Thermodynamics

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Chemical Thermodynamics and Energetics
Prof.Jadhav Swapnil. S.
Thermo = Heat dynamics = science of motion or transfer
Introduction-
Energy – is defined as its capacity to do work. A system with higher energy can do more work.
Different forms of energy
Kinetic Energy Energy due to motion.
Potential Energy Energy due to position and state of matter.
Heat Energy Thermal energy
Mechanical Energy Heat energy converted into mechanical energy in engine.
Electrical Energy Energy produced in the galvanic cell.
Chemical Energy Energy produced in the chemical reaction.
All these forms of energy are intercovertible.
Thermodynamics:-
“Thermodynamics is branch of science that deals with the different forms of energy,
qualitative relation between them and energy changes occur in physical and chemical processes.”
Scope of Thermodynamics:-
1) It provides information regarding the feasibility of a chemical reaction.
2) It tells us whether a particular reaction is possible under the given set of condition. (T, V, C)
3) It tells us the direction in which the reaction tends to occurs.
4) Important laws of physical chemistry (ie. lowering of vapour pressure, phase rule, distribution law)
can be derived from the law of thermodynamic.
Limitations of Thermodynamics:-
1) Thermodynamics does not gives information regarding the rates of physical or chemical process.
Ie. It does not bother about time factor.
2) It is concerned only with the initial and final state of system and not the mechanism of process.
3) Thermodynamics is applicable to macroscopic system consisting matter in bulk.
4) It does not applicable to microscopic system of individual atoms or molecule.
5) It ignores the internal structure of atoms and molecules.

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Basic concepts in Thermodynamics:-
1) System:- The part of universe which is under thermodynamic
study is called system.
2) Surrounding:- The remaining part of universe excluding
system is surroundings.
3) Boundery:- The real or imaginary surface separating the
system from the surroundings.
Types of system:-
Open system Closed system Isolated system
1) Open system:-
2) Closed system:-
3) Isolated system:-
Boundary
A system that can exchange both the matter and energy with its surroundings
is called an open system.
Eg. Hot tea in an open cup.
A system that can exchange only energy but not matter with its surroundings
is called a closed system.
Eg. Hot tea in a cup covered with saucer.
A system that cannot exchange either energy or matter with its surroundings
is called an isolated system.
Eg. Hot tea in thermos flask.

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Properties of system:-
Extensive Property:-
The property which depends upon the quantity
of matter is known as Extensive Property.
Eg. Mass, Volume, Work, Entropy, Enthalpy,
Surface area, Internal energy etc.
Note. (1) Extensive Properties are additives. (2) Intensive Properties are not additives. (3) The ratio
of two Extensive Properties is an intensive property.
State and state function:-
A known set of measurable properties like temperature, pressure,
volume define a state of system.
When these variables are changes, then system is said to be in different state.
i.e
State function:-
Intensive Property:-
The property which depends upon the quantity
of matter is known as intensive Property.
Eg. Temperature, Pressure, Concentration, Specific
gravity etc.
P1 , V1 , T1 P2 , V2 , T2 Change in state
State- I State- II
Any property of the system whose value depends only on the current state of
system and does not depend on the path fallowed by system is called state function.
ΔP
ΔV
A (P1 ,V1 )
B(P2 ,V2 )
Consider a system at point ‘A’ P = P1 , V = V1
After condition pressure and volume are changed,
at point ‘B’ P = P2 , V = V2
The change in pressure ΔP = P2 — P1 and The
change in volume ΔV = V2 — V1 does not depend
upon the path fallowed by system from point ‘A’ to
‘B’.
Hence ΔP and ΔV are state function.
P
V
Thermodynamic Equilibrium:- A system is said to be in thermodynamic equilibrium when
the state functions of the system do not change with time.
1) Thermal Equilibrium 2) Chemical Equilibrium 3) Mechanical Equilibrium
Temperature constant Chemical composition constant Mechanical properties (or Pressure) constant

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Types of processes:-
Process: - The transition from one state to another state is called process.
Isothermal process:-
A process in which temperature of the system remain constant throughout the process is
called an isothermal process.
For isothermal process, ΔT = 0 , ΔU = 0 U = internal energy
Isobaric process:-
A process in which pressure of the system remain constant throughout the process is
called an isobaric process.
For isobaric process, ΔP = 0 This process is carried out in open vessel.
Isochoric process:-
A process in which volume of the system remain constant throughout the process is called
an isochoric process.
For isochoric process, ΔV= 0 This process is carried out in closed vessel.
Adiabatic process:-
A process in which no heat can exchange between system and surroundings is called an
adiabatic process.
For adiabatic process, q = 0
Isothermal process Adiabatic process
A process in which temperature of the system
remain constant throughout the process is called
an isothermal process.
A process in which no heat can exchange between
system and surroundings is called an adiabatic
process.
Temperature of the system remains constant.
Ie ΔT = 0
Temperature of the system remains changes.
Ie ΔT ≠ 0
Exchange of heat between system and
surroundings.
No exchange of heat between system and
surroundings.
Internal energy of the system remains constant.
Ie ΔU = 0
Internal energy of the system remains changes.
Ie ΔU ≠ 0
System is not thermally isolated. System is thermally isolated.
Eg. Fusion of ice. Eg. Expansion of gas in vacuum.

