introduction,First law.Zerothlaw of thermodynamics

1. 1
Thermodynamics
Part - I
Mr.M.RAGU
Assistant Professor of Chemistry
Vivekananda College
Tiruvedakam West – 625234
Madurai -Tamilnadu

2. INTRODUCTION TO THERMODYNAMICS
Scope and Importance of Thermodynamics
The term ‘thermodynamics’ literally means ‘flow of
heat’. Heat is one form of energy and thermodynamics
deals with the interconversion of various kinds of energy
(heat, light, mechanical and electrical energy) in physical
and chemical processes. Thermodynamics is a useful guide
to understand the energy changes as given below:
i) It predicts the feasibility of a physical process or chemical
reaction under the given conditions of temperatures and
pressure.
ii) It predicts whether a chemical reaction would occur
spontaneously or not under a given set of conditions.
iii) It helps to determine the extent to which a reaction
would take place.
iv) It introduces the concept of temperature and provides
methods to measure it.
v) It helps to deduce important generalisations including
Van’t Hoff’s law, Raoult’s law and the Phsae rule.

3. Terms used in Thermodynamics
System and Surroundings
⚫ Any part of the universe which is chosen
for thermodynamic study is called a
‘system’. The rest of the universe in its
neighbourhood is known as the
‘surroundings’. The system is separated
from its surroundings by a real or imaginary
boundary through which exchange of
energy or matter may take place.

4. Types of Systems
⚫1. Homogeneous system: A system is said to be
‘homogeneous’ if it consists of only one phase and
uniform throughout.e.g. Air, a solution of sugar in
water.
⚫2. Heterogeneous system: When a system consists
of two or more phases and is not uniform throught. It
is called ‘heterogeneous’.e.g. ice in water, two
immiscible liquids (chloroform-water).

5. 3. Macroscopic system: A system which contains
measurable quantities of atoms, molecules or ions
is known as a ‘macroscopic system’.
4. Microscopic system: A system is said to be
“microscopic” if it contains microscopic quantities
of atoms, molecules or ions whose amount can not
be measured.
5. Open system: A system which can exchange
both matter and energy with its surroundings is
called an “open system”.
6. Closed system: A system which can exchange
energy but not matter with its surroundings
through a ‘diathermal wall’ is known as a ‘closed
system’.e.g. Water in stoppered bottle.
7. Isolated system: It is a system in which neither
energy nor matter can exchange with its
surroundings. It contains an ‘adiabatic wall’
(diathermal wall with insulation) as boundary.e.g.
Milk in a thermos flask.

7. Properties of a system
⚫ The properties associated with a macroscopic system are called
thermodynamic properties or variables. They divided into two classes:
⚫ i) Extensive properties: These are thermodynamic properties which
depend on the quantity of matter specified in the system.e.g. mass,
volume, energy etc.
⚫ Ii) Intensive properties: These are thermodynamic properties which
depend upon the characteristics of matter but independent of its amount
in the system e.g. pressure, temperature, viscosity, density, surface
tension, melting point, boiling point etc.

8. State of a system
⚫ The system associated with definite values of
thermodynamic properties is said to be in a definite ‘state’. If
these properties change , the system will attain a new state.
⚫ The thermodynamic properties which depend only on the
initial and final states of the system are called “state functions”
or “state variables”.e.g. pressure, volume, temperature,
composition, energy (internal energy, enthalpy, entropy) etc.
⚫ There are some physical properties (e.g heat and work)
which depend on the path by which the system changes
from the initial to the final state. These are known as
“path functions”.

9. Thermodynamic equilibrium
⚫ A system is said to be in thermodynamic equilibrium when its observable properties
do not change with time. There are simulataneous existence of three types of
equilibrium.
⚫ i) Thermal equilibrium: A system is said to be in thermal equilibrium if the
temperature remains the same throughout the system.
⚫ ii) Chemical equilibrium: For a system consisting of more than one substance , if the
composition does not vary with time, it is said to be in chemical equilibrium.
⚫ Iii) Mechanical equilibrium: If the pressure is uniform in all parts of the system,
there will be no macroscopic movement within the system. Now, the system is said to
be in mechanical equilibrium.

10. Thermodynamic processes
⚫ The method by which the state of a system is
changed is called a “process”. It can be effected by
changing any one of a state variables viz P, T etc.
Some common types of processes are:
⚫1. Isothermal process: It is a process in which
exchange of heat takes place between the system and
the surroundings so that the temperature of the
system remains the same as that of its surroundings.
It means it is carried out at constant temperature.

