System and its types, chemical thermodynamics, thermodynamics, first law of thermodynamics, enthalpy, internal energy, standard enthalpy of combustion, formation, atomization, phase transition, ionization, solution and dilution,second law of thermodynamics, gibbs energy change, spontaneous and non-spontaneous process, third law of thermodynamics

1. CLASS:11-Thermodynamics
CBSE
SHORT NOTES
BY
SAK

2.  Concepts of System and types of systems,
surroundings, work, heat, energy, extensive and
intensive properties, state functions.
 First law of thermodynamics ‐internal energy and
enthalpy, measurement of ΔU and ΔH, Hess's
law of constant heat summation, enthalpy of
bond dissociation, combustion, formation,
atomization, sublimation, phase transition,
ionization, solution and dilution.
Second law of Thermodynamics (brief introduction)
Introduction of entropy as a state function, Gibb's
energy change for spontaneous and
nonspontaneous processes.
 Third law of thermodynamics (brief introduction).

3.  Introduction:
 The branch of science which deals with the
study of different forms of energy and the
quantitative relationship between them is
known as Thermodynamics.
 How do we determine the energy changes
involved in a chemical reaction/process?
 What drives a chemical reaction /process?
 To what extent do the chemical reactions
proceed?

4.  Some basic terms and concepts:
System and Surrounding : A system in
thermodynamics refers to that part of universe
in which observations are made .
The remaining portion of the universe
,excluding the system, is called surrounding.
Ex: If we are studying the reaction between two
substances A and B kept in a beaker, the beaker
containing the reaction mixture is the system
and the room where the beaker is kept is the
surroundings.
The wall that separates the system from the
surroundings is called boundary.

5.  Types of the system:
 Open ,closed and isolated system
Open system : A system is said to be an open
system if it can exchange both matter and
energy with the surrounding.

6.  For ex: If some water is kept in an open
vessel or if some reaction is allowed to take
place in an open vessel ,exchange of both
matter and energy takes place between the
system and surrounding.
 Closed system : If a system can exchange
only energy with the surrounding but not
matter it is called a closed system.
 Ex: The presence of reactants in a closed
vessel made of conducting material e.g
copper or steel is an example of a closed
system.

7.  If the reaction is exothermic ,heat is given by
the system to the surrounding.
 If the reaction is endothermic ,heat is given
by surrounding to the system.
 Isolated system :If a system can neither
exchange matter no energy with the
surrounding ,it is called an isolated system.
 Ex: The presence of reactants in a
thermoflask or any other closed insulated
vessel is an example of an isolated system.

8.  State function:
 A physical quantity is said to be state
function if its value depends only upon the
state of the system and does not depend
upon the path by which this state has been
attained.
 For ex: Physical quantities which are state
function include pressure ,volume,
temperature ,internal energy enthalpy,
entropy ,free energy etc.
 Work and heat are path functions.

9.  We can describe the state of a gas by quoting
its pressure(p), volume (V), temperature(T)
 Variables like p,v,T, are called state variables
or state functions, because their values depend
only on the state of the system and not on how
it is reached.
 Thermodynamic does not deal with the
properties of the individual atoms and
molecules but deals with the matter in bulk.
Properties of the macroscopic systems like
temperature ,pressure ,volume, density
,melting point ,boiling point are
called macroscopic properties or
thermodynamic properties.

10. The Internal energy as a state function:
When we talk about our chemical system losing
or gaining energy, we need to introduce a
quantity which represents the total energy of the
system .
It may be chemical, electrical, mechanical or any
other type of energy, the sum of all these energy
of the system.
In thermodynamics we call it the internal energy
U of the system .
The energy thus stored within a substance is
called its internal energy.
It is represented by symbol U or E.

11.  The internal energy may change when
Heat passes into or out of the system
Work is done on or by the system
Matter enters or leaves the system.

12.  Internal energy as a state function:(Work)
 Internal energy is a state function i.e. depends only
upon the state of the system and is independent of
the method by which this state has been attained.
 Let us change the internal energy of the system by
doing work called adiabatic work. This can be done
in two ways:
 (1) By doing mechanical work by rotating a set of
small paddles.
 (2) By doing electrical work with the help of an
immersion rod.

13.  AB
 2040
 UU1
 ABEndothermic
 B-AExothermic

14.  Sign of ΔU
 ΔU = U2 – U1 = Wad
 (1) If U1 > U2 , the extra energy possessed by
the system in the initial state would be given
out and ΔU will be negative.
 (2) If U1 < U2 , energy will be absorbed in the
process and ΔU will be positive.ΔU is negative
if energy is evolved and ΔU is positive if
energy is absorbed.

15. (b)Internal energy change by transfer of Heat:
 We can change the internal energy of a system
by transfer of heat from the surroundings to
the system or vice-versa.
 We take water at temperature T1 in a container
having thermally conducting walls, say made
up of copper and enclose it in a huge heat
reservoir at temperature T2.
 The change in internal energy u=q when no
work is done at constant volume.