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Reversible process:-
A process in which the driving force is infinitesimally greater than the opposing
force and direction of the process can be reversed by slight increase in the opposing force
is called Reversible process.
Features of Reversible process:-
(1) Driving force and opposing force are slight different from each other. (2) It takes infinite time.
(3) It takes place in infinite number of steps. (4) The process can be reversed at any point.
(5) Reversible process is occurs at a very slow rate. (6) Maximum work is obtained.
Nature of heat and work:-
Change of energy in closed system occur (1) by transfer of energy as
work and (2) by transfer of energy as heat
Work (W):- In mechanics, work is product of force and displacement, W = f × d
where f = force in the direction of displacement d = displacement
Pressure – volume work
Work
Maximum work
Pressure – volume work:-
The work is done due to expansion or compression of a gas against an external
opposing pressure P is called Pressure – volume work.
Expression:-
Consider a gas having pressure( P), Volume (V1 ) and
Temperature (T) enclosed in a cylinder fitted with frictionless
Piston having area (A) as shown in fig.
As gas expand against external pressure Pext , the
Piston moves in upward direction through a distance ‘d’
Then work done (W) by gas is,
W = Opposing force × displacement W = — f × d …... (1)
Pext
P
A
Ideal Gas

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Pressure is force per unit area
Pext = f / A ===> f = Pext × A …….…. (2)
From eq. (1) and (2), W = — Pext × A × d ………… (3)
But, A × d = ΔV = V2 — V1 V1 = initial volume V2 = final volume
Eq. (3) becomes,
W = — Pext × ΔV W = — Pext × (V2 — V1 )
Or ……….. (4)
Free Expansion:- Expansion against zero opposing force is called Free Expansion.
When gas expand in vacuum, there is no opposing force i.e Pext = 0
W = — Pext × ΔV  W = 0 × ΔV  W = 0
Hence, no work is done when gas expand in vacuum.
Note= For getting work done opposing force is necessary.
Maximum Work:- When work is done there are two forces [1] acting force (driving force)(P)
and [2]Opposing force ( Pext )
When opposing force is zero, no work is done and the opposing force is increased more,
more work will be obtained. Ie. W = — Pext × ΔV  W = 0 × ΔV  W = 0
If opposing force = driving force, Pext = P then ΔV = 0.
Ie. W = — Pext × ΔV  W = — Pext × 0  W = 0
Maximum work done is obtained, when driving force is slightly greater than opposing force.
Ie. P > — Pext
If P— Pext = ΔP  Pext = P—ΔP
W = — Pext × ΔV
W = — (P—ΔP) × ΔV = (ΔP—P) × ΔV Work is maximum when ΔP is smallest.
Condition for maximum work:
Maximum work is obtained when-
(1) Process is thermodynamic ally reversible. (2) Change takes place in infinite number of steps.
(3) Driving force is slightly greater than opposing force. (4) System is in mechanical equilibrium with
its surrounding.

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Expression for Maximum Work:- ( Isothermal reversible expansion )
Fig.
Consider ‘n’ moles of an ideal gas enclosed in a Cylinder fitted with frictionless Piston
having area (A). Let, P = pressure, V = volume and T = temperature of the gas. Let the gas
expand isothermally and reversibly. The opposing pressure ( Pext ) is slightly less than
pressure of gas (P) hence, Pext = P — dP ………(1)
Suppose gas expand by a small volume dV, then work done dW Is given by
dW = — Pext × dV
= — (P — dP) × dV
= — P.dV + dP . dV
dP . dV isvery small hence neglected
dW = — P.dV ………….. (3)
Now, total work done when gas expand from initial volume V1 to final volume V2 is
obtained by integrating eq. (3) between limits V1 and V2
Wmax = ∫ ܹ݀ ୚ଶ
୚ଵ = ∫ — P. dV ୚ଶ
୚ଵ …………….(4)
For ‘n’ moles ideal gas eq. is PV = nRT ===> P = nRT / V
Wmax = ∫ — ୬ୖ୘
୚ଶ
୚ଵ ܸ݀
௏
= — nRT ∫ — ଵ
୚ଶ
୚ଵ ܸ݀
௏
= — nRT [loge V] ୚ଶ
୚ଵ
V + dV
dv
V
Cylinder
P—dP
P n

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= — nRT [loge V2 — loge V1 ] = — nRT loge ௏ଶ
௏ଵ
According to Boyle’s law, P1 V1 = P2 V2  ௏ଶ
௏ଵ
= ௉ଵ
௉ଶ

