, pressure and specific heat) do not undergo any
change with time is called thermodynamic equilibrium.
(v) Thermal equilibrium If there is no flow of heat from one portion of the system to
another, the system is said to be in thermal equilibrium.
(vi) Mechanical equilibrium If no mechanical work is done by one part of the system on
another part of the system. it is said to be in mechanical equilibrium. Such a condition exists
when pressure remains constant.
Types of Systems
(i) Open system The system in which energy and matter both can be exchanged with the
surroundings freely (walls both diathermal and permeable).
(ii) Closed system The system in which only energy can be exchanged with the surroundings
(walls impermeable but diathermal).
(iii) Isolated system The system in which neither energy nor matter can be exchanged with
the surroundings (walls impermeable and diathermal).
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Thermodynamics Properties
Thermodynamic properties are defined as characteristic features of a system, which are
capable of specifying the system's state. They can generally be classified into two states
namely;
1. Intensive Properties
Properties of the system which depend only on the nature of matter but not on the quantity of
matter are called Intensive properties, e.g., pressure, temperature, specific heat, etc
2. Extensive Properties
Properties of the system which are dependent on the quantity of matter are called extensive
properties, e.g., internal energy, volume, enthalpy, etc.
State of System
When microscopic (properties of matter at the atomic level) properties have definite value,
the conditions of existence of the system is known as state of system.
State functions When values of a system is independent of path followed and depend only
on initial and final state of the system, it is known as state function,e.g., Δ U, Δ H, Δ G etc.
Path functions These depend upon the path followed, e.g., work, heat, etc.
The Zeroth Law of Thermodynamics and temperature
Also known as the law of thermal equilibrium, the law states that if two systems (A and B)
are in thermal equilibrium with a third system (C) then they are also in thermal equilibrium
with each other. Temperature is used here to know if the system is in thermal equilibrium or
not.
Stated differently; There exists for every thermodynamic system in equilibrium a property
called temperature. Equality of temperature is a necessary and sufficient condition for
thermal equilibrium. The Zeroth Law thus defines a property (temperature) and describes its
behaviour.
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Schematic representation of the zeroth law
First Law of Thermodynamics
The total amount of energy in the form of heat transferred into a system is equal to the change
in internal energy of the system minus the work done by the system. The first law basically
introduces us to the concept of internal energy. The first law is represented mathematically
as, ΔU = q + W,
where, ΔU = internal energy change
q = heat added to system
W = work added to system
Work and heat are the two modes by which energy is transferred or transformed.
Sign convention
(i) q is + ve = heat is supplied to the system
(ii) q is – ve = heat is lost by the system
(iii) W is + ve = work done on the system
(iv) W is – ve =work done by the system
Internal Energy (E or U)
We do not have a rigorous definition of energy as physicist still do not know exactly what
energy is. A lose definition of energy is given as the capacity to do work. Thermodynamics
deals with energy transfer which is the basic theme of the first law of thermodynamics.
Internal energy of a system is the total energy within the system. It is the sum of many types
of energies like vibrational energy, translational energy. etc. The total energy of a system is
given by:Etotal= K + V + U;
Where K = kinetic energy of the system, V = potencial energy of the system and U = internal
energy which is a summation of the translational, vibrational and rotational energies.
When the system is at rest K + V = 0; and so Etotal = U or ∆E = ∆U where ∆U is the change in
internal energy. It is an extensive property and state function. Its absolute value cannot be
determined but experimentally change in internal energy (ΔU) is given by ΔU = U 2 – U1 or
ΔU = Uf – Ui or ΣUprod – ΣUreact
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U is a product of the random motion of molecules in the system. This is primarily a function
of the temperature of the system. Two major contributors to U are (i) heat transfer into the the
system and (ii) work done on the system.
For exothermic process, ΔU = -ve, whereas for endothermic process ΔU = +ve
U depends on temperature, pressure, volume and quantity of matter.
Modes of Transference of Energy
Heat (q)
Heat is energy transferred due to temperature differences only.
1.
2.
3.
4.
Heat transfer can alter system states;
Bodies don't ``contain'' heat; heat is identified as it comes across system boundaries;
The amount of heat needed to go from one state to another is path dependent;
Adiabatic processes are ones in which no heat is transferred.
It occurs when there is a difference of temperature between system and surroundings. It is a
random form of energy and it is path dependent. Its units are joule or calorie. Heat depends
on the following factors:-temperature change; the higher the temperature difference between two bodies the higher the
difference in heat.
-amount of substance;
-nature of substance; this is represented as the specific heat capacity of the substance.
Heat is given by the fomular q = nCm ∆T. where Cm is molar heat capacity.
Also, q can be written in terms of mass as q = mc ∆T. where m is mass and c is specific heat
capacity.
HEAT CAPACITY
When heat flows into a system, provided no phase change occurs, its temperature will
increase. The increment dT by which T increases is proportional to the amount of heat flow
dq.
 dq 


The ratio  dT  is called the heat capacity [Units J/K].
Definition: The heat capacity (denoted by C) is the increment in heat dq required to increase
the temperature by an amount dT.
C is an extensive quantity but molar heat capacity Cm = C/n is an intensive quantity
(property). Usually subscript m is omitted. The quoted units then define whether C or Cm is
intended. The term specific heat capacity, i.e. heat capacity per gram, is also used.
The value of the heat capacity however depends on the circumstances under which the system
is heated. So in order to have a defined heat capacity one must specify the condition.
1). Isochoric (constant volume), heat capacity is abbreviated as “CV”. If heat is supplied at
constant volume (dV = 0), the system can do no PV work (assuming the system does no other
kind of work). Then, from first law of thermodynamics
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dq V
 dU
 q 
 U 
By definition  CV     

 T V  T V
dU
 U 

 
From an ideal gas  T V dT
CV 
dU = dq + dW; but dW = -pdV
i.e. dU = dq – pdv; dV = 0
therefore
-pdv = 0;
It follows that dU = dqv = nCvdT
Integrating we get
∆U = qv = nCv∆T
dU
 dU  CV dT
dT
2). Isobaric (constant pressure) heat capacity is abbreviated as “Cp”. If the heat is supplied at
constant pressure, usually the sample expands doing work against the external pressure. Thus
in addition to raising the temperature some extra heat is required for the expansion work.