16.  The q is positive when heat is transferred from the
surroundings to the system and q is negative
when heat is transferred from system to the
surroundings.
 (c)As the change of state is brought about both by
doing work and by transfer of heat. We write
change in internal energy for this case
 The sum total of all the energies possessed by the
system is internal energy.
 U=q+w will depend only on initial and final
state. It will be independent of the way the change
is carried out.
 If there is no transfer of energy as heat or as work
(isolated system) if w=0 and q=0, then U=0.

17.  First Law Of Thermodynamics
 Energy can neither be created nor destroyed
although it may be converted from one form to
another.
 The total energy of the universe remains
constant, although it may undergo
transformation from one form to the other.
 The energy of an isolated system is constant.

18.  Mathematical formulation of the first law of
thermodynamics:
 The internal energy of a system can be increased in
2 ways:
 (1) By supplying heat to the system
 (2) By doing work on the system
 Suppose the initial internal energy of the system =
U1
 If it absorbs heat q, its internal energy will become
= U1 + q
 If work w is done on the system, the internal
energy will further increase and become = U1 +q
+w

19.  Final internal energy = U2
Then
 U2 = U1 + q +w
 U2 -U1 = q +w
 ΔU = q +w
 If the work done is the work of expansion, then
w = – PΔV
 ΔU = q – PΔV
 q = ΔU + PΔV
 Neither q nor w is a state function, yet the
quantity (q + w )is a state function.

20.  Internal energy of an ideal gas is a function of
temperature. For an ideal gas undergoing an
isothermal change, ΔU= 0.Hence q = -w i.e.
the heat absorbed by the system is equal to
the work done by the system.

21.  Work of expansion or compression or
Pressure- volume work
 It is the work done when the gas expands or
contracts against the external pressure.
 Consider a gas enclosed in a cylinder fitted
with a frictionless piston.
 Area of cross -section of cylinder= a sq cm

23.  Pressure on the piston= P
 Distance through which gas expands =dl cm
 Pressure is force per unit area, force acting
on the Piston will be f = P × a
 Work done by the gas =Force × Distance =
f ×dl = P × a × dl
 But a × dl= dV, a small increase in the
volume of the gas.The small amount of work
( δw) by the gas can be written as:
 δw= P × dV

24.  If the gas expands from initial volume V1 to the
final volume V2 ,then the total work done (w) will
be given by:
 w= ∫ PdV
 If the gas expands against constant external
pressure we get
 w= P ∫ dV= P ( V2 – V1 ) = P .ΔV
 If the external pressure is slightly more than the
pressure of the gas, the gas will contract i.e. the
work will be done by the surrounding on the
system.
 P is the external pressure and hence it is written
as Pext.
 w =Pext × ΔV

25. (1) w is taken as positive if work is done on the
system i.e. for compression.
(2) w is taken as negative if work is done by the
system i.e. for work of expansion.
w = – Pext × ΔV
w = – Pext × ( V2 – V1)
w = – Pext × (Vf – Vi)
Vf and Vi represents the final and initial volume.
The above expression applies for work of expansion
as well as work of compression. This is because
(1) For expansion, V2 > V1 so that ( V2 – V1 ) is
positive and hence w is negative.
(2) For compressions V2 < V1 so that (V2 -V1) is
negative and negative multiplied by negative will be
positive.

26.  Work done in isothermal reversible expansion
of an ideal gas

27.  Work done in isothermal reversible expansion
of an ideal gas:

28.  Reversible work of expansion ( wrev ) is the
maximum work
 For reversible expansion, Pext , should be
infinitesimally smaller than Pint. Pext is the
maximum possible value of the pressure. As
w = – Pext ΔV ,therefore for a given change
of volume, w is maximum.
 Thus wrev = wmax
 Free expansion of an ideal gas i.e. expansion
against vacuum

29.  If an ideal gas expands against vacuum ,the
irreversible expansion is called free
expansion.
 As Pext =0 for reversible as well as irreversible
expansion, work done =0
 wirrev = – Pext ΔV = 0 × ΔV = 0
 wirrev = – ∫ Pext dV =0

31.  Enthalpy
According to first law of thermodynamics,
q= ΔU + PΔV
If the process is carried out at constant volume
, ΔV=0
qv = ΔU
Where V indicates constant volume.

32.  Internal energy change is the heat absorbed
or evolved at constant volume.
 ΔU is a state function, therefore qv ,is also a
state function.
 If the process is carried out at constant
pressure, the work of expansion is given by
 w = –P ΔV
 where ΔV is the increase in volume and P is
the constant pressure.
 q=ΔU – w
 where q is the heat absorbed by the
system, ΔU is the increase in internal energy
of the system and w is the work done by the
system.