Path function:-
Work (W)and heat (q) are path functions
A) Path dependence nature of work
Wmax = — 2.303 nRT log10 ࡼ૚
1. When gas expand in vacuum work done is zero
2. When gas expand from V1 to V2 work done is W = — Pext × (V2 — V1 )
3. When gas expand isothermally and reversibly, work done is Wmax = — 2.303 nRT log10 ୚ଶ
୚ଵ
B) Path dependence nature of heat
1. For adiabatic process q = 0
2. For isothermal process q ≠ 0
Unit of energy and work:- Joule (J)
1 J = 1 kgm3 /s2 1 atm = 101.3 × 10—3 kgm3 /s2
1 J = 1 pa.m3 1 L.atm = 101.3 kgm3 /s = 101.3 J
1 J = 107 ergs 1 J = 0.239 cal. 1 cal. = 4.184 J 1 kcal. = 4.184 J
Sign convention of work done, heat and energy:
Adiabatic q = 0
+W work done on system heat absorbed +q
—w work done by system heat released —q
+ q : heat is absorbed by system from surrounding.
— q: heat is released by system from surrounding.
+ w: work is done on system by surrounding. — w: work is done by system on surrounding.
ࡼ૛
Wmax = — 2.303 nRT log10 ࢂ૛
ࢂ૚
System
A
B
Isothermal q ≠ 0
A (V1 )
B (V2 )
1
3
2

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Internal Energy (U):-
“The energy associated with a substance or system is called internal energy.”
(1) Internal energy is a sum of kinetic energy and potential energy.
(2) Internal energy is a state function. (3) Internal energy is an extensive property.
(4) Internal energy depends on temperature hence for isothermal process, (ΔT = 0) ΔU = 0
(5) Change in Internal energy ΔU = U2 — U1
(6) —ΔE = energy is evolved by system, +ΔE = energy is absorbed by system.
U = Utran. + Urot. + Uvibr. + Uele. + Uintra. + Uinter.
Note:- Kinetic energy = Translation + Rotational + Vibrational + Electronic
Potential energy = Intramolecular + Intermolecular
First law of thermodynamics:- It is simply law of conservation of energy.
Statements
1. Energy can neither be created nor destroyed, although it can be converted from one form to
another
2. The total internal energy of an isolated system is constant.
3. The total quantity of energy of the universe is constant.
4. Whenever a one form of energy disappears, exactly an equivalent amount of other form must
appear.
5. It is impossible to construct perpetual motion machine which produce work continuously without
consuming energy.
Mathematical equation of First law of thermodynamics:-
Consider U1 = initial internal energy. Suppose ‘q’ amount of energy supplied to system
which produce ‘W’ amount of work done.
Thus, final internal energy U2 is given by, U2 = U1 + q + W ==> U2 — U1 = q + w
ΔU = q + w ………… (A)
ΔU = increase in internal energy of system.
Eq. (A) represents equation of First law of thermodynamics.
For infinitesimal change eq. (A) becomes dU = dq + dW

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Modified First law of thermodynamics:-
According to Einstein theory, mass can be covered into energy. (ie. Energy produced due to
loss of mass.)
Einstein eq. is E = mC2
Where E = energy produced by loss of mass ‘m’. C = velocity of light = 3 × 108 m/s
Hence modified First law of thermodynamics is stated as, “total mass and energy of an isolated
system (or universe) remain unchanged.”
First law of thermodynamics for various processes:-
We have ΔU = q + w …….. (A)
1) For isothermal process:- ΔT = 0 and ΔU = 0
Eq. (A) becomes, 0 = q + W
Ie. q = — W or — q = W
+ W : work done by system on surroundings. — W : work done by surroundings on system.
2) For isobaric process:- ΔP = 0
ΔU = qp + W
= qp + ( — Pext × ΔV )
ΔU = qp — Pext × ΔV or qp = ΔU + Pext × ΔV
Heat used to increase internal energy of system.
3) For isochoric process:- ΔV = 0
ΔU = qV + W
= qV + ( — Pext × ΔV )
ΔU = qV + (0) or ΔU = qV
Change in internal energy of system is due to heat transfer.
4) For adiabatic process:- q = 0
Eq. (A) becomes, ΔU = 0 + W
Ie. ΔU = W or — ΔU = — W
1. work done by system, internal energy increases.
2. Work done by surroundings internal energy decreases.

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Second law of thermodynamics:-
Statements
1) “The spontaneous flow of heat is always unidirectional from higher temperature to lower
temperature.”
2) “Heat cannot be completely converted into an equivalent amount of work without
producing permanent changes either in the system or its surroundings.”
3) “No machine has yet been made that has an unit efficiency.”
4) “The entropy of the system is constantly increasing.”
Enthalpy (H):-
It is defined as “the sum of internal energy of the system and PV energy.”
(PV energy is energy related with the pressure and volume.)
Mathematically, the enthalpy is defined as H = U + PV ………… (1)
Since P, V and U are state functions, H is also state function.
Hence change in enthalpy (ΔH) is given by ΔH = H2 — H1 …………(2)
If H1 = U1 + P1 V1 and H2 = U2 + P2 V2
Then from eq. (2) we have ΔH = U2 + P2 V2 — ( U1 + P1 V1 )
= ( U2 — U1 ) + ( P2 V2 — P1 V1 )
= ΔU + ( P2 V2 — P1 V1 )
At constant pressure i.e P1 = P2 = P then
ΔH = ΔU + ( P V2 — P V1 ) = ΔU + P ( V2 — V1 )
………… (3)
ΔH = ΔU + PΔ V
Hence, for isobaric process enthalpy of system is the sum of increase in internal energy the
work of expansion.
Also we have, qp = ΔU + Pext × ΔV
If Pext = P then qp = ΔU + PΔV ……….. (4)
From eq. (3) & (4) we write ΔH = qp
since, qP is state function.
Hence, for isobaric process enthalpy change is the heat absorbed by system.
Note:- Enthalpy is Heat content of the system. Heat evolved or absorbed at constant
pressure is expressed in terms of enthalpy.