Hence, more heat will be required to raise the system temperature under constant pressure
condition than under constant volume condition.
Thus
CP > CV
dU = dqp + dW; but dW = pdV
i.e. dU = dqp – pdv; pdV =
nRdT and qp = nCp∆T;
therefore
dU = nCpdT - nRdT;
It follows that dU = (Cp R)ndT
But Cp – R = Cv;
Substituting we get
dU = nCvdT and
∆U = nCv∆T
st
From 1 law of thermodynamics dU = dq + dw and dw= -PdV, then
dU  dq  P dV
dq  dU  P dV
dq U  P dV  H 

CP  p  


  
dT  T
P T P
For an ideal gas:
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dH
 H 

 
 T  P dT
 CP 
dH
 dH  C P dT
dT
In order to obtain U when an ideal gas is heated from temperature T1 to T2 at constant
volume, the integrated form of dU  CV dT is to be used.
T2
U   CV dT
T1
where CV is the heat capacity at constant volume.
Similarly the enthalpy change H when an ideal gas is heated from temperature T1 to T2 at
T2
H   C P dT
T1
constant pressure is given by
.
where CP is the heat capacity at constant pressure.
Note if CV and Cp are given as molar heat capacities then;
T2
U   nCV dT
T2
H   nC P dT
T1
and
.
For an ideal gas it can be shown that Cp – Cv = R
T1
HEAT CAPACITIES OF GASES
The heat capacity of a gas at constant pressure Cp exceeds the heat capacity of that gas at
constant volume, Cv, because the gas performs work on its surroundings as it expands at
constant pressure. If the volume is held constant such expansion work by the gas is
impossible.
From the kinetic molecular theory of gases, the energy, U, of one mol of ideal monatomic gas
3
U

2 RT
as a function of temperature is given as:
U 
Cv     32 R
T v
Therefore
and Cp = Cv, + R =
5
2
R since, the enthalpy of an ideal gas is
3
5
greater than its internal energy by PV (= RT). So that H = 2 RT + RT = 2 RT
H 
C p     52 R
T p
The heat capacity of a perfect gas is independent of temperature.
Predicting Heat Capacities of gases: The equipartition theorem
We can predict the heat capacities of monoatomic, diatomic or polyatomic gases based on
equipartition theorem, simply by counting the degrees of freedom. The mean energy per
1
degree of freedom per mol of monatomic ideal gas is 2 RT. Note however that the classical
equipartition theorem holds in the high-temperature limit.
The internal energy of 1 mol of monatomic ideal gas is
possess, on average,
3
2 kBT
3
2 RT.
This means that each particle
units of energy. Monatomic particles have only three translational
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degrees of freedom, corresponding to their motion in three dimensions. They possess no
internal rotational or vibrational degrees of freedom. Thus, the mean energy per degree of
freedom per mol of monatomic ideal gas is
1
2
RT.
For rigid (non-vibrating) linear molecules and rigid diatomic molecules, the internal energy U
3
2 RT
=
+ RT =
(rotational energy).
5
2 RT;
where RT is the contribution of rotational degrees of freedom
U 
Cv     52 R
T v
Therefore giving
for rigid diatomic molecules.
P = linear momentum = mv
I = moment of Inertia = Σmiri2
J = angular momentum = Iω
ω = angular velocity
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Real Gases: Temperature Dependence of Heat Capacities:
In figure 1 the dependence of constant pressure molar heat capacity, Cp, on temperature is
shown for a number of gases. In general the more complex the molecule, the greater is its
molar heat capacity and the greater is the increase of heat capacity with rising temperature. It
has been customary to express the molar heat capacity of a gas empirically as a power series
in the temperature, either as:
CP    T  T 2  ...
C  a  bT  cT 2  ...
or P
The value of constant volume molar heat capacity, Cv, can then be found from Cp at any
temperature by using the relation: Cp = Cv + R.
Example
1. One mol of methane gas originally at 298 K and 1 atm is heated at constant pressure until
the volume is doubled. Assuming the gas behaves ideally, calculate U and H for the
process. The molar heat capacity of methane at constant pressure is Cp = 22.34 + 48.1 x
10-3T J mol-1 K-1.
Answers:
10587 and 13065 J.
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Work (W)
Any other means for changing the energy of a system is called work. We can have push-pull
work (e.g. in a piston-cylinder, lifting a weight), electric and magnetic work (e.g. an electric
motor), chemical work, surface tension work, elastic work, etc. In defining work, we focus on
the effects that the system (e.g. an engine) has on its surroundings. Thus we define work as
being positive when the system does work on the surroundings (energy leaves the system). If
work is done on the system (energy added to the system), the work is negative.
Work can be defined as the energy being transferred to a system as a result of a (generalized)
force acting over a (generalized) distance.
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Consider a simple compressible substance, for example, a gas (the system), exerting a force
on the surroundings via a piston, which moves through some distance, (Figure). The work
done on the surroundings Won surr is given by;
dWon surr = force on surr × dl
multiply equation by Area/Area
dWon surr = force on surr/Area × (Area × dl); but F/A = P, and Area × dl is V
and so dWon surr = P.V
𝑊𝑜𝑛 𝑠𝑢𝑟𝑟 = −𝑃𝑒𝑥𝑡 𝑑𝑉
Example; a cylinder contains 7.0 g of nitrogen gas. How much work must be done to
compress the gas at a constant temperature of 80 °C until the volume is halved.
(Nitrogen gas is N2, with molar mass Mmol = 28 g/mol so 7.0 g is 0.25 mol of gas. The
temperature is T = 353 K. Vf = 1/2Vi.)
If the system involves gaseous substances and there is a difference of pressure between
system and surroundings. work is referred to as pressure – volume work (W).
It is convenient to use expansion work to explain the difference between work that is done
reversibly and that which is done irreversibly.
Consider a cylinder fitted with a massless, frictionless, rigid perfectly fitting piston of area A.
if external pressure is equal to Pext, at equilibrium, the pressure of the gas in the cylinder is
equal to the external pressure (Pgas = Pext), the force acting on the internal face of the piston
(F) therefore is given by F = Pext × A. when the system expands through a distance dx, work
done by the system = Pext × Adx, Adx is change in volume, therefore work done dW by the
system is = -Pextdv.