11. 2. Adiabatic process: A process in which there is no exchange
of heat between the system and the surroundings is known as an
“adiabatic process”. During an adiabatic change, the system and
the surroundings are at different temperatures.
3. Isobaric process: It is a process that is carried out at constant
pressure.
4. Isochoric process: If the process occurs at constant volume.
It is called “isochoric process”.
5. Cyclic process: If a system undergoes a series of changes in
such a way that it return to its initial state, it is referred to as a
“cyclic process”.

12. 6. Reversible process: It is a process which takes place
infinitesimally slowly so that the system is in thermodynamic
equilibrium at any stage of the change. Since the process is carried
out extremely slowly, the properties of the system remain virtually
unchanged and the direction may be reversed by a small change in
a variable like temperature, pressure etc. The driving force is
greater than the opposing force only by an infinitesimal quantity
and hence the process would require infinite time for its
completion.
7. Irreversible process: It is a process which takes place rapidly or
spontaneously so that it is not in equilibrium with its surroundings.
The driving force differs from the opposing force by a large amount
and hence it can not be reversed unless some external force is
applied. All natural processes are irreversible. They take finite time
for their completion.

13. Reversible process Irreversible process
i) Proceeds extremely
slowly
Proceeds rapidly
ii) Carried out in an
infinite number of
steps
Carried out in a single step
iii) The system remains in
equilibrium with its
surroundings
The system does not
remain in equilibrium with
its surroundings
iv) Can be reversed by a
small change in a
variable
Can not be easily reversed
v) Takes infinite time for
completion
Takes finite time for
completion
vi) Ideal process Real process
vii) Maximum work is
obtainable
Only a small amount of
work is obtainable
Differences between Reversible and Irreversible processes

14. Work, Energy and Heat
⚫Work: If an object is lifted through a distance ‘L’
against a force, F then the amount of work which has
to be done is defined as
⚫ W = F x L
⚫ The unit of work in SI system is the Joule(J) , which
is equal to one Newton metre
⚫ 1J = 1Nm
⚫ The work done can be expressed as the product
of intensity factor and capacity factor

15. Work Intensity factor Capacity factor
i) Gravitational
force(mgh)
mg h
ii) Mechanical
work(PΔV)
P ΔV
iii) Electrical
work(QV)
Potential
difference(V)
Quantity of
electricity (Q)

16. Energy:
⚫ The energy of a system may be defined as “any property
which is capable of doing work”. There are several forms of
energy – thermal energy(heat), mechanical energy, electrical
energy, chemical energy etc. Energy can be quantitatively
converted into work and can be produced from work.
⚫ The unit of energy in C.G.S. system are:
⚫ i) erg ii) calorie iii) Joule
⚫ They are related as
⚫ 1Joule = 107 erg
⚫ 1 calorie = 4.184J
⚫ = 4.184 x 107 erg
⚫ The SI unit of energy is the Joule(J).

17. Heat
⚫ Heat is one form of energy and can be produced from
work. However, it is not completely convertible into work.
It can only partly be transformed into work. In this respect
heat differs from all other forms of energy.
⚫Convention of expressing heat and work
⚫ Heat is expressed as q, work is expressed as w
⚫q is +ve when heat is absorbed by the system
⚫q is –ve when heat is given out by the system
⚫w is +ve when work is done by the system
⚫w is –ve when work is done on the system

18. Work done by an ideal gas during
isothermal irreversible expansion
⚫ When a gas expands irreversibly under isothermal
conditions against an external pressure p1 the volume
changes considerably from V1 to V2 in a single step. Now
the gas performs mechanical work which is also known as
“work of expansion” or “pressure – volume work”.
⚫Work done by the gas=External pressure x Volume
change
⚫ w= P(V2 - V1)
⚫ = PΔV
⚫ Since (V2 - V1) is positive, the value of w will be positive

19. Work done by an ideal gas during isothermal reversible expansion
⚫ Isothermal reversible expansion is a multistep process. Since the external pressure
is only infinitesimally lower than the pressure of the gas , the process occurs
extremely slowly. In each step of expansion, there will be an infinitesimal change of
volume (dV).
⚫ δw = Pext dV
⚫ = (P – dP)dV
⚫ = PdV – dP Dv
⚫ Neglecting the very small second order product dPdV we get
⚫ δw = PdV
⚫ where P = Pressure of the gas = External pressure
⚫ The total work done by the gas is obtained.
⚫ On integration,
⚫ ʃəw = ʃPdV
⚫ w = v1 ʃv2 PdV
⚫ For an ideal gas
⚫ PV = nRT
⚫ ჻ P = nRT/V

20. Substituting the value of P
w = nRT v1 ʃv2 dV/V
= nRT ln V2/V1
= 2.303 nRT log V2/V1
Also P1V1 = P2V2
჻V2/V1= P1/P2
჻ w = 2.303 nRT log P1/P2

21. Work done by a van der Waals’ gas during
isothermal reversible expansion
⚫ For one mole of a van der Waals’ gas
⚫ [P+a/V2](V-b) = RT
⚫ or P = [RT/(V-b)-(a/V2)]
⚫ w = v1 ʃv2 PdV
⚫ = v1 ʃv2 [RT/(V-b)-(a/V2)]Dv
⚫ = v1 ʃv2 [RT/(V-b)]dV-v1ʃv2(a/V2) dV
⚫ = RT v1 ʃv2 dV/(V-b) –a v1ʃv2dV/V2
⚫ = RT[ln(V-b) v1]v2 – a[-1/V v1]v2
⚫ = RT ln (V2-b)/(V1-b) - a[-(1/V2 ) – (1/V1)]
⚫ = RT ln (V2-b)/(V1-b) + a[(1/V2 ) – (1/V1)]
⚫ = 2.303 RT log (V2-b)/(V1-b)+a[(1/V2 )–(1/V1)]
⚫

22. Zeroth Law of Thermodynamics
⚫ The law states that “if two systems A and B are in
thermal equilibrium with the system C, then A and B
are also in thermal equilibrium with each other.

23. Significance of Zeroth law:
Zeroth law of thermodynamics introduces the concept of
“temperature” and provides methods to measure it. A hot body is said to
have a higher temperature than a cold body. If a hot body is placed in
contact with a cold body. If a hot body is placed in contact with a cold
body, heat transforms from the former to the later till they attain thermal
equilibrium. The temperature difference can be measured with a
“thermometer”. Suppose a thermometer is placed in contact with a hot
body untill thermal equilibrium is reached, the temperature of the body is
represented by the position of the mercury column in the thermometer.

24. The first law of thermodynamics
⚫ The first law of thermodynamics stated in two forms:
⚫ i) as the law of conservation of energy:
⚫ “Energy can neither be created nor destroyed,
although it may be converted from one form into
another”. It implies that total energy of a system and
its surroundings remains constant.

25. ii) in terms of increase in Internal energy:
Internal energy is the total energy content of the
system. It is due to the translational, vibrational and
rotational motions of the molecules and their mutual
attraction in a system. The internal energy changes
when the system changes from one state to another.
The energy change ΔE depends on the initial and final
state of the system and hence it is a state function.
The internal energy change of a system is equal to
the quantity of heat absorbed minus the mechanical
work done by the system.
dE = δq – δw
Since, work = Pressure x Volume change
δw = PdV
჻ dE = δq - PdV

26. Heat content or Enthalpy
⚫ When work is done on a system, an equivalent amount of heat is evolved.
For example, when a system absorbs heat at constant volume, no mechanical
work is done.
⚫ q = ΔE + w
⚫ = ΔE + PΔV
⚫ or qw = ΔE (since ΔV = 0)
⚫ When the system absorbs heat at constant pressure, the volume of the
system changes and performs mechanical work.
⚫ qp = ΔE + PΔV
⚫ The heat changes of a system at constant pressure is designated as ΔH. Thus,
⚫ ΔH = ΔE + PΔV
⚫ or H = E + PV
⚫ Where H is a thermodynamic function called “heat content” or ”enthalpy” of
the system. Chemical reactions are isoboric processes. The quantity of heat
absorbed or liberated in the course of a reaction is measured in terms of
“enthalpy change”.
⚫ HProducts – HReactants = ΔH
⚫ ΔH is +ve for endothermic reactions (heat absorbed)
⚫ ΔH is -ve for exothermic reactions (heat liberated)

27. Internal Energy of an ideal gas is a function of its
temperature but not of its volume
⚫ E = f(T, V)
⚫ dE = (ƏE/ƏT)V dT + (ƏE/ƏV)T dV
⚫ For an ideal gas dE = 0
(ƏE/ƏV)T dV = - (ƏE/ƏT)