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Relation between ΔH and ΔU for chemical reaction:-
Que. Obtain relationship betn ΔH and ΔU for chemical reaction.
Ans. We have, ΔH = ΔU + PΔ V …………. (1)
For solid and liquid, ΔV = 0 then ΔH = ΔU + P(0)
ΔH = ΔU ………………… (2)
For gas, ΔV is not neglected.
ΔH = ΔU + P ( V2 — V1 )
= ΔU + PV2 — PV1 …………… (3)
V1= initial volume (volume of gaseous reactant) , V2 = final volume (volume of gaseous
product)
According to ideal gas eq. PV = nRT
Suppose, n1 = moles of gaseous reactant. n2 = moles of gaseous product.
PV1 = n1 RT and PV2 = n2 RT
From eq. (3) ΔH = ΔU + ( n2 RT — n1 RT )
= ΔU + RT ( n2 — n1 )
ΔH = ΔU + RT.Δn
Work done in chemical reaction:-
We have W = — Pext . ΔV ……… (1)
Suppose Pext = P then W = — P . ΔV = — P ( V2 — V1 )
= — PV2 + PV1 …………….. (2)
According to ideal gas eq. PV = nRT
If n1 = moles of gaseous reactant. n2 = moles of gaseous product.
Then PV1 = n1 RT and PV2 = n2 RT
Eq. (2) becomes, W = — n2 RT + n1 RT = — RT ( n2 — n1 )
W = — RT . Δn

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Condition under which ΔH = ΔU
1) When reaction is carried out in a closed vessel ΔU = 0, ΔH = ΔU + PΔV  ΔH = ΔU
2) When reaction involves solid and liquid ΔV = 0, ΔH = ΔU + PΔV  ΔH = ΔU
3) When reaction in which n1 = n2 , Δn = 0 ΔH = ΔU + RTΔn  ΔH = ΔU
Sign conventions of ΔH
1) When heat is absorbed by system (ie. for endothermic reaction) ΔH is positive.
2) When heat is evolved by system (ie. for exothermic reaction) ΔH is negative.
Enthalpies of physical changes:-
A) Enthalpy of phase transition:-
Whenever there is a change in the state of matter, the process is called phase transition or
change.
It is accompanied by the change in enthalpy (Heat content) of the system.
Commonly observed phase transitions are:-
Solid liquid fusion (melting)
Liquid vapour vaporization
Solid vapour sublimation
1) Enthalpy of fusion (Heat of fusion):- ( Δfus. H)
The enthalpy change when one mole of a solid substance is converted into liquid at
its melting point at constant pressure is called its enthalpy of fusion.
H2 O (s) H2 O (l) Δfus. H = + 6.01 kJ/mol at 00 C
¤ The reverse of fusion is freezing of liquid.
2) Enthalpy of vaporization:- ( Δvap. H)
The enthalpy change when one mole of a liquid is converted into vapour at its
boiling point at constant pressure is called its enthalpy of vaporization.
H2 O (l) H2 O (g) Δvap. H = + 40.7 kJ/mol at 1000 C
¤ The reverse of vaporization is condensation of vapour.

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3) Enthalpy of sublimation:- ( Δsub. H)
The enthalpy change when one mole of a solid is directly converted into vapour at
constant temperature and pressure is called its enthalpy of sublimation.
H2 O (s) H2 O (g) Δsub. H = + 51.08 kJ/mol at 00 C
Relationship between Δfus. H , Δvap. H and ΔSub. H
Δsub. H = Δfus. H + Δvap. H
Eg for water,
51.08 = 6.01 + 45.07
B) Enthalpy of atomic or molecular changes:-
1) Enthalpy of ionization:- (Δion. H)
Vaporization
Fusion
Sublimation
Gas
Liquid
Solid
The enthalpy change when removal of an electron from each atom or ion in one
mole of a gaseous atom or ion called enthalpy of ionization.
Na (g) Na+ (g) + e— Δion. H = 494 kJ/mol
# 494 kJ / mol of heat is required to ionize one mole of Na atom.
Note:-
Ca (g) Ca+ (g) + e— Δion. H = 590 kJ/mol …… first ionization enthalpy
Ca+ (g) Ca++ (g) + e— Δion. H = 590 kJ/mol …… second ionization enthalpy
The reverse of ionization is electron gain process. Hence electron gain enthalpy is defined as,
“The enthalpy change when one mole of a gas phase atom of an element accept electron so
as to form gaseous anion.”
Cl (g) + e— Cl— (g) Δeg. H = — 349 kJ/mol
2) Enthalpy of atomization:- (Δatm. H)
The enthalpy change when the dissociation of all the molecule in one mole of a gas
phase substance into gaseous atoms is called enthalpy of atomization.
OR The enthalpy change when one mole of a gas phase substance dissociates into its
gaseous state constituent atoms is called enthalpy of atomization.
Cl2 (g) Cl (g) + Cl (g) Δatm. H = 242 kJ/mol
CH4 (g) C (g) + 4 H (g) Δatm. H = 1660 kJ/mol