33.  Under conditions of constant pressure, putting
w= –P ΔV and representing the heat absorbed by
qp we get,
 qp = ΔU + PΔV
 When the system absorbs qp joules of heat, its
internal energy increases from U1 to U2 and the
volume increases from V1 to V2.Then, we have,
 ΔU = U2 – U1
 ΔV = V2 -V1
 qp = ( U2 – U1 ) + P ( V2 -V1)
 qp= ( U2 + PV2 ) – ( U1 + PV1)
 The thermodynamic quantity U + PV is called heat
content or enthalpy of the system and is
represented by symbol H.

34.  H= U + PV
 If H2 is the enthalpy of the system in the final
state and H1 is the value in the initial state,
then
 H2 = U2 + PV2
 H1 = U1 + PV1
 qp =H2 -H1
 qp =ΔH
 Enthalpy change of a system is equal to the
heat absorbed or evolved by the system at
constant pressure.

35.  As most of the reactions are carried out at
constant pressure ,the measured value of the
heat evolved or absorbed is the enthalpy change
enthalpy.
 ΔH= ΔU + PΔV
 Enthalpy change accompanying a process may
be defined as the sum of the increase in internal
energy of the system and the pressure- volume
work done i.e. the work of expansion.
 The energy stored within the substance or the
system that is available for conversion into heat
is called the heat content or enthalpy of the
system or substance.
 The absolute value of heat content or enthalpy
of a substance or a system cannot be measured.
 U and V are extensive properties therefore
the enthalpy is also an extensive property.

36.  Relation between heat of reaction at constant
pressure and at constant volume
 qp =ΔH and qv = ΔU
 ΔH= ΔU + PΔV
 ΔH= ΔU + P(V2 – V1)
 ΔH= ΔU + (PV2 – PV1)
 where V1 is the initial volume and V2 is the
final volume of the system.

37.  For ideal gas,
 PV = nRT
 PV1 =n1 RT
 PV2 =n2 RT
 where n1 is the number of moles of gaseous
reactants and n2 is the number of moles of
gaseous products.
 ΔH= ΔU + (n2RT -n1RT)
 ΔH= ΔU + (n2 – n1 )RT
 ΔH= ΔU + Δng RT
 where Δng = n2 -n1 is the difference between
the number of moles of the gaseous products
and those of the gaseous reactants.

38.  qp = qv – ΔngRT
 Conditions under which qp =qv or ΔH = ΔU
 1) When the reaction is carried out in closed
vessel so that volume remains constant, i.e. ΔV=
0.
 2)When reaction involves only solids or liquids or
solution but no gaseous reactant and product.
This is because the volume changes of the solids
and liquids during a chemical reaction and
negligible .
 3)When reaction involves gaseous reactants and
products but their number of moles are equal in
the reactions.

39.  Intensive and extensive property:
 An intensive property is one that does not depend
on the mass of the substance or system.
 Temperature (T), pressure (P) and density (r) are
examples of intensive properties.
 Extensive Property:
 An extensive property of a system depends on the
system size or the amount of matter in the
system.
 If the value of the property of a system is equal to
the sum of the values for the parts of the system
then such a property is called extensive property.
Volume, energy, and mass are examples of
extensive properties

40. Intensive property Extensive property
Independent property Dependent property
Size does not change Size changes
It cannot be computed It can be computed
Can be easily
identified
Cannot be easily
identified
melting point, colour,
ductility, conductivity,
pressure, boiling
point, lustre, freezing
point, odour, density,
Ex: Length, mass, weight,
volume

41.  Measurement Of Change In Internal Energy
and Enthalpy
 The experimental technique of measuring
energy changes accompanying any chemical
or physical process is called calorimetry.
 These measurements are generally carried
out under 2 condition
 1) at constant volume ( ΔU or qv )
 2) at constant pressure ( ΔH or qp)

42.  Measurements of ΔU:

43.  Internal energy change is measured
experimentally using an apparatus
called Bomb calorimeter.
 It consists of a strong steel vessel which can
stand high pressure. It is surrounded by a
bigger vessel which contains water and is
insulated. A thermometer and a stirrer are
suspended in it.

44.  Procedure
 1) A known mass of compound is taken in the
platinum cup. Oxygen under high pressure is
introduced into the bomb. A current is passed
through the filament immersed in the
compound. Combustion of the compound takes
place. The increase in the temperature of water
is noted. From this, the heat capacity of the
apparatus can be calculated.
 2) The experiment is repeated as in step 1.As
the reaction is carried out in a closed vessel,
therefore, heat evolved is the heat of combustion
at constant volume and hence is equal to
internal energy change.

45.  The value of ΔU can be calculated using the
formula
 ΔU = (Q × Δt × M ) / m
 where Q =heat capacity of the calorimeter
 Δt = rise in temperature
 m= mass of the substance taken
 M =molecular mass of the substance

46.  Measurement of ΔH

47.  Taking the example of neutralisation of HCl with
NaOH.
 1) A known volume of HCL of known
concentration ( 100cm3 of 0.5 N ) is taken in one
beaker and an equal volume of NaOH of the same
concentration (100cm3 of 0.5 N ) is taken in
another beaker.
 2) Both the beakers are kept in water bath till the
solution attain the same temperature.
 3) HCl solution kept in the first beaker is
transferred into the polythene bottle.Immediately
NaOH solution kept in the second beaker is
added into the polythene bottle.Stirring is done
to mix HCl and NaOH.The highest temperature
attained is noted.

