Thermodynamic terms • Properties of the system directly related to the fundamental laws of

thermodynamics are known as fundamental properties.

• A system in thermodynamics refers to that part of universe in which
observations are made. Examples: internal energy and entropy
• remaining universe constitutes the surroundings. • The Internal Energy (U):When we talk about our chemical system losing
Thermodynamics • In an open system, there is exchange of energy and matter between or gaining energy, we need to introduce a quantity which represents
system and surroundings. the total energy of the system, the internal energy, U of the system,
By • In a closed system, there is no exchange of matter, but exchange of energy which may change, when
is possible between system and the surroundings. heat passes into or out of the system,
Dr.puspanjali Mishra • In an isolated system, there is no exchange of energy or matter between work is done on or by the system,
Asst.Prof. in Chemistry
the system and the surroundings. matter enters or leaves the system.
PMEC, Berhampur • The State of the System: The state of a thermodynamic system is described • Entropy (S) is a thermodynamic quantity whose value depends on the
by its measurable or macroscopic (bulk) properties such as its pressure (p),
volume (V), and temperature (T ) as well as the composition of the system. physical state or condition of a system. In other words, it is a
thermodynamic function used to measure the randomness or disorder.

Module–II: FREE ENERGY IN CHEMICAL EQUILIBRIA (9 Hours) • Properties of the system that have specific relations and include
Concepts of Entropy, Entropy in Physical and Chemical Changes, Free Energy Thermodynamic Properties: combinations of measured and derived properties are known as derived
Concepts, Gibbs Helmholtz Equation, Free Energy Change and Criterion of
Spontaneity of Chemical Equation and Chemical Equilibrium, Van’t Hoff Equation.
Thermodynamic properties are defined as characteristic features of a properties.
L1-Terminology of thermodynamics, First law of thermodynamics, Isothermal & adiabatic system, capable of specifying the system’s state. Thermodynamic Examples: Enthalpy, Gibbs, free energy.
processes, properties may be extensive or intensive. • Enthalpy (H): H = U + PV, ∆H = ∆U + P∆V
L2-concept of Entropy, Entropy in Physical and Chemical Changes, entropy changes in An extensive property is a property whose value depends on the ∆H is negative for exothermic reactions which evolve heat during the
reversible and irreversible processes.
L3-Entropy changes of ideal gases.
quantity or size of matter present in the system. For example, mass, reaction and ∆H is positive for endothermic reactions which absorb heat
L4-Entropy changes in solids and liquids, entropy changes in phase changes. volume, internal energy, enthalpy, heat capacity, etc. are extensive from the surroundings.
L5-Entropy of mixing, Entropy of a reaction, Numerical problems. properties. • Free Energy(G) is the energy available to do useful work and a property
L6-Free Energy Concepts, Variation of free energy change with temperature and pressure of Those properties which do not depend on the quantity or size of that provides a convenient measure of driving force of the reaction.
an ideal gas, Gibb’s Helmoltz equation.
L7-Free Energy Change and Criterion of Spontaneity of Chemical Equation and Chemical matter present are known as intensive properties. For example The Gibb’s free energy change ∆G is related to enthalpy change ∆H and
Equilibrium temperature, density, pressure etc. are intensive properties. entropy change ∆S by the relation ∆G = ∆H – T∆S.
L8-Free energy and chemical equilibrium, Numerical problems. • The fundamental properties and derived properties cannot be