Indicator diagram shows work done for irreversible pathway for isothermal expansion work.
Work done is equal to the area of the rectangle under the graph.
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Reversible expansion (Pext = Pgas at all times) (reversible process – system is always at
equilibrium)
δq = δw = -PdV = -nRTdV/V per mole of gas Integration between states 1 and 2 gives;
𝑉2
𝑑𝑉
𝑊𝑟𝑒𝑣 = −𝑛𝑅𝑇 ∫
𝑉1 𝑉
𝑉2
𝑞 = 𝑤 = −𝑛𝑅𝑇𝑙𝑛 ( )
𝑉1
𝑃1
𝑞 = 𝑤 = −𝑛𝑅𝑇𝑙𝑛 ( )
𝑃2
In the case of reversible expansion, maximum work is done and the work is represented by
the area under the graph.
Also, for Isothermal compression, the irreversible work done can be represented as follows;
The work done for this process is represented by the shaded box. This implies that maximun
work is done for the irreversible process of compression. In the case of the reversible
pathway however, minimum work is done. Same work is done for the compression work as it
is for the expansion work in the reversible process.
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Energy transfer (changes in internal energies)
Energy transfer can be achieved in thermodynamics through various pathways. The types of
processes that lead to energy transfer in thermodynamics are listed below:
(i)
Isothermal process
(ii)
Isochoric process
(iii) Isobaric process
(iv)
Adiabatic process
(i) Isothermal work: An isothermal process is a constant temperature process. Any heat flow
into or out of the system must be slow enough to maintain thermal equilibrium For ideal
gases, if ΔT is zero, ΔU = 0 Therefore, Q = W. Any energy entering the system (Q) must
leave as work (W)
(ii) Isochoric process: An isochoric process is a constant volume process. When the volume
of a system doesn’t change, it will do no work on its surroundings. W = 0
From dU = dq + dw; w = P. dV
V is constant therefore
dV = 0; so w = P.dV = 0
this implies that
ΔU = qv
As earlier stated, q = nCv ∆T
∆U = qv = nCv ∆T
1 𝑑𝑈
𝐶𝑣 = ( )
𝑛 𝑑𝑇 𝑣
Heating a gas in a closed container is an isochoric process
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(iii) Isobaric process: An isobaric process is a constant pressure process. ΔU, W, and Q are
generally non-zero, but calculating the work done by an ideal gas is straightforward.
W = P·ΔV
Water boiling in a saucepan is an example of an isobar process
When P = constant (isobaric process) w = P∫dV = P(V2 – V1) and the 1st law takes form:
from dU = δqp - δw
U2 – U1 = qp – P(V2 – V1) or (U2 + PV2) – (U1 + PV1) = qp
qp is heat added at constant pressure.
Isothermal process
For isothermal processes, ∆T = 0, therefore,
dU = nCv∆T = 0 (internal energy of an ideal gas is a function only of T).
also, ∆H = nCp∆T = 0
from (dU = δq + δw)
=> q = -W; (meaning that any heat flow is equal in magnitude and opposite in sign to work)
Work done depends on the path, i.e. how the external pressure is
changing during the transformation. For example: Free expansion (no external pressure): w
= 0.
𝑉 𝑑𝑉
1 𝑉
𝑊𝑟𝑒𝑣 = −𝑛𝑅𝑇 ∫𝑉 2
Wirrev = -Pext ∆V
qirrev = -W
;
𝑉 𝑑𝑉
1 𝑉
𝑞 = +𝑛𝑅𝑇 ∫𝑉 2
qirrev = +Pext ∆V
Question:
A 400 L sample of H2 gas at 15.00 atm and 25 oC is allowed to expand to a final pressure of
1.00 atm. Calculate q, w, ∆U and ∆H if the gas expands (i) isothermally and reversibly, (ii)
isothermally and irreversibly.
(iv) Adiabatic process: In an adiabatic process, the system is isolated (transfers no heat),
therefore Q = 0
ΔU = Q –W; ΔU = W
=> energy change happens only through work.if the system does work, then the energy
requirements for the work come from the internal energy and consequently a drop in internal
energy of the system is expected. This also implies a drop in temperature of the system.
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Reversible work; adiabatic process
dU = dq + dw; dq = 0
dU = dw; dU = nCvdT and dw = -Pext dV
TF, nCvdT = -Pext dV = Pgas dv; (Pext=Pgas)
nCvdT = nRTdV/V; collect like terms
dT/T = -nR/Cv dV/V; R = Cp – Cv
dT/T = -((Cp – Cv)/Cv). dV/V
=> dT/T = -(Cp/Cv – 1)dV/V; Cp/Cv = γ
dT/T = -(γ – 1)dV/V;
integrating both sides we get
lnT2/T1 = -(γ – 1)lnV2/V1
take exponents of both sides
exp(lnT2/T1 = -(γ – 1)lnV2/V1)
𝑇2
𝑉2 −(𝛾−1)
=( )
𝑇1
𝑉1
 When a system expands adiabatically, W is positive (the system does work) so ΔU is
negative.
 When a system compresses adiabatically, W is negative (work is done on the system)
so ΔU is positive.
T and P
𝑇2
𝑃2 (
=( )
𝑇1
𝑃1
P and V
𝛾
𝛾−1
)
𝛾
𝛾
𝑃2 𝑉2 = 𝑃1 𝑉1
For irreversible adiabatic process, the T-V
relationship is given as follows:
Thermodynamic processes for tranformation
Process
W
Isothermal
𝑉 𝑑𝑉
𝑊𝑟𝑒𝑣 = −𝑛𝑅𝑇 ∫𝑉 2 𝑉
1
Wirrev = -Pext ∆V
Isochoric ∆U = q
Q
q = -W
𝑉 𝑑𝑉
𝑊𝑟𝑒𝑣 = +𝑛𝑅𝑇 ∫𝑉 2 𝑉
1
Wirrev = +Pext ∆V
qv = nCv∆T
∆U
∆U =
nCv∆P
0
∆H
∆H = nCp∆T
0
∆U =
nCv∆P
∆H = nCp∆T
∆U =
∆H = nCp∆T
nCv∆T
qv = nCv∆T qv = nCp∆T
Isobaric
0
Wirrev = -Pext ∆V
qp = nCp∆T
Adiabatic
W = ∆U = nCv∆T
q = nCv∆T
15
Wirrev = -Pext . dV
T2 is very important
Question:
A 400 L sample of H2 gas at 15.00 atm and 25 oC is allowed to expand to a final pressure of
1.00 atm. Calculate q, w, ∆U and ∆H if the gas expands (i) adiabatically and reversibly, (ii)
Adiabatically and irreversibly.