48.  Suppose the initial temperature of the acid and
the base = t1 ° C
 Final temperature of the solution after mixing
= t2 ° C
 Rise in temperature= ( t2 – t1) °C
 Total mass of the solution = 200 g
 Heat produced = mass × specific heat × rise in
temperature
 Heat produced by neutralisation of 1000 cc of 1
N HCl
 This give the heat of neutralisation per
equivalent.
 For exothermic reaction, ΔH is negative whereas
for endothermic reaction, ΔH is positive.

49.  Heat of reaction or Enthalpy of reaction or
Enthalpy change of reaction:
 The amount of heat evolved or absorbed in a
chemical reaction when the number of moles of
reactants as represented by the chemical equation
have completely reacted, is called the heat of
reaction or enthalpy of reaction or enthalpy
change of reaction.
 It is represented by Δr H.
 For example
 CH4 ( g ) + O2 ( g ) ————> CO2 ( g ) + 2 H2O ( l
) ΔH = – 890.4 KJ mol-1
 C ( s ) + H2O ( g ) ———> CO ( g ) + H2 ( g ) ΔH =
131 Kj mol-1

50.  1 mole of methane combines completely with
2 moles of Oxygen gas, 890.4 KJ of heat is
produced.
 1 mole of solid carbon react completely with
one mole of steam ,131.4 KJ of heat is
absorbed.
 Enthalpy of reaction, ΔH = Sum of enthalpies
of products – Sum of enthalpies of reactants
 ΔH = ∑ai H (products) – ∑bi H (reactants)

51.  If the reaction is reversed , the sign of ΔH
changes
 H2 ( g ) + ½ O2 ( g ) ———> H2O ( l ) ΔH = –
285.8 KJ mol-1
 H2O ( l ) ——–>H2 ( g ) + ½ O2 ( g ) ΔH =
285.8 KJ mol-1
 In terms of S.I. unit, 1 atm=101.325 KPa
whereas 1 bar=105 Pa .
 The enthalpy change of a reaction in the
standard state is represented by the
symbol ΔH°

52.  Standard enthalpy of reaction:
 The enthalpy change of a reaction depends upon
the condition under which the reaction is carried
out. Hence it is essential to specify some standard
conditions.
 The standard enthalpy of a reaction is the enthalpy
change accompanying the reaction when all the
reactants and products are taken in their standard
state.
 A substance is said to be in standard state when it
is in the purest and most stable form at 1 bar
pressure and the specified temperature.
 This temperature is usually taken as to 298 K.
 Standard state of pure ethanol at 298 k is pure
liquid ethanol at 298 K and 1 bar pressure.

53.  Enthalpy of fusion:
 Enthalpy of fusion is the enthalpy change
accompanying the transformation of one mole of a
solid substance into its liquid state at its melting
point. it is also called molar enthalpy of fusion.
 The molar enthalpy of fusion ( Δ fus H ) of ice is +6
KJ mol-1.
 H2O ( s ) ———–> H2O ( l ) Δfus H = + 6 KJ mol-1
 The enthalpy of freezing has same value as
enthalpy of fusion but has the opposite sign.
 H2O ( l ) ———–> H2O (s) Δ freezing H = − 6 KJ mol-
1

54.  Enthalpy of vaporisation:
 It is the amount of heat required to convert 1
mole of a liquid into its vapour state at its
boiling point.
 It is also called molar enthalpy of vaporisation.
 The molar enthalpy of vaporisation of water into
its vapour at the boiling point of water is 40.7
KJ.
 H2O ( l ) ———–> H2O ( g ) Δ vap H = + 40.7 KJ
mol-1
 Δ 𝑣𝑎𝑝 is the standard enthalpy of vapourisation.
𝐻0

55.  As condensation is reverse of vaporisation,
the enthalpy of condensation has the same
value as the enthalpy of vaporisation but has
opposite sign.
 H2O ( g ) ———–> H2O ( l ) Δfus H = − 40.7
KJ mol-1

56.  Enthalpy of sublimation
 Sublimation is a process in which a solid on
heating changes directly into gaseous state
below its melting point.
 Enthalpy of sublimation of a substance is the
enthalpy change accompanying the conversion
of one mole of a solid directly into vapour phase
at a given temperature below its melting point.
 The enthalpy of sublimation of iodine is 62.39 KJ
mol-1
 I2 ( s ) ————> I2 ( g ) Δ sub H = + 62.39 KJ
mol-1
 Solid CO2= Δ sub H =25.2kj/mol, Napthalene

57.  The magnitude of the enthalpy change
depends on the strength of the intermolecular
interactions in the substance undergoing the
phase transformations.
 Ex:H2O the strong hydrogen bonds between
water molecules hold them tightly in liquid
phase.
 In acetone the intermolecular dipole-dipole
interaction are significantly weaker. Thus it
requires less heat to vaporize 1mol of acetone
than it does to vaporize 1 mol of water.