L9-Maxwell’s relations, Van’t Hoff Equation.
measured.
 • It is definite that this cannot be explained by considering the decrease in enthalpy
alone. There are some additional contributory factors like entropy to the
spontaneity of these reactions.
• Entropy is a property that is used to express the extent of disorder or randomness
of a system and may be defined as the thermodynamic property that is a measure
Thermodynamics of the randomness or disorder of the molecules of a system.
• Entropy is expressed by the symbol ‘S’. Like internal energy and enthalpy, entropy is
also a state function and therefore change in entropy depends only on the initial
Thermodynamic Process: By and final states of the system.
A system undergoes a thermodynamic process when there is some energetic • Change in entropy is given by
Dr.puspanjali Mishra ΔS = S (final state)−S (initial state).
change within the system that is associated with changes in pressure,
Asst.Prof. in Chemistry For a reversible process at equilibrium, the change in entropy is expressed as
volume and internal energy. 𝑑𝑞
PMEC, Berhampur 𝑑S = 𝑟𝑒𝑣 , Unit of entropy is Joule per Kelvin (J K -1).
There are four types of thermodynamic processes that have their unique 𝑇
• Thus, entropy change may be defined as the amount of heat absorbed by the
properties, and they are: system in a reversible manner divided by the absolute temperature at which the
Adiabatic Process – A process where no heat transfer into or out of the heat is absorbed.
system occurs.
Isochoric Process – A process where no change in volume occurs and the
system does no work. Module–II: FREE ENERGY IN CHEMICAL EQUILIBRIA (9 Hours) • A process occurs because the randomness or entropy increases.
Isobaric Process – A process in which no change in pressure occurs. Concepts of Entropy, Entropy in Physical and Chemical Changes, Free Energy
• For a spontaneous process in an isolated system, entropy change is positive, that is, ΔS > 0.
Isothermal Process – A process in which no change in temperature occurs. Concepts, Gibbs Helmholtz Equation, Free Energy Change and Criterion of
Spontaneity of Chemical Equation and Chemical Equilibrium, Van’t Hoff Equation. • However, if the system is not isolated, the entropy changes of both the system and
Cyclic Process- It is a process in which the final state of the system is equal to L1-Terminology of thermodynamics, First law of thermodynamics, Isothermal & adiabatic
the initial state. As we know, change in internal energy is a state function, so, processes, surroundings have to be taken into account. Then
in this case, ∆U = 0. L2-concept of Entropy, Entropy in Physical and Chemical Changes, entropy changes in ΔStotal = ΔSsystem + ΔSsurroundings ≥ 0
reversible and irreversible processes.
L3-Entropy changes of ideal gases. For a reversible process, ΔS total must be zero. ΔStotal = ΔSsystem + ΔSsurroundings = 0
L4-Entropy changes in solids and liquids, entropy changes in phase changes.
For a irreversible (spontaneous) process, ΔS total must be positive, that is,
L5-Entropy of mixing, Entropy of a reaction, Numerical problems.
L6-Free Energy Concepts, Variation of free energy change with temperature and pressure of • ΔStotal = ΔSsystem + ΔSsurroundings > 0.
an ideal gas, Gibb’s Helmoltz equation.
L7-Free Energy Change and Criterion of Spontaneity of Chemical Equation and Chemical • The system and the surroundings together constitute the universe.
Equilibrium
• Therefore, for a spontaneous change, ΔS universe > 0.
L8-Free energy and chemical equilibrium, Numerical problems.
L9-Maxwell’s relations, Van’t Hoff Equation. • Hence, according to the second law of thermodynamics ‘in any natural process the energy
of the universe is conserved but the entropy of the universe always increases’.