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3) Enthalpy of solution:- ( Δsol. H )
The enthalpy change when one mole of a substance is dissolved in a specified
quantity of solvent at a given temperature is called enthalpy of solution.
Note:- The Δsol. H is the sum of three kinds of interactions.
i) Solute – solute interaction (ie crystal lattice energy) ΔL H
ii) Solvent – solvent interaction
ii) Solute – solvent interaction (ie enthalpy of hydration) Δhyd. H
1) Lattice enthalpy:- ( ΔL H )
The enthalpy change when one mole of a ionic solid breaks down into its ions is
called Lattice enthalpy.
2) Enthalpy of hydration :-( Δhyd. H )
The enthalpy change involved in the hydration of one mole of a gaseous ions is
called enthalpy of hydration .
So,
NaCl (s) + H2 O (l) 1 dm3 NaCl (aq.) Δsol. H = + 4 kJ/mol
XY (s) X+ (g) + Y— (g) or X+ (g) + Y— (g) XY (s) ΔL. H = + ve kJ/mol
X+ (g) + Y— (g) X+ (aq.) + Y— (aq.)
Δsol. H
XY X+ (aq.) + Y— (aq.)
X+ (g) + Y— (g)
ΔL H
Δhyd. H
Ie. Δsol. H = ΔL H + Δhyd. H

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Born – Haber cycle:-
Maw Born & Fritz Haber proposed the method to calculate lattice enthalpy.
Eg. Formation of NaCl. Na (s) + ଵ
1) Sublimation of Na metal:- Na (s) Na (g) Δsub H
2) Ionization of Na atom:- Na (g) Na+ (g) + e— (g) Δion H
3) Dissociation of ଵ
ଶ
mole Cl2 :- ଵ
4) Electron gain by Cl atom:- Cl (g) + e— Cl— (g) Δeg. H
5) Combination of Na+ and Cl— atom:- Na+ (g) + Cl— (g) NaCl (s) ΔL. H
4) Enthalpy of dilution:-
ଶ
Cl2 NaCl (s) Δf H
ଶ
Cl2 (g) Cl (g) Δa H
Δf H = Δsub H + Δion H + ଵ
The enthalpy change that occurs when a solution of one concentration is diluted to
form the solution of another concentration.
Thermochemistry:-
Enthalpy of chemical reactions (Heat of reaction):-
The enthalpy change occur by chemical reaction is called Enthalpy reaction.
Consider a rean aA + bB cC + dD
The enthalpy change for rean is ΔH = (c HC + d HD ) — (a HA + b HB )
Where HA , HB , HC and HD are the enthalpies of A, B, C and D respectively.
Hence enthalpy of chemical reaction is the difference between the sum of enthalpies
of product and sum of enthalpies of reactant.
ଶ Δa H + Δeg. H + ΔL H
HCl (g) + 50 H2 O (finite amount) HCl (50 H2 O ) Δs. H1 = — 73.26 kJ
HCl (g) + H2 O (infinite amount) HCl (aq. ) Δs. H2 = — 75.14 kJ
ΔH = Δ H2 — Δ H1 = (— 75.14 )— (— 73.26 ) = — 1.88 kJ enthalpy of dilution.
Thermochemistry deals with the study of enthalpy changes in
chemical reactions.
ΔH = Σ HProduct — Σ HReactant

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Exothermic and Endothermic reactions:-
Exothermic reactions Endothermic reactions
1. The reaction in which heat released in the
surrounding is called Exothermic reaction.
1. The reaction in which heat absorbed from
surrounding is called Endothermic reaction.
2. Exothermic reactions are occurs by evolution
of heat.
2. Endothermic reactions are occurs by
absorption of heat.
3. Σ HProduct < Σ HReactant 3. Σ HProduct > Σ HReactant
4. ΔH is negative 4. ΔH is positive
5. Products are more stable than reactants. 5. Reactants are more stable than Products.
6. i) C (s) + O2 CO2 ΔH = — 393.5 kJ/mol
ii)Making ice cubes, iii) condensation of water
vapour
6. i) N2 + 2O2 2NO2 ΔH = + 66.4 kJ/mol
ii) Melting of ice, iii) Evaporation of water.
Standard Enthalpy of chemical reactions:- (ΔH0 )
“The standard Enthalpy of chemical reactions is the enthalpy change for a reaction
when all the substances involved are in their standard states.”
Note: The standard state of substance is the pure form in which it is most stable at pressure
1 atm & 25 0C temperature.
Standard states of element = H2 (g) , Hg (l) , Na (s) , C (S)
Standard states of substance = C2 H5 OH (l), H2 O (l) , CaCO3 (S) , CO2
Thermochemical equation:-
1) The equation must be balanced
2) The physical state of reactants and products are indicated by placing the letters (s) for
solid, (l) for liquid and (g) for gas.
3) The numerical value and appropriate sign of enthalpy change must be shown on right
hand side of equation.
4) A thermodynamic eq. can be reversed, if the sign of ΔH is changed.
C (s) + O2 CO2 ΔH = — 393.5 kJ/mol CO2 C (s) + O2 ΔH = + 393.5 kJ/mol
5) The enthalpy of excess of water shown by (aq) is taken as zero.
NH4 Cl (s) + aq. NH4 Cl (aq) ΔH0 = + 15.1 kJ