58.  Enthalpy of formation
 The enthalpy of formation of a substance is
defined as the heat change i.e. heat evolved or
absorbed when 1 mole of the substance is
formed from its elements under given
conditions of temperature and pressure.
 It is usually represented by ΔfH.
 The condition of temperature and pressure
usually chosen as 298 K and 1 bar pressure.
This is called standard state.

59.  Standard enthalpy of formation of a substance is
defined as the enthalpy change accompanying
the formation of 1 mole of the substance in the
standard state from its elements, also taken in
the standard state. It is usually
represented ΔfH°.
C ( s ) + O2 ( g ) —————> CO2 ( g ) ΔfH°= -393.5
KJ mol-1
When one mole of CO2 is formed from its
elements, C ( s ) and O2 ( g ) , 393.5 KJ of heat is
produced.
 Enthalpy of reaction,
 ΔH = Sum of enthalpies of products – Sum of
enthalpies of reactants
 ΔH = ∑ai H (products) – ∑bi H (reactants)

60.  H2(g) + Br2(l)2HBr(g)
 ΔfH°=72.8kj/mole
 For the reaction enthalpy change is not standard
enthalpy of formation, ΔfH° for HBr(g). Here two
moles, instead of one mole of the product is
formed from the elements
 ΔrH°=2 ΔfH°
 Therefore, by dividing all coefficients in the
balanced equation by 2 expression for enthalpy of
formation of HBr(g) is written as

1
2
𝐻2(𝑔)+
1
2
𝐵𝑟2(𝑙)HBr(g)
 ΔfH°=-36.4kj/mol

61. CaCO3(s)CaO(s)+CO2(g)
Wecanmakeuseofstandardenthalpyofformationandcalculatetheenthalpychangeforthe
reaction.

62. The decomposition of CaCO3 is an endothermic
process and have to heat it for getting the
desired products.

63. Thermochemical equations:
When a balanced chemical equation not only
indicates the quantities of the different
reactants and products but also indicates the
amount of heat evolved or absorbed, it is
called thermochemical equation.

64.  It would be necessary to remember the
following conventions regarding
thermochemical equations.
1. The co-efficients in a balanced
thermochemical equation refer to the
number of moles(never molecules) of
reactants and products involved in the
reaction.
2. The numerical value of ΔrH° refers to the
number of moles of substances specified by
an equation.

65. Fe2O3(s) +3H2(g)2Fe(s) +3H2O(l)
ΔfH°(H2O,l)=-285.83kj/mol
ΔfH°(Fe2O3, s)=-824.2kj/mol
Also ΔfH°(Fe, s)=0
ΔfH°(H2, g)=0
Then
ΔrH°=3(-285.83kj/mol)-1(-824.2kj/mol)
=(-857.5+824.2)kj/mol
=-33.3kj/mol
The coefficients used in these calculations are
pure numbers, which are equal to the respective
stoichiometric coefficients.

66. If we had balanced the equation differently
1
2
Fe2O3(s) +
3
2
H2(g)Fe(s) +
3
2
H2O(l)
 ΔrH°=
𝟑
𝟐
(-285.83kj/mol)-
𝟏
𝟐
(-824.2kj/mol)
 =(-428.7+412.1)kj/mol
 =-16.6kj/mol
It shows that enthalpy is an extensive quantity.
3. When a chemical equation is reversed, the
value of ΔrH° is reversed in sign.
 N2(g)+3H2(g)2NH3(g) ΔrH°=-91.8kj/mol
 2NH3(g)N2(g)+3H2(g) ΔrH° =+91.8kj/mol

67.  Hess’s law of heat summation
According to this “total energy change for a
reaction is same, whether reaction takes place
in one step or in many steps.
(or)
The law states that the change in enthalpy for a
reaction is the same whether the reaction takes
place in one or a series of steps.

68.  For example, consider following two paths for
the preparation of methylene chloride
Path I :
CH4​(g)+2Cl2​(g)→CH2​Cl2​(g)+2HCl(g)
ΔH10​=−202.3kJ
CH4(g)+Cl2(g)CH3Cl(g)+HCl(g)
ΔH2
0
= -98.3kJ-------------1
CH3Cl(g) +Cl2(g)CH2Cl2(g)+HCl(g)
ΔH3
0
= −104.0kJ------------2
Adding two steps
CH4(g)+2Cl2(g)CH2Cl2(g) +2HCl(g)
ΔH10​=−202.3kJ

69.  Enthalpies for different types of reactions:
Enthalpy of combustion:
The enthalpy of combustion of a substance is
defined as the heat change when 1 mole of
substance is completely burnt or oxidised in
oxygen.