Cyclic process
Cyclic process A cyclic process is a sequence of processes that leaves the system in the same
state in which it started. When a system is carried through a cyclic process its initial and final
internal energies are equal. So the total internal-energy change in any cyclic process is zero.
Apply the first law for a cyclic process
dU = δq + δw = 0
q = -w
The work done by the system in a cyclic transformation is equal to the heat absorbed by the
system.
THERMOCHEMISTRY
Thermochemistry is the area of chemical thermodynamics which deals with the study of the
heat (enthalpy (qp = ∆H)) changes that accompany chemical reactions at constant pressure.
There are different types of reactions whose enthalpy change values have been determined
and internationally agreed on. Enthalpy values that are referenced are reported under standard
conditions of pressure. Values that are measured under a temperature of 298 K and a pressure
of 1 atm (101 kPa) and a concentration of 1.00 mol dm–3 are referred to as standard enthalpy
change values of reactions. The superscript (θ) is used to denote standard enthalpy values.
Enthalpy or heat of reaction is the amount of heat released or absorbed by a chemical
reaction at constant pressure (as one would do in a laboratory. We use the symbol ∆H to
indicate enthalpy.
Relationship between ∆H and ∆U
∆H = ∆U + P∆V – enthalpy - state function (since U,P,V are state functions) H2 – H1 = ΔH =
qp – change in enthalpy equals to heat added to the system at constant pressure
dU = dqp + dW;
i.e. dU = dqp – pdv;
qp = H ; dU = dH – pdv
dH = dU + pdv
dU = dH – pdV; pdV = ∆ngasRT;
therefore
∆Uo = ∆Ho - ∆ngasRT
also dH= qp = nCp∆T;
1 𝑑𝐻
+∆H indicates that heat is being absorbed in
the reaction (it gets cold) endothermic.
–∆H indicates that heat is being given off in the
16
𝐶𝑝 = 𝑛 ( 𝑑𝑇 )
𝑝
reaction (it gets hot) exothermic.
Standard Enthalpy = ∆H° (° is called a “not”).
Thermochemical equations are balanced chemical equation that include the physical states
of the reactants and products (either solid, liquid or gas) and the energy change (∆H)
accompanying the reaction.
The following are examples of thermochemical equations:
(i)
(ii)
(iii)
H2O(s) ⟶ H2O(l)
H2O(l) ⟶ H2O(g)
CH4(g)+2O2(g)→CO2(g)+2H2O(l)
∆Hfus = +6.00 kJ
∆Hvap = +40.65 kJ
ΔHC = –890kJ
Enthalpy or Heat of Reaction (ΔrxnH)
It is the change in enthalpy that accompanies a chemical reaction represented by a balanced
chemical equation.
ΔrxnH = ΣHprod. – ΣHreact.
Enthalpy of reaction expressed at the standard state conditions is called standard enthalpy of
reaction (ΔH).
Factors affecting enthalpy of reaction
(i) Physical state of reactants and products.
(ii) Allotropic forms of elements involved.
(iii) Chemical composition of reactants and products.
(iv) Amount of reactants.
(v) Temperature.
Various Forms of Enthalpy of Reaction
1. Enthalpy of Formation (ΔHf)
It is heat change when one mole of compound is obtained from Its constituent elements.
Enthalpy of formation at standard state is known as standard enthalpy of formation ΔfH° and
is taken as zero by convention. It also gives the idea of stability.
2. Enthalpy of Combustion
It is the Enthalpy change taking place when one mole of a compound undergoes complete
combustion In the presence of oxygen (ΔHc.) ΔHc is negative because process of combustion
is exothermic.
3. Enthalpy of Solution
It is the Enthalpy change when one mole of a substance is dissolved in large excess of
solvent, so that on further dilution no appreciable heat change occur.
4. Enthalpy of Hydration
It is the enthalpy change when one mole of anhydrous substances undergoes complete
combustion. It is an exothermic process.
5. Enthalpy of Fusion
It is the enthalpy change that accompanies melting of one mole of solid substance.
6. Enthalpy of Vaporisation
It is the enthalpy change that accompanies conversion of one mole of liquid substance
completely into vapours.
17
7. Enthalpy of Neutralisation
It is the enthalpy change that takes place when 1 g-equivalent of an acid (or base) is
neutralised by 1 g-equivalent of a base (or acid) in dilute solution.
Enthalpy of neutralisation of strong acid and strong base is always constant, i.e., 57.1 kJ.
[Enthalpy of neutralisation of strong acid and weak base or weak acid and strong base is not
constant and numerically less than 57.1 kJ due to the fact that here the heat is used up in
ionisation of weak acid or weak base. This is known as enthalpy of ionisation of weak acid/or
base.]
8. Enthalpy of Transition
It is the enthalpy change when one mole of the substance undergoes transition from one
allotropic form to another.
9. Enthalpy of Atomisation
It is the enthalpy change occurring when one mole of the molecule breaks into its atoms.
10. Enthalpy of Dilution
It is the enthalpy change, when one mole of a substance is diluted from one concentration to
another.
11. Enthalpy of Sublimation
It is the enthalpy change, when one mole of a solid substance sublines.
12. Lattice Enthalpy
It is the enthalpy change, when one mole of an ionic compound dissociates into its ions in
gaseous state.
Laws of Thermochemistry
1. Lavoisier Laplace Law
The enthalpy change during a reaction is equal in magnitude to the enthalpy change in the
reverse process but it is opposite in sign.
2. Hess’s Law of Constant Heat Summation
The standard enthalpy of a reaction. which takes place in several steps, is the sum of the
standard enthalpIes of the intermediate reactions into which the overall reactions may be
divided at the same temperature.