Entropy is a state function

Entropy In the definition of entropy, the quantity qrev is not a state function and depends on
• It is known that a state with minimum energy is the most stable state of a the path of the reaction, whereas the (dqrev /T ) is a perfect differential. This can be
system. shown as follows:
• Therefore, in this universe there is a natural tendency of all systems to Suppose 1 mole of an ideal gas is undergoing a reversible expansion.
stabilise by acquiring minimum energy. From the first law, dqrev = dU − dw
• It was believed that all processes in which energy decreases (exothermic, For work expansion, − dw = PdV hence dqrev = dU + PdV -----------(1)
having negative value of ΔH) occur spontaneously.
𝜕𝑈
• Hence, decrease in enthalpy is the driving force behind the spontaneous As P=RT/V and Cv=( ) or dU= Cv dT,
𝜕𝑇 v
processes.
𝑅𝑇
• However, there are many endothermic reactions (ΔH positive) that are dqrev = Cv dT + dV ------------------(2)
𝑉
spontaneous.
• For example, Water absorbs energy and is evaporated (endothermic dqrev = Cv dT+
𝑅𝑇
dV
𝑉
process) but it is spontaneous in nature.
As dqrev is not an exact differential, it can not be integrated. This is confirmed by
• H2O (l) → H2O (g) ΔH = + 44.0 kJ mol-1 𝑅𝑇
observing the right hand side of Eq. (2) where dV can not be evaluated unless
• Melting of ice is endothermic but spontaneous. 𝑉
we specify the path or in other words the relation between T and V is known.
The value of V will be different for different values of T. Thus, dqrev depends on the If the entropy, volume and temperature of the system in the initial state are S1,V1 and T1 and
path by which the gas is expanded. in the final state are S2, V2 and T2 respectively, then integrating Eq. (3) between these limits
we get
Dividing Eq. (2) by T on both sides we get 𝑆2 2 dq 𝑇2 C 𝑉2 𝑅
𝑆1
𝑑𝑆= 1 rev = 𝑇1 v dT + 𝑉1 dV
𝑇 𝑇 𝑉
dqrev Cv RT 1 Cv
𝑇
=
𝑇
𝑑𝑇 +
𝑉 𝑇
dV =
𝑇
𝑑𝑇 +
𝑅
𝑉
dV = Cv ln T+ R ln V+constant. Thermodynamics Assuming CV to be independent of temperature over the temperature range considered, for 1
mole of the gas, we obtain
dqrev
It follows that can be evaluated; hence, dqrev/T is an exact differential. T2 V
𝑇
S2 – S1= ∆𝑆=Cv ln + R ln 2, ------------------------------(4)
Hence, we may conclude that entropy change dS is a perfect differential and hence By T1 V1
entropy S is a state function.
T2 V
Thus, if a system changes from the initial state 1 to the final state 2, the entropy For n moles, ∆𝑆=nCv ln +nR ln 2 ------------------------------(5)
Dr.puspanjali Mishra T1 V1
change is expressed by Asst.Prof. in Chemistry An alternate form of Eq. (4 and 5) involving pressure terms can be obtained in the following
2 2 dq 2 2 dq
1
𝑑𝑆= 1 rev or 1 𝑑𝑆=S2 – S1= ∆𝑆= 1 rev, PMEC, Berhampur manner. If P1 is the pressure of the system in the initial state and P2 in the final state, then
𝑇 𝑇
where S1 and S2 are the entropies of the system in initial and final states from the gas equation
P1V1 P2V2 V PT
respectively. = or 2 = 1 2
T1 T2 V1 P2T1

Module–II: FREE ENERGY IN CHEMICAL EQUILIBRIA (9 Hours) T2 V T PT T T P

S2 – S1= ∆𝑆=Cvln + R ln 2 =Cvln 2+ R ln 1 2 =Cvln 2+ R ln 2 +R ln 1
Concepts of Entropy, Entropy in Physical and Chemical Changes, Free Energy T1 V1 T1 P2T1 T1 T1 P2
Concepts, Gibbs Helmholtz Equation, Free Energy Change and Criterion of T P T P
Spontaneity of Chemical Equation and Chemical Equilibrium, Van’t Hoff Equation. Or, ∆𝑆 = (Cv+ R) ln 2 − R ln 2 = CP ln 2 − R ln 2.
T1 P1 T1 P1
L1-Terminology of thermodynamics, First law of thermodynamics, Isothermal & adiabatic
T2 V T P
processes, For 1 mole, ∆𝑆=Cvln + R ln 2 = CP ln 2 − R ln 2 ------------------------------(6)
L2-concept of Entropy, Entropy in Physical and Chemical Changes, entropy changes in T1 V1 T1 P1
reversible and irreversible processes. T2 V T P
And for n moles, ∆𝑆= nCvln + nR ln 2 = nCP ln 2 − nR ln 2 ------------------------------(7)
L3-Entropy changes of ideal gases. T1 V1 T1 P1
L4-Entropy changes in solids and liquids, entropy changes in phase changes. From Eqs (4), (5), (6) and (7), entropy change for an ideal gas can be calculated.
L5-Entropy of mixing, Entropy of a reaction, Numerical problems. These equations take different forms under different conditions.
L6-Free Energy Concepts, Variation of free energy change with temperature and pressure of
(i) For isothermal process When T1 = T2,
an ideal gas, Gibb’s Helmoltz equation.
V P
L7-Free Energy Change and Criterion of Spontaneity of Chemical Equation and Chemical ∆ST= R ln 2 = R ln 1 ,
V1 P2
Equilibrium V P
For n moles, ∆ST = nR ln 2 = nR ln 1 ------------------------------(8)
L8-Free energy and chemical equilibrium, Numerical problems. V1 P2
L9-Maxwell’s relations, Van’t Hoff Equation. As for expansion, V2 > V1 and P1 > P2 , ∆ST is positive (entropy increases).
And for isothermal contraction, ∆ST is negative (entropy decreases).