70.  Cooking gas in cylinders contains mostly
butane(C4H10).During complete combustion of
one mole of butane, 2658KJ of heat is
released.
 C4H10(g)+
13
2
O2(g)4CO2(g)+5H2O(l)
ΔcHo=-2658.okJ/mol
Similarly, combustion of glucose gives out
2802KJ/mol of heat.
C6H12O6(g)+6O2(g)6CO2(g)+6H2O(l)
ΔcHo=-2802kJ/mol

71.  Enthalpy of Atomization:
 When 1 mole of a given substance dissociates into
gaseous atom ,the enthalpy change accompanying
the process is called enthalpy of atomisation.
 It is represented by the symbol Δa H°
H2(g)2H(g) Δa H°=435kJ/mol
The H atoms are formed by breaking H-H bonds in
dihydrogen.
The enthalpy of atomization is also the bond
dissociation enthalpy.
CH4(g)C(g)+4H(g) Δa H°=1665kJ/mol

72. Bond Enthalpy
Bond energy is the amount of energy released
when 1 mole of bonds are formed from the
isolated atoms in the gaseous state or the amount
of energy required to dissociate one mole of bonds
present between the atoms in the gaseous
molecules.
It is represented by Δ b H or Δbond H0.
For diatomic molecules like H2, O2, N2, Cl2, HCl, HF
etc. the bond energies are equal to their
dissociation energies. These may also be called as
their enthalpy of atomisation.

73.  In methane , all the four C-H bonds are identical
in bond length and energy. However, the
energies required to break the individual C-H
bonds in each successive step differ:
 CH4(g)CH3(g)+H(g) Δbond H0=+427KJ/mol
 CH3(g)CH2(g)+H(g) Δbond H0=+439KJ/mol
 CH2(g)CH(g)+H(g) Δbond H0=+452KJ/mol
 CH(g)C(g)+H(g) Δbond H0=+347KJ/mol
 CH4(g)C(g)+4H(g) Δa H°=1665kJ/mol
 In such case we use mean bond enthalpy of
C-H bond.

74.  The standard enthalpy of reaction , ΔrH0 is related
to bond enthalpies of the reactants and products in
gas phase reactions.
ΔrH0=bond enthalpies(reactants)- bond
enthalpies(products)
Enthalpy of solution
The enthalpy of solution of a substance in a
particular solvent is defined as the enthalpy change
when 1 mole of the substance is dissolved in a
specific amount of the solvent.
However, if such a large volume of the solvent is
taken that further addition of the solvent does not
produce any more heat change, it is called enthalpy
of solution at infinite dilution.
Water is usually used as a solvent and the symbol aq
is used to represent it at large dilution.

76.  When an ionic compound dissolves in a
solvent, the ions leave their ordered positions
on the crystal lattice.
 These are now free in solution. But solvation
of these ions also occurs at the same time.
 The enthalpy of solution of AB(s), ΔsolH 0 in
water is therefore, determined by the
selective values of the lattice enthalpy,
ΔlatticeH 0 and enthalpy of hydration of ions,
ΔhydH 0 as
ΔsolH 0 = ΔlatticeH 0 + ΔhydH 0

77.  For most of the ionic compounds ΔsolH 0 is
positive and the dissociation process is
endothermic. Therefore the solubility of most
salts in water increases with rise of temperature.
Lattice enthalpy:
The lattice enthalpy of an ionic compound is the
enthalpy change which occurs when one mole of
an ionic compound dissociates into its ions in
gaseous state.
NaCl(s) Na+ (g) + Cl- (g) ΔlatticeH 0 =+788kJ/mol.

78.  Max born and Fritz Haber in 1919 put
forward a method for the calculation of lattice
enthalpy and hence for predicting the
stability of the ionic compounds formed.
 The method is known as Born-Haber cycle.
The heat of formation of NaCl ( s ) is found to
be – 411 KJmol-1.

80. 1.Crystalline sodium metal is sublimed to form
gaseous atom
Na ( s ) ———> Na ( g ) ΔH = Δsub H = S
2)The gaseous sodium atoms are ionised to form
gaseous sodium ions
Na ( g ) ———-> Na + ( g ) + e– Δ i H = I.E.
3) one half mole of gaseous Cl2 molecule is
dissociated to form one mole of gaseous atom
½ Cl2 ( g ) ————-> Cl ( g ) ΔH =½ Δdiss H = ½D
4) The gaseous chlorine atoms are converted into
gaseous chloride ions by adding electrons.
Cl ( g ) + e– ———> Cl‾ ( g ) Δ eg H = E.A.

81.  The gaseous Na + and Cl‾ ions combine to
form 1 mole of crystalline sodium chloride.
Na + ( g ) + Cl‾ ( g ) —–> NaCl ( s ) Δlattice H = U
Greater the lattice enthalpy more stable is ionic
compounds.
Greater the lattice enthalpy of an ionic
compound, less is its solubility in water.