According to Hess’s law
ΔH = ΔH1 + ΔH2 + ΔH3
Applications of Hess’s law are
(a) In determination of beat of formation.
(b) In determination of heat of transition.
(c) In determination of heat of hydration.
(d) To calculate bond energies.
We will treat this in more detail.
3. Trouton’s Rule
According to this law, “The ratio of enthalpy of vaporization and normal boiling point of a
liquid is approximately equal to 88 J per mol per kelvin. i.e.,
ΔHvap / T = 88 J / mol / K
4. Dulong and Petit Law
This law states “The product of specific heat and molar mass of any metallic element is equal
to 6.4 cal/ mol/ °C. i.e.,
5. kirchhoff’s Equation
ΔCp = ΔH2 – ΔH1 / T2 – T1
and ΔCv = ΔE2 – ΔE1 / T2 – T1
18
we’ll treat this also in more details.
6. Clausius – Clapeyron Equation
– 2.303 log p2 / p1 = ΔHv / R (T2 – T1 / T1 T2)
where, ΔHv = molar heat of vaporisation.
Bond Enthalpy
It is the average amount of energy required to break one mole of bonds in gaseous molecules.
Bond Dissociation Enthalpy
The energy required to break the particular bond in a gaseous molecule is called bond
dissociation enthalpy. It is definite in quantity and expressed in kJ mol-1.
In diatomic molecule, bond dissociation enthalpy = Bond enthalpy In polyatomic molecule,
bond dissociation enthalpy ≠ Bond Enthalpy
ΔH = [sum of bond enthalpies of reactants] – [sum of bond enthalpies of products]
Factors affecting bond enthalpy
i)
Size of atoms
ii)
Electronegativity
iii)
Bond length
iv)
Number of bonding electrons
Limitations of First Law of Thermodynamics
 The limitation of the first law of thermodynamics is that it does not say anything
about the direction of flow of heat.
 It does not say anything regarding whether the process is a spontaneous process or
not.
 The reverse process is not possible. In actual practice, the heat doesn’t convert
completely into work.
The first law does not have a criteria for spontaneity and hence, the need for a second law.
Second Law of Thermodynamics
The entropy of the universe is always Increasing in the course of every spontaneous or
natural change.
Or
All spontaneous processes or natural change are thermodynamically irreversible without the
help of an external work. i.e., heat cannot flow itself from a colder to hotter body.
Clausius statement. It is impossible for heat to move by itself from a lower temperature
reservoir to a higher temperature reservoir. That is, heat transfer can only occur
spontaneously in the direction of temperature decrease. For example, we cannot construct a
refrigerator that operates without any work input.
Kelvin-Planck statement. It is impossible for a system to receive a given amount of heat
from a high-temperature reservoir and provide an equal amount of work output. While a
system converting work to an equivalent energy transfer as heat is possible, a device
converting heat to an equivalent energy transfer as work is impossible. Alternatively, a heat
engine cannot have a thermal efficiency of 100%.
The second law of thermodynamics leads to a deduction of a state function, entropy, that for
isolated systems, the equilibrium position corresponds to maximun entropy.
19
The Heat Engine
Consider a machine that does a cyclic work through a reversible isothermal process, the
machine does maximum work when it expands from Vi to Vf and work done is given by
W = -nRTlnV2/V1;
In order to return the system to its original position, reversible compression work has to be
done on the system for this to be possible. The work here is given by;
W = nRTlnV2/V1
The complete cycle requires work done to be equal to Wcyc = Wrev expansion + Wrev
compression which is equal to 0. This means that, in such a system, no meaningful work will
be done.
No useful work done.
The French engineer, Sadi Carnot was able to show that for an engine to produce continues
mechanical work it must exchange heat with two bodies at different temperatures. i.e, useful
work can be obtained only when the machine is modified to contain a low temperature
reservoir where some of the heat from the source tank can be transferred.
Now let us consider a device that uses heat transfer to do work, such a device is called a heat
engine, and one is shown schematically below. There are two types of heat engines. The
gasoline, diesel engines and jet engines are examples of internal combustion engines. the
steam turbines an example of the external heat engines that do work by using part of the heat
transfer from some source. Heat transfer from the hot object (or hot reservoir) is denoted
as QH, while heat transfer into the cold object (or cold reservoir) is QL, and the work done by
the engine is W. The temperatures of the hot and cold reservoirs are TH and TL, respectively.
Basic schematic of a heat engine
The Carnot cycle is a reversible cyclic process (or engine) made of the following four steps:
1. It starts with an adiabatic process which raises the temperature of the working
material of the engine to, say, TH
2. This is followed by an isothermal process, taking in heat from the reservoir at TH.
20
3. The next step is an adiabatic process which does some amount of work and lowers the
temperature of the material to TL.
4. The final step is isothermal, at the lower temperature TL, dumping some amount of
heat into a colder reservoir, with the material returning to the thermodynamic state at
the beginning of the cycle.
This is an idealized engine, no real engine can be perfectly reversible. The utility of the
Carnot engine is to give the framework and logic of the arguments related to the second law
of thermodynamics. We may say it is a gedanken engine. The processes involved in the
Carnot cycle may refer to compression and expansion if the material is a gas; in this case, the
cycle can be illustrated in a P-V diagram as shown below. But any other pair of
thermodynamic variables will do as well. We can think of a Carnot cycle utilizing
magnetization and magnetic field, or surface tension and area, or one could consider an
electrochemical cell.
Let the amount of heat taken in at temperature TH be QH and let the amount of heat given up
at the lower temperature TL be QL. Since this is an idealized case, we assume there is no loss
of heat due to anything like friction. Thus the amount of work done, according to the first law
is
W = QH – QL
process
B-C
Isothermal
Expanssion
C-D adiabatic
expanssion
D-A
Isothermal
compression
A-B adiabatic
compression
Work, heat, ∆U, and ∆H in the P-V diagram of the Carnot Cycle
W
Q
∆U
∆H
𝑉2
𝑉2
0
0
−𝑛𝑅𝑇 𝑙𝑛 ( )
𝑛𝑅𝑇𝑙𝑛 ( )
𝐻
𝑉1
𝑛𝐶𝑣 (𝑇𝐿 − 𝑇𝐻 )
𝑉
𝑉1
𝑉
−𝑛𝑅𝑇𝐿 𝑙𝑛 (𝑉4 )
𝑛𝑅𝑇𝐿 𝑙𝑛 (𝑉4 )
𝑛𝐶𝑣 (𝑇𝐻 − 𝑇𝐿 )
0
3
𝑛𝐶𝑝 (𝑇𝐿
− 𝑇𝐻 )
𝑛𝐶𝑣 (𝑇𝐿 − 𝑇𝐻 )
0
0
0
3
𝑛𝐶𝑣 (𝑇𝐻 − 𝑇𝐿 )
21
𝑛𝐶𝑝 (𝑇𝐻
− 𝑇𝐿 )
𝑉
Full cycle
−𝑛𝑅𝑇𝐻 𝑙𝑛 ( 2 ) −
𝑉1
𝑉4
𝑛𝑅𝑇𝐿 𝑙𝑛 (𝑉 )
3
𝑉
𝑛𝑅𝑇𝐻 𝑙𝑛 ( 2 ) +
0
0
𝑉1
𝑉4
𝑛𝑅𝑇𝐿 𝑙𝑛 (𝑉 )
3
It can be shown that
𝑉
𝑞𝑐𝑦𝑐𝑙𝑒 = 𝑛𝑅(𝑇𝐻 − 𝑇𝐿 )𝑙𝑛 (𝑉2 )
1
W = -q;
The efficiency of the engine is given by the amount of work done when a given amount of
heat is supplied divided by the amount of the heat supplied (Eff = W/ QH) (The
heat QL which is dumped into the reservoir at lower temperature is not usable for work.) The
efficiency η for a Carnot cycle is thus
𝑄𝐻 −𝑄𝐿
𝜂=
𝑄𝐻
𝑄
= 1 − 𝑄 𝐿 ; QL is always going to have a negative sign, so expression can be
𝐻
written as;
𝑄𝐻 −(−𝑄𝐿 )
𝜂=
𝑄𝐻
𝑄
= 1 + 𝑄𝐿
𝐻
The importance of the Carnot cycle is due to its idealized nature of having no losses and
because it is reversible. This immediately leads to some simple but profound consequences.
For carnot cycle, we can also show that efficiency is also given by the expression;
𝜂=
𝑇𝐻 −𝑇𝐿
𝑇𝐻
𝑇
𝑄
= 1 − 𝑇𝐿 = 1 + 𝑄𝐿
𝐻
𝐻
Subtract 1 from both sides, we get
𝑇
𝑄
1 − 𝑇𝐿 = 1 + 𝑄𝐿
𝑇
𝐻
𝑄
𝐻
− 𝑇𝐿 = 𝑄𝐿 ;
𝑄𝐿
𝑄𝐻
𝐻
𝐻
𝑇𝐿
= −𝑇
𝐻
Collecting like terms
𝑄𝐿
𝑄
𝑄
𝑄
= − 𝑇 𝐻 ; then 𝑇 𝐿 + 𝑇 𝐻 = 0
𝑇
𝐿
𝐻
𝐿
𝐻
The expression, Q/T is a state function and it is known as the ENTROPY of the system.
Entropy (S)
It is the measurement of randomness or disorder of the system. It is a state function and an
extensive property and it is defined as the reversible heat flow devided by temperature.
𝑞
𝑑𝑠 = 𝑟𝑒𝑣
𝑇
Units : jK-1 mol-1
The change in entropy during a process is mathematically given as
ΔrS° = ΣS°(products) – ΣS°(reactants) = dqrev / T = ΔH / T
Where, qrev heat absorbed by the system in reversible manner
T = temperature
22
But this entropy has been worked out for reversible process. For irreversible processes
however, if any process is irreversible then the efficiency of the engine is less than the
efficiency of Erev.
Eirrev < Erev
This implies that:.
𝑄
𝑄
1 + 𝑄 𝐿𝑖𝑟𝑟𝑒𝑣 < 1 + 𝑄 𝐿𝑟𝑒𝑣
𝐻𝑟𝑒𝑣
𝐻𝑖𝑟𝑟𝑒𝑣
𝑄𝐿𝑖𝑟𝑟𝑒𝑣
𝑄
< 𝑄 𝐿𝑟𝑒𝑣
𝑄𝐻𝑖𝑟𝑟𝑒𝑣
𝐻𝑟𝑒𝑣
𝑄
Write 𝑄 𝐿𝑟𝑒𝑣 in terms of temperature.
𝑄𝐿𝑖𝑟𝑟𝑒𝑣
𝑄𝐻𝑖𝑟𝑟𝑒𝑣
𝑄𝐿𝑖𝑟𝑟𝑒𝑣
𝑇𝐿
𝐻𝑟𝑒𝑣
𝑇
< − 𝑇𝐿
𝐻
<−
𝑄𝐻𝑖𝑟𝑟𝑒𝑣
𝑇𝐻
This implies that
𝑄𝐿𝑖𝑟𝑟𝑒𝑣
+
𝑇𝐿
𝑄𝐿𝑖𝑟𝑟𝑒𝑣
𝑄𝐻𝑖𝑟𝑟𝑒𝑣
𝑇𝐻
< 0 for this to hold;
∮ 𝑇 <0
𝐿
A cycle requires only one irreversible process to become irreversible. Consider a cycle with
two steps, step 1 is reversible while step 2 is irreversible. The entropy of the cycle could be
written as follows:
2 𝑑𝑞𝑖𝑟𝑟𝑒𝑣
∫1
𝑇
1 𝑑𝑞𝑟𝑒𝑣
+ ∫2
𝑇
<0
Inverting the limits for the reversible process changes the sign to negative and therefore;
2 𝑑𝑞
2 𝑑𝑞
− ∫1 𝑇𝑟𝑒𝑣 < 0
∫1 𝑖𝑟𝑟𝑒𝑣
𝑇
2 𝑑𝑞𝑖𝑟𝑟𝑒𝑣
∫1
𝑇
2 𝑑𝑞
2 𝑑𝑞𝑟𝑒𝑣
< ∫1
𝑇
; but
𝑑𝑞𝑟𝑒𝑣
𝑇
= 𝑑𝑆 ; so
2
< ∫1 𝑑𝑆
∫1 𝑖𝑟𝑟𝑒𝑣
𝑇
This implies that:.
ds > dqirr/T
general implications
for permissible process, ds ≥ dq/T (Clausius inequality). The Clausius inequality for a closed
system states that the process is irreversible if TdS > dQ, and the process is reversible
if TdS = dQ. three points are implied here;
i.
if TdS = dq, this is when the process is reversible, spontaneous process.
ii.