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Standard Enthalpy of formation:- (ΔfH0 ) (Standard heat of formation ):-
The Standard Enthalpy of formation of a compound is defined as, the enthalpy
change that take place when one mole of pure compound under standard condition is
formed from its constituent elements which is in their standard states.
Eg. Formation of one mole of liquid water, H2 (g) + ½ O2 (g) H2 O (l) ΔfH0 = —286 kJ
Calculation of Standard Enthalpy of reaction from Standard Enthalpy of
formation:-
Consider the reaction aA + bB cC + dD
Standard Enthalpy of reaction is given by ΔH0 = Σ Δf ܪ௉
଴ — Σ Δf ܪோ
଴
= {c ΔfH0 (C)+ d H0ΔfH0 (D)} — {a ΔfH0 (A) + b ΔfH0 (B)}
Where a, b, c and d are stiochiometric coefficient.
Eg. Formation of CO2 (g) C (s) + O2 CO2 ΔfH = — 393.5 kJ/mol
ΔH0 = Σ Δf ܪ௉௥௢ௗ௨௖௧
଴ = { ΔfH0 (CO2 )} — { ΔfH0 (C) + ΔfH0 (O2 )}
଴ — Σ Δf ܪோ௘௔௖௧௔௡௧
By convention enthalpies of C (s) and O2 (g) is zero
ΔH0 = ΔfH0 (CO2 ) — {0 + 0} = ΔfH0 (CO2 )
Hence enthalpy of CO2 is equal to its enthalpy.
Standard Enthalpy of combustion:- (ΔCH0 )
The Standard Enthalpy of combustion of a substance is the enthalpy change that
takes place when one mole of pure substance in the standard state reacts completely with
oxygen.
Eg. Complete combustion of 1 mole of CO is,
CO (g) + ½ O2 (g) CO2 (g) ΔcH0 = —283 kJ
Combustion of ethyne C2 H2 (g) + 5/2 O2 (g) 2CO2 (g) + H2O (l) ΔcH0 = — 1300 kJ

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Bond Enthalpy (Bond energy):- (D)
“The amount of energy required to break a particular covalent bond in 1 mole of
gaseous molecule to produce gaseous atom or radical is called bond enthalpy.”
# Bond energy is always positive because energy is always needed to break the bond.
Eg. 1) H2 (g) H (g) + H (g) ΔH0 = + 436.4 kJ
Ie. to break H—H bond in H2 , 436.4 kJ energy is required. Ie. DH—H = 436.4 kJ or
ΔH0 (H—H)= + 436.4 kJ
2) HCl (g) H (g) + Cl (g) ΔH0 = + 431.9 kJ Ie ΔH0 (H—Cl) = + 431.9 kJ
# The bond enthalpy for diatomic molecule is the same as enthalpy of atomization.
# In polyatomic molecule, every covalent bond has its own specific bond enthalpy. Hence
average bond enthalpy has to be considered.
Eg. Dissociation of water  Step i) H2 O (g) OH (g) + H (g) ΔH0 = 499 kJ
Step ii) OH (g) O (g) + H (g) ΔH0 = 428 kJ
Overall Rean H2 O (g) 2H (g) + O (g) ΔH0 = 927 kJ
Average bond enthalpy of O—H bond = ଽଶ଻
ଶ
= 463.5 kJ hence ΔH0 (O—H) = 463.5 kJ
Reaction enthalpy from bond enthalpy:-
Que. Define bond enthalpy. How is it useful to calculate reaction enthalpy?
Ans.
In a chemical reaction, in reactants bonds broke while in product bond formed. For
enthalpy calculation we assume all the bonds are identical and bond enthalpy independent
of nature of substance.
The enthalpy of chemical reaction is the difference between the sum of bond
enthalpies of reactants and the sum of bond enthalpies of product.
ΔH0 (Reaction) = Σ ΔH0 (Reactants broken bonds) — Σ ΔH0 (Product formed bonds)
Eg. H2 (g) + I2 HI (g) ΔH0 ={ [ ΔH0 (H—H) + ΔH0 (I—I) ] — 2ΔH0 (H—I) }

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Hess’s Law of constant heat summation:-
The law states that, “the enthalpy change for a reaction is the same whether the
reaction takes place in one step or a series of steps.” OR
“The enthalpy change for a reaction is same regardless of the path of the reaction.”
Proof / illustration:- Consider, a conversion of A to C
i) Conversion of A to C in single step
A C ΔH0 = ΔH1 ………. (1)
ii) Conversion of A to C in two steps
A B ΔH0 = ΔH2
B C ΔH0 = ΔH3
B
ΔH2 ΔH3
A C
Overall A C ΔH0 = ΔH2 + ΔH3 ……. (2)
According to Hess’s Law,
Eg. Formation of CO2 from C and O2
1) Formation of CO2 in single step C (s) + O2 (g) CO2 ΔH1 = — 394 kJ
2) Formation of CO2 in two step C (s) + ଵ
ଶ
ΔH1
O2 (g) CO ΔH2 = — 83 kJ
ΔH1 = ΔH2 + ΔH3
CO (g) + ଵ
ଶ
O2 (g) CO2 ΔH3 = — 311 kJ
According to Hess’s Law, ΔH1 = ΔH2 + ΔH3  — 394 =— 83 — 311  — 394 = — 394
Thus heat evolved in both methods is same. Hence Hess’s Law is proved experimentally.
Application of Hess’s Law:-
1) The thermodynamic equations can be added, subtracted or multiplied by numeral like
simple algebraic equation.
2) It can be used to calculate enthalpy of reaction which does not occurs directly.
3) The enthalpy of formation of any compound from its elements can be calculated.
4) It can be used to calculate enthalpy of formation, combustion, neutralization etc.
5) Enthalpy of formation of slow reaction can be calculated.
6) Enthalpy of reaction which is not possible to determine experimentally can be calculated.

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Spontaneous Process (Irreversible process):-
The process that takes place on its own accord without any external influence is
called spontaneous process. It is also called natural process.
Eg. (1) All natural process are spontaneous i) cliff jumping, ii)
(2) Flow of gas from higher pressure to lower pressure. (3) Flow of heat from hot body to
cold body
(4) Flow of water from higher level to lower level
(5) Dissolution of KCl in water
KCl (s) + H2O KCl (aq.)
(6) Acid – Base neutralization reaction
NaOH (s) + HCl NaCl + H2O
Note:- Spontaneity:- The tendency of process to occur naturally is called spontaneity.
Criteria for spontaneity:-
(1) A spontaneous change is one-way or unidirectional. Ie. for reverse change work is required.
(2) A spontaneous change is time is no factor. Ie. Spontaneous reaction may be slow or rapid.
(3) A spontaneous process is continuing till the system attains the equilibrium state.
(4) Once system in equilibrium, it does not undergo any further spontaneous change.
Energy and Spontaneity:-
Generally Spontaneous process are exothermic (ΔH0 = — ve) ie. Exothermility favors
Spontaneity. But there are many processes that are endothermic but Spontaneous.
Eg. 1) Melting of ice Ice H2O ΔH = + 4 kJ
2) Dissolution of NaCl in water NaCl (s) + H2O (aq) Na+ (aq) + Cl— (aq)
ΔH = + 3.9 kJ
3) Conversion of N2O4 to NO2 above 400 k N2O4 (g) NO2 (aq) ΔH = + 57 kJ
From above example it is evident that energy is not only criteria for spontaneity.
Hence with energy there must be another factor which governs spontaneity. That factor is
entropy. 2nd law of thermodynamics introduces entropy (S).

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Entropy (S):-
The thermodynamic property which measures the molecular disorder
or randomness is called entropy.
Ordered State
Disordered State
Highly disordered State
Greater the disorder of a system, higher is the entropy and vice versa. Hence solid
state (ice) has lowest entropy while gaseous state (steam) has highest entropy.
Entropy is a state function. Hence change in entropy ΔS is ΔS = Sfinal — Sinitial
1) I2 (s) (ordered state) I2 (aq) (disordered state ) ΔS is +ve
H2 O
2) 2 H2 (g) + O2 2 H2 O (l) ΔS is —ve
More disordered state disordered state
Quantitative definition of entropy:-
The entropy change (ΔS) of a system in a process is the ratio of amount heat
transferred in a reversible manner to the temperature at that heat transfer (T).
ΔS =
୯୰ୣ୴
୘
(qrev /T)
Unit of ΔS SI unit = J/k or J/mol.k CGS unit = cal/k or cal/mol.k
Entropy and spontaneity:- (2nd law of thermodynamics)
A process occur by an increase in entropy tend to be spontaneous.
Consider two spontaneous process 1) 2H2O2 (l) 2 H2O (l) + O2 (g) ΔS = +ve
2) 2H2 (g) + O2 (g) 2 H2O (l) ΔS = —ve
From above two examples it is clear that in spontaneous process the entropy may
increase or decrease. This discrepancy is explained by second law of thermodynamics.

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2nd law of thermodynamics:-
“The total entropy of the system and surroundings increases in a spontaneous process.”
Thus,
ΔSuniverse = ΔSTotal = ΔSsys. + Δ Ssurr. > 0 , for spontaneous process
OR The total entropy of the system is constantly increasing.
(1) ΔSTotal > 0 , the process is spontaneous. (2) ΔSTotal < 0 , the process is non-spontaneous.
(3) ΔSTotal = 0 , the process is at equilibrium.
Gibbs Energy (Gibbs Function):- (G)
For most of the chemical reactions there are changes in both enthalpy and entropy.
Hence to decide the spontaneity of process in terms of both enthalpy and entropy change
we define new thermodynamic function, called Gibbs energy (G).
Gibbs free energy is defined as , G = H — TS
Where, H is enthalpy and S is entropy of system. T is absolute temperature.
H, S and T are state functions, G is also state function. Hence change in Gibbs Energy (ΔG)is
defined as , ΔG = ΔH — TΔS
Unit of ΔG SI unit = J or kJ or kJ/mol CGS unit = cal. Or k.cal or k.cal/mol
Gibbs Energy and spontaneity:-
We have ΔSTotal = ΔSsys. + Δ Ssurr.  ΔSTotal = ΔS + Δ Ssurr. ……… (1)
If + ΔH is the enthalpy change of system, then enthalpy change of surrounding is —ΔH.
ΔS = qrev. / T = qsur. / T = — ΔH / T …….. (2)
From eq. (1) & (2), total entropy ΔSTotal is, ΔSTotal = ΔS — Δୌ
୘
Multiplying above eq. by — T — TΔSTotal = — TΔS + ΔH
Ie — TΔSTotal = ΔH— TΔS ……… (3)
The change in Gibbs Energy (ΔG)is ΔG = ΔH — TΔS .……… (4)
From eq. (3) & (4) ΔG = — TΔSTotal
Ie. ΔG = T ( —ΔSTotal ) because T is always positive.
Hence if ΔSTotal > 0 then ΔG < 0
(1) ΔG < 0, the process is spontaneous (2) ΔG > 0, the process is non-spontaneous
(3) ΔG = 0, the process is at equilibrium.

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¤ The spontaneity of reaction with respect to sign dependence of ΔH, ΔS and ΔG
ΔH ΔS ΔG Spontaneity of reaction
—
(Exothermic)
+ — Reactions are spontaneous at all the temperatures.
—
(Exothermic)
— — Or + Reactions become spontaneous at low temperatures. When
ІT.ΔSІ<ІΔHІ
+
(Endothermic)
+ — Or + Reactions become spontaneous at high temperatures. When
T.ΔS > ΔH
+
(Endothermic)
— + Reactions are non-spontaneous at all the temperatures.
Temperature of equilibrium:- We have ΔG = ΔH — TΔS ………….. (1)
At equilibrium, ΔG = 0 Ie. process is neither spontaneous nor non-spontaneous.
Eq. (1) becomes, 0 = ΔH — TΔS ie. ΔH = TΔS T = Δୌ
Δୗ
ΔG and equilibrium constant:-
The change in Gibbs Energy (ΔG) of a reaction is related to standard Gibbs Energy
change (ΔG0 ) by the relationship
ΔG = ΔG0 + RT lnQ = ΔG0 + 2.303 RT log10 Q ………..(A)
Where, ΔG0 = Standard Gibbs energy change R = Gas constant (8.314 kJ/mol)
Q = Reaction quotient T = Temperature
For Q consider a reaction aA + bB cC + dD
[ C ]c [ D ]d OR ܲ஼
௖
× ܲ஼
௖
QC = Q P =
[ A ]a [ B ]b ܲ஼
௖ × ܲ஼
௖
Where [] indicate concentration and P indicate partial pressure
At equilibrium Q = k and ΔG = 0 K = equilibrium constant
Eq. (A) becomes 0 = ΔG0 + 2.303 RT log10 k Ie. ΔG0 = — 2.303 RT log10 k

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Third law of thermodynamics:-
It states that, “at absolute zero temperature, the entropy of perfectly ordered
crystalline substance is zero.”
Thus, S = 0 at T = 0 for perfectly ordered crystalline substance. Substance contain impurities
S >0
# The entropy of solid greater than zero (S > 0) at T = 0 & is called residual entropy of
substance.
Importance of 3rd law of thermodynamics:-
(1) It gives absolute starting place from which to measure entropy.
(2) It allows determining the absolute entropy of any substance at any desired temperature.
(3) The standard molar entropy (S0 ) of pure substance can be measured.
(4) Standard entropy change ΔS0 for a reaction can be calculated.
(5) Spontaneity of reaction can be predicted.
Que. How is absolute entropy measured for a pure crystalline substance at high
temperature? Or why is it impossible for any substance to have an absolute entropy zero
at temp. greater than 0 k?
Ans:- For perfectly ordered crystalline substance at absolute zero temperature, the
entropy of is zero S0 = 0. When this substance is heated to the temp. T its entropy will
increase to ST then increase in entropy is given by
ΔS = ST — S0 since S0 = 0 then ΔS = ST
Thus the value of absolute entropy at different temperature can be calculated. Hence
at temp. greater than absolute zero, it is impossible for any substance to have an absolute
entropy zero.
்
଴ Where CP is heat capacity of solid.
Note: ST is determined by ST = ∫ ஼௣.ௗ்
்
Standard molar entropy (S0 ):-
The absolute entropy of one mole of pure substance at 1 atm. Pressure and 25 0C
temperature is called Standard molar entropy of the substance.
Usefulness of Standard molar entropy:-
(1) To determine the Δ S0 for a reaction. (2) It is used to compare entropies of different
substance under same condition of temperature and pressure. (3) Spontaneity of the
process can be predicted.

Thermodynamics

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