82.  A process which can take place by itself or has
an urge or tendency to take place is
called spontaneous process.
 A spontaneous process is simply a process
which is feasible.
 The rate of the process may vary from
extremely slow to extremely fast.

83. Examples of processes which take place by
themselves :
1) Dissolution of common salt in water
2) Evaporation of water in an open vessel
3) Flow of heat from hot end to cold end or from
a hot body to a cold body
4) Flow of water down a hill
5) Combination of nitric oxide and oxygen to
form Nitrogen dioxide
2NO ( g ) + O2 ( g ) ———–> 2NO2 ( g )

84. Examples of processes which take place on
initiation:
1) Lightning of candle involving burning of wax
2) Heating of calcium carbonates to give
calcium oxide and carbon dioxide
CaCO3 ( s ) ————–> CaO ( s ) + CO2 ( g )

85.  Is decrease in enthalpy a criterion for
spontaneity?
Tendency for minimum energy:
For example
1) A stone lying at height has a tendency to fall
down so as to have minimum potential energy
2) Water flows down a hill to have minimum energy
3) A wound watch spring has tendency to unbind
itself to decrease its energy to minimum
4) Heat flows from hot body to cold body so that
heat content of the hot body becomes minimum
All the processes are spontaneous because they
have tendency to acquire minimum energy.

86.  Consider the following exothermic reaction, all of
which are spontaneous :
1) H2 ( g ) + ½ O2 ( g ) ———–> H2O ( l ) ΔrH° = -
285.8 KJ mol-1
2) N2 ( g ) + 3H2 ( g ) ———–> 2NH3 (g) ΔrH° = -
92.2 KJ mol-1
3)C ( s ) + O2 ( g ) ————> CO2 ( g ) ΔrH° = -285.8
KJ mol-1
All these reactions are accompanied by evolution of
heat. The heat content of the products is less than
those of the reactants. These reactions
are spontaneous because they are accompanied by
decrease of energy.
A tendency to attain minimum energy i.e. a negative
value of enthalpy change, might be responsible for a
process or a reaction to be spontaneous or feasible.

87. limitation of the criteria for minimum energy
A number of reactions are known which are
endothermic i.e. for which ΔH is positive but still
they are spontaneous.
1) Evaporation of water or melting of ice
It takes place by absorption of heat from the
surrounding.
H2O ( l ) ———-> H2O ( g ) Δvap H° = 40.8 KJ
mol-1
H2O ( s ) ———-> H2O ( l ) Δfus H° = 6 KJ mol-1
2) Dissolution of salts like NH4Cl , KCl
NH4Cl (s) + aq ——–> NH4
+(aq) + Cl– (aq )
ΔrH° = 15.1 KJ mol-1

88. 3) Decomposition of calcium carbonate on
heating
 CaCO3 ( s ) ————> CaO (s ) + CO2 ( g )
ΔrH° = 177.8 KJ mol-1
4) Decomposition of N2O5 at room temperature
 2N2O5 ( g ) ———–> 4NO2 ( g ) + O2 ( g )
ΔrH° = 219 KJ mol-1
5) Decomposition of mercuric oxide on heating
 2HgO ( s ) ———–> 2Hg ( l ) + O2 ( g )
ΔrH° = 90.8 KJ mol-1

89.  A number of reactions are known for which ΔH
is zero but still they are spontaneous
 Reaction between acetic acid and ethyl alcohol
CH3COOH ( l ) + C2H5OH ( l ) ———>
CH3COOC2H5 ( l ) + H2O ( l )
2) Expansion of an ideal gas into vacuum
3) Even those reactions for which ΔH is negative,
rarely proceed to completion even though ΔH
remains negative throughout.
4) Reversible reactions also occur.
 H2 ( g ) + I2 ( g ) ——>2HI (g )
 2HI (g ) ———-> H2 ( g ) + I2 ( g )

90.  The energy factor or enthalpy factor cannot be the
sole criterion for predicting the spontaneity or the
feasibility of a process.
Tendency for maximum randomness:
 Suppose the two gases are enclosed in bulbs A and B
connected to each other by a tube and kept separated
by a stopcock. If the stock cock is opened the two
gases mix completely. The gases which were confined
to bulbs A and B separately are no longer in order. A
disorder has come in or the randomness of the system
has increased.
 A spontaneous process for which ΔH = 0 is the
spreading of a drop of ink in a beaker filled with
water.
 Second factor which is responsible for the spontaneity
of a process is the tendency to acquire maximum
randomness.