If TdS > dq, this refers to an irreversible, spontaneous process
23
iii.
TdS < dq is not allowed because it implies that the efficiency of the engine is >
than the reversible engine.
Criteria for spontaneity for isolated systems (where dq = 0)
Since dq is equal to 0 dS ≥ dq/T = 0, therefore dS ≥ 0. The criteria for spontaneity then is dS
≥ 0. If dS ≥ 0, entropy increases with spontaneous change in an isolated system.
If dS > 0: irreversible, spontaneous process.
If dS = 0 reversible spontaneous process
If dS < 0: not allowed
Entropy exchange
of an open system
In an isolated system dS represents the produced entropy dSprod and this is a good criterion for
spontaneity. Of course the requirement that the system is isolated is very restrictive and
makes the criterion as good as useless... What happens in a system that can exchange heat
with the rest of the universe? We do have entropy changes in that case, but part of them may
have nothing to do with production, because we also have to consider the heat that is
exchanged.
dS = dSprod+dSexchange
IF the process is reversible (that is completely non-spontaneous) we are dealing with δqrev so
that dSexchange =δqrev /T, but that is also what dStot is equal to (by definition). This leaves no
room for entropy production.
So we have:
Notice that this demonstrates that for non-isolated systems entropy change is not a good
criterion for spontaneity at all... In the case the heat exchange is irreversible part of the
entropy is entropy production by the system:
Generalizing the isolated, irreversible and reversible cases we may say
dS ≥ δqT
this is the Clausius Inequality
The Universe as an isolated system
If universe = isolated system
Then; dSuniv ≥ 0
dSuniv = dSsys + dSsurr ≥ 0
(this is the second law of
thermodynamics)
24
For heat flow from system to surrounding, the criteria from the second law is that Tsurr must
be lower that Tsys.
Criteria for heat flow; dSuniv must be posive
𝑄
dSuniv = dSsurr + dSsys ≥0; 𝑇
𝑠𝑢𝑟𝑟
𝑄
+ (− 𝑇 ) ≥ 0
𝑠𝑦𝑠
for dSuniv to be positive, Tsurr must be less than Tsys.
Tsurr < Tsys.
When dS = 0; system is in thermal equilibrium
Consider again the heat engine that converts all heat from heat reservoir to work. The ∆Ssys
for a cyclic process is 0, ∆Ssurr is equal to -QH/TH (since heat is removed from surrounding).
This implies that ∆Suniv = 0 + (-QH/TH);
∆Suniv is negative and this is not allowed.
Calculation of ∆Ssys for:i)
Isochoric process (constant volume)
dqrev = dqv = nCvdT;
𝑑𝑞
𝑛𝐶 𝑑𝑇
𝑑𝑆 = 𝑇𝑟𝑒𝑣 = 𝑇𝑣 ; integrating both sides we get;
𝑇 𝑛𝐶𝑣 𝑑𝑇
∆𝑆 = ∫𝑇 2
𝑇
1
𝑇
If Cv is constant; then ∆𝑆 = 𝑛𝐶𝑣 𝑙𝑛 𝑇2
1
ii)
𝑇 𝑎+𝑏𝑇+𝑐𝑇 2
If Cv varies with temperature; then ∆𝑆 = 𝑛 ∫𝑇 2
1
Isobaric process (constant pressure)
dqrev = dqv = nCpdT;
𝑑𝑆 =
∆𝑆 =
𝑑𝑞𝑟𝑒𝑣
=
𝑛𝐶𝑝 𝑑𝑇
𝑇
𝑇
𝑇2 𝑛𝐶𝑝 𝑑𝑇
∫𝑇
𝑇
1
𝑇
𝑑𝑇
; integrating both sides we get;
𝑇
If Cv is constant; then ∆𝑆 = 𝑛𝐶𝑝 𝑙𝑛 𝑇2
1
iii)
𝑇 𝑎+𝑏𝑇+𝑐𝑇 2
If Cv varies with temperature; then ∆𝑆 = 𝑛 ∫𝑇 2
𝑇
1
Isothermal process (constant temperature)
dqrev = -dWrev = -(-PgasdV) = -(-nRT/V)dV
𝑑𝑞
𝑛𝑅𝑇
𝑛𝑅
𝑑𝑆 = 𝑇𝑟𝑒𝑣 = 𝑇𝑉 𝑑𝑉 = 𝑉 𝑑𝑉 integrating, we get;
2 𝑑𝑉
∆𝑆 = 𝑛𝑅 ∫1
𝑉
𝑑𝑇
𝑉
; ∆𝑆 = 𝑛𝑅𝑙𝑛 𝑉2
1
Suppose we are dealing with a process in which everything is changing (i.e. the
temperature and the pressure) and the whole process might require 2 separate
steps (e.g isothermal and isochoric) to complete, then, ∆S is given by:
∆𝑆 = 𝑆1 + 𝑆2
25
∆𝑆 = 𝑛𝑅𝑙𝑛
iv)
𝑉2
𝑉1
+ 𝑛𝐶𝑝 𝑖𝑛
𝑇2
𝑇1
Adiabatic process (reversible)
𝑑𝑞
For adiabatic processes, dqrev = 0, 𝑑𝑆 = 𝑇𝑟𝑒𝑣 = 0
Gas (T1 & V1)
(T2 & V2)
In order to move from the initial conditions of the gas to the second state, the first
step can be an isothermal process (constant temperature).
Gas (T1 & V2)
2 𝑑𝑉
𝑉
∆𝑆 = 𝑛𝑅 ∫1 𝑉 ; ∆𝑆 = 𝑛𝑅𝑙𝑛 𝑉2
1
The second step would then be an isobaric process and T1 changes to T2. We get:
𝑇
∆𝑆 = 𝑛𝐶𝑝 𝑖𝑛 𝑇2
1
Entropy for the whole process is
∆𝑆 = 𝑆1 + 𝑆2
𝑉
𝑇
∆𝑆 = 𝑛𝑅𝑙𝑛 𝑉2 + 𝑛𝐶𝑣 𝑖𝑛 𝑇2
1
v)
𝑑𝑞
vi)
1
Reversible phase change
For phase changes temperature remains constant throughout duration of change
(e.g. melting of ice).
2
2 𝑑𝑞
∫1 𝑑𝑆 = ∫1 𝑇𝑟𝑒𝑣 temperature is constant so;
𝑛∆𝐻
∆𝑆 = 𝑇𝑟𝑒𝑣 = 𝑇
Mixing of innert gasses at constant T & P
Consider an insulated rigid container of gas separated into two halves by a heat
conducting partition so the temperature of the gas in each part is the same. One
side contains air, the other side another gas, say argon, both regarded as ideal
gases. The mass of gas in each side is such that the pressure is also the same. The
entropy of this system is the sum of the entropies of the two parts:
∆𝑆𝑠𝑦𝑠 = 𝑆1 + 𝑆2
Suppose the partition is taken away so the gases are free to diffuse throughout the
volume. For an ideal gas, the energy is not a function of volume, and, for each
gas, there is no change in temperature. (The energy of the overall system is
unchanged, the two gases were at the same temperature initially, so the final
temperature is the same as the initial temperature.) The entropy change of each
gas is thus the same as that for a reversible isothermal expansion from the initial
specific volume Vi to the final specific volume, Vf . For a mass
of ideal gas,
𝑉2
the entropy change is ∆𝑆 = 𝑛𝑅𝑙𝑛 𝑉 (V2 = Vtot for both gases). The entropy change
1
of the system is
𝑉
𝑉
∆𝑆 = 𝑆1 + 𝑆2 = 𝑛1 𝑅𝑙𝑛 𝑉𝑡𝑜𝑡 + 𝑛2 𝑅𝑖𝑛 𝑉𝑡𝑜𝑡
1
1
Δ S > 0, Increase in randomness, heat is absorbed
Δ S < 0, Decrease in randomness, heat is evolved.
Entropy of even elementary substances are not zero.
Joule-Thomson Effect
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The phenomenon of cooling of a gas when it is made to expand adiabatically from a region of
high pressure to a region of extremely low pressure is known as Joule-Thomson effect. This
effect is zero when an ideal gas expands in vacuum.
[When an ideal gas undergoes expansion under adiabatic condition in vacuum, no change
takes place in its internal energy, i.e., (∂E / ∂V)T = 0 where, (∂E / ∂V)T is called the Internal
pressure.]
Joule-Thomson Coefficient
The number of degrees of temperature change produced per atmospheric drop in pressure at
constant enthalpy when a gas is allowed to expand through a porous plug is called JouleThomson coefficient. It is given as
μ = dT / dp
where, μ = Joule-Thomson coefficient
dT = change in temperature
dp = change in pressure.
Inversion Temperature
The temperature below which a gas becomes cooler on expansion is known as the inversion
temperature. It is given as
Ti = 2a / Rb
where, a and b = van der Waals’ constant
At inversion temperature Ti, the Joule Thomson coefficient μ = 0, i.e., the gas is neither
heated nor cooled.
Entropy Change During Phase Transition
The change of matter from one state to another state is called phase transition.
The entropy changes at the time of phase transition:
Spontaneous Process
The physical or chemical process which proceeds by its own in a particular direction under
given set of conditions without outside heir is called spontaneous process. It cannot be
reversed.
All natural processes are spontaneous process.
Spontaneous process where no initiation is needed
(i) Sugar dissolves in water.
(ii) Evaporation of water.
(iii) Nitric oxide (NO) reacts with oxygen.
Spontaneous process where some initiation is required
(i) Coal keeps on burning once initiated.
(ii) Heating of CaCO3 to give calcium oxide and CO2 is initiated by heat.
Enthalpy Criterion of Spontaneous Process
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All the processes which are accompanied by decrease of energy (exothermic reactions,
having negative value of ΔH) occur spontaneously. It fails when some endothermic reactions
occur spontaneously.
Entropy Criterion of Spontaneous Process
A process is a spontaneous if and only if the entropy of the universe increases.
For a process to be Spontaneous
(ΔSuniv > 0 or ΔSsys + ΔSsurr > 0)
At equilibrium state, ΔS = 0,
Limitations of ΔS criterion and need for another term
We cannot find entropy change of surroundings during chemical changes. So we need
another parameter for spontaneity viz Gibbs’ energy of system (G).
Gibbs Energy or Gibbs Free Energy
It is the energy available for a system at some conditions and by which useful work can be
done. It is a state function and extensive property.
Mathematically,
G = H – TS
Change in Gibbs energy during the process 1S given by Gibbs Helmholtz equation.
(ΔG = G2 – G1 = ΔH – TΔS)
where, ΔG = Gibbs free energy
H = enthalpy of system
TS = random energy
ΔGsystem = – TΔStotal
The Gibbs energy criterion of spontaneity
ΔG > 0, process is non-spontaneous
ΔG < 0, 0, process is spontaneous
ΔG = 0, process is in equilibrium state
Effect of Temperature on Spontaneity
Now an exothermic reaction which is non-spontaneous at high temperature may become
spontaneous at low temperature. Similarly, endothermic reactions which are non-spontaneous
at low temperature may become spontaneous at high temperature.
Standard Free Energy Change (Δ G)
It is the change in free energy which takes places when the reactants are converted into
products at the standard states, i.e., (1 atm and 298 K) where, ΔG°f = standard energy of
formation Standard energy of formation of all free elements is zero.
Gibbs Energy Change and Equilibrium
Criterion for equilibrium,
⇒ We also know that
Relation between ΔG° and EMF of the Cell
Third Law of Thermodynamics
This law was formulated by Nernst in 1906. According to this law, “The entropy of a
perfectly crystalline substance at zero K or absolute zero is taken to be zero”. We can find
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absolute entropies of pure substances at different temperature. where, Cp = heat capacities T
= temperature between 0 K and T K
This law is on1y applicable for perfectly crystalline substances. If there is imperfection at 0
K, the entropy will be larger than zero.
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