Calculation of Entropy For an isobaric process: P1= P2

1. Entropy changes for an ideal gas: T T
∆SP = CP ln 2, and for n moles, ∆SP = nCP ln 2 ------------------------------(9)
Consider a system consisting of n moles of an ideal gas occupying a volume V at a pressure T1 T1
P and temperature T. If dqrev amount of heat is absorbed by the system reversibly, then the For an isochoric process When V1 = V2,
increase in entropy of the system dS is given by
T2
dq ∆SV = Cvln ------------------------------(10)
dS= rev -------------(1)
𝑇
T1
• From the first law of thermodynamics, for a reversible process in which only P-V work is Examples:
involved, dqrev = dU − dw 1. Calculate the change in entropy accompanying the isothermal expansion
𝑅𝑇
of 3 moles of an ideal gas at 35oC until its volume has increased 10 times.
• For work expansion, − dw =PdV , hence dqrev = dU + PdV = Cv dT + dV -----------(2)
𝑉

𝜕𝑈
V2=10V1, R= 8.314 J mol-1K-1, n=3
• As P=RT/V and Cv=( )v or dU= Cv dT.
𝜕𝑇
V2
dqrev Cv C ∆ST = nR ln =3x 8.314 x ln 10 = 57.44 J/K
• dS=
𝑅𝑇 𝑅
= dT + dV = v dT + dV ------------------(3) V1
𝑇 𝑇 𝑉𝑇 𝑇 𝑉
2. Calculate the entropy change involved in expanding 1 mole of an ideal gas from 20 L at
2.5 atm pressure to 75 L at 1 atm pressure.( CP = 7.42 cal deg−1mol−1 ). dq mC dT
Entropy change dS= rev =
𝑇 𝑇
𝑆2 2 dqrev 2 mC dT
𝑆1
𝑑𝑆= 1 = 1
𝑇 𝑇
Thermodynamics T
S2-S1= ∆𝑆= mC ln 2 --------------(11)
T1
Assuming that S remains constant within the temperature limits T1
By and T2.
Similarly when the molar heat capacity value is known,
Dr.puspanjali Mishra
T
∆𝑆= nC ln 2, Where n= number of moles and C= molar heat
Asst.Prof. in Chemistry
T1
PMEC, Berhampur
capacity.

3. Calculate the entropy change when 3 moles of an ideal gas (Cv = 7.88 cal deg–1 mol–1) Module–II: FREE ENERGY IN CHEMICAL EQUILIBRIA (9 Hours)
Entropy change during phase transition
are heated from a volume of 200 L at 50 °C to a volume of 300 L at 150 °C (CV = 7.88 cal Concepts of Entropy, Entropy in Physical and Chemical Changes, Free Energy
deg−1mol−1 ) Concepts, Gibbs Helmholtz Equation, Free Energy Change and Criterion of When matter changes from one phase to another like from the solid
Spontaneity of Chemical Equation and Chemical Equilibrium, Van’t Hoff Equation. phase to the liquid phase or from liquid to gaseous phase, then it is
L1-Terminology of thermodynamics, First law of thermodynamics, Isothermal & adiabatic termed as the phase transition. Melting of solid or vaporisation of
processes,
L2-concept of Entropy, Entropy in Physical and Chemical Changes, entropy changes in liquid occurs at constant temperature as the two phases are in
reversible and irreversible processes. equilibrium at all times. The entropy change for these reactions may
L3-Entropy changes of ideal gases. be calculated as
L4-Entropy changes in solids and liquids, entropy changes in phase changes. dq
L5-Entropy of mixing, Entropy of a reaction, Numerical problems. dS= rev
𝑇
L6-Free Energy Concepts, Variation of free energy change with temperature and pressure of
an ideal gas, Gibb’s Helmoltz equation.
where q is the heat evolved or absorbed during transition and T is
L7-Free Energy Change and Criterion of Spontaneity of Chemical Equation and Chemical the temperature. For transformation of 1 mole of a substance at
Equilibrium constant pressure, qrev is equal to the molar enthalpy change for
L8-Free energy and chemical equilibrium, Numerical problems. that transformation, that is, qrev = ΔHtrans. Let us consider some
L9-Maxwell’s relations, Van’t Hoff Equation.
examples:

Practice problems Entropy change in heating a solid or a liquid

1. 1 mole of an ideal gas (CV = 12.55 J K-1 mol-1) is transferred from 298 K and 2 atm to 233 K and Entropy of fusion Entropy of fusion is the entropy change during
0.4 atm. Calculate the value of entropy change in the system. When a solid or a liquid is heated without producing change in state,
the conversion of 1 mole of the solid substance into liquid form at its
[Ans 10.294 J K-1 mol-1] melting point. For example, when ice
the heat change is given by
2. 1 mole of an ideal gas (CV = 12.471 J K-1 mol-1) is heated from 300 to 600 K. Calculate melts, Water (s)⇌ Water (l)
entropy change when (a) volume is kept constant, (b) pressure is kept constant. Quantity of heat absorbed = mass × heat capacity × temperature rise
The change in entropy is given by
[Ans (a) 8.645 J K-1 mol-1(b) 14.408 J K-1 mol-1] ∆𝐻𝑓𝑢𝑠𝑖𝑜𝑛
= number of moles × molar heat capacity × temperature rise. Swater-Sice= ∆𝑆𝑓𝑢𝑠𝑖𝑜𝑛= , where ∆𝐻𝑓𝑢𝑠𝑖𝑜𝑛is the enthalpy of
3. Calculate the entropy change when 5 moles of an ideal gas undergoes isothermal expansion at 𝑇𝑓
20 °C from a pressure of 10 atm to a pressure of 2 atm. [Ans 66.909 J K-1 ] Thus, if m g of a solid or a liquid of specific heat or heat capacity C fusion and 𝑇𝑓 is the fusion temperature.
4. Calculate the entropy change when 2 mole of an ideal gas is allowed to expand from a volume of Entropy of vaporisation Entropy of vaporisation is the change in
is heated reversibly through temperature dT, the amount of heat
1 L to a volume of 10 L at 27 °C. [Ans 38.294 J K-1 ]
entropy when 1 mole of a liquid changes into vapor at its boiling
5. Calculate the entropy change when 1 mole of an ideal gas is heated from 20 °C to 40 °C at a
absorbed is given by point. The entropy of vaporisation of a liquid at its boiling point is
∆𝐻𝑣𝑎𝑝
constant pressure. The molar heat at constant pressure of the gas over this temperature range is dqrev = m C dT ∆𝑆𝑣𝑎𝑝= , where ∆𝐻𝑣𝑎𝑝is the enthalpy of vaporisation and 𝑇𝑏 is
𝑇𝑏
6.189 cal deg-1. the boiling point.
Examples
1. Calculate the molar entropy of vaporisation of 1 mole of water at
100°C. Latent heat of vaporisation of water at 100 °C is 540 cal/g.
2. ΔSvap of acetone is 93.0 J K–1 mol–1. If the boiling point of acetone is
56°C, calculate the heat required to vaporise 1 g of acetone.
3. Calculate the entropy change accompanying the freezing of one
mole of steam at 100oC to ice at -20o C; given that
∆𝐻𝑓𝑢𝑠𝑖𝑜𝑛=6.0kJ/mol , Cice=36.82 J K-1mol-1 & Cwater=75.31 J K-1mol-1.

Standard Entropies:
Entropy of 1 mole if a substance in pure state at I atm pressure and 298 K , is
termed as Standard Entropy of the substance and is denoted by S0.
In a reaction involving the reactants and the products at their standard states,
the standard entropy change is given by
∆S0 = S0products - S0reactants