91.  A system at higher temperature has greater
randomness in it than one at lower
temperature.
 The entropy change is inversely proportional
to the temperature, s is related with q and T
for a reversible reaction.
 s=
𝑞
𝑇

92. For these processes, we have to consider the total
entropy change of the system and the surrounding.
ΔStotal = ΔSsystem + ΔSsurrounding
For the process to be spontaneous, ΔStotal must be
positive.
For all spontaneous processes, the total entropy
change ( ΔStotal ) must be positive.
ΔStotal = ΔSsystem + ΔSsurrounding . >0
The randomness and hence the entropy keeps on
increasing till ultimately an equilibrium is reached.
The entropy of the system at equilibrium is
maximum and there is no further change in entropy
i.e. ΔS =0.

93. 1)If ΔStotal or ΔSuniverse is positive, the process is
spontaneous.
2)If ΔStotal or ΔSuniverse is negative, the direct
process is non-spontaneous whereas the
reverse process may be spontaneous
3)If ΔStotal or ΔSuniverse is zero, the process is in
equilibrium.

94. Second Law Of Thermodynamics
All spontaneous processes are thermodynamically
irreversible.
or
Without the help of an external agency, a
spontaneous process cannot be reversed.
For Ex: Heat cannot by itself flow from a colder to
hotter body.
or
The complete conversion of heat into work is
impossible without leaving some effect elsewhere
or
All spontaneous processes are accompanied by a
net increase of entropy i.e. for all the spontaneous
processes, the total entropy change of the system
is positive.

95. Another thermodynamic quantities that helps in
predicting the spontaneity of a process is Gibbs
free energy or Gibbs energy of Gibbs function.
It is denoted by G and is given by the equation
G=H -TS
where H is the heat content ,T is the absolute
temperature and S is the entropy of the system.
G1 = H1 -TS1 for the initial state
G2 = H2 -TS2 for the final state
G2 -G1 =( H2 -H1) -T (S2 -S1)
ΔG = ΔH -TΔS

96. ΔG =G2 -G1 is the change in Gibbs free energy
of the system
ΔH= H2 -H1 is the enthalpy change of the
system
ΔS =S2 -S1 is the entropy change of the system.
ΔG = ΔH -TΔS is known as Gibbs – Helmoholtz
equation.
Gibbs free energy is that thermodynamic
quantity of a system the decrease in whose
value during a process is equal to the maximum
possible useful work that can be obtained from
the system.

97.  The total entropy change when the system is
not isolated from the surrounding is given by
ΔS = ΔSsystem + ΔSsurrounding
 Consider a process being carried out at
constant temperature and pressure. Suppose
the heat is lost by the surrounding and gained
by the system.If the heat lost by the
surrounding is represented by qp then by
definition of entropy change.

98. ΔSsurrounding = − qp /T
At constant pressure,
qP = ΔH
ΔSsurrounding = − ΔH /T
ΔStotal = ΔSsystem − ΔH /T
ΔStotal = ΔS − ΔH /T
Multiplying throughout by T ,we get
TΔStotal = TΔS – ΔH
ΔG = ΔH -TΔS
TΔStotal = – ΔG
ΔG = − TΔStotal
1) If ΔStotal is positive, the process is spontaneous .
2) If ΔStotal is zero, the process is in equilibrium.
3) If ΔStotal is negative ,the direct process is non
spontaneous, the reverse process may be spontaneous.

99. The criteria in terms of free energy change for
the spontaneity of the process will be as follow:
1) If ΔG is negative, the process will be
spontaneous.
2) If ΔG is zero, the process is in equilibrium.
3)If ΔG is positive, the direct process is non
spontaneous, the reverse process may be
spontaneous.

100. Free energy change and chemical equilibrium
 A reaction is spontaneous if ΔG is negative. In
a reversible reaction, the forward as well as
backward reaction takes place. ΔG is negative
for the forward as well as for the backward
reaction.
 The free energy change of the reaction in any
state ΔG is related to the standard free
energy change of the reaction,ΔG° according
to the equation
 ΔrG = ΔrG° + RT lnQ

101. where Q is the reaction quotient. It is equal to
Qp if the products and reactants are gaseous
and equal to Qc if they are in the solution.
When equilibrium is attained, there is no further
free energy change i.e. ΔG =0 and Q becomes
equal to K.
0 = ΔrG° + RT lnQ
ΔrG° = − RT lnQ
ΔrG° = −2.303 RT log K

102. According to Gibbs energy equation,
ΔrG° = ΔrH° – T ΔrS°
RT ln K = T ΔrS° – ΔrH°
ln K = ( T ΔrS° – ΔrH° ) / RT
 For strongly endothermic reaction, ΔH° is large
and positive. Hence the term on the right hand
side have a small value and so will be ln K and
hence K.
 For strongly exothermic reaction, ΔH° is large
and negative. Hence the term on the right hand
side will have large value and so will be ln K and
hence K.

104.  Third law of thermodynamics
 The entropy of a pure substance increases with
increase of temperature because molecular
motion increases with increase of temperature.
 The entropy of a perfectly crystalline solid
approaches zero as the temperature
approaches absolute zero.
 The entropy of all perfectly crystalline solids
may be taken as zero at absolute zero of
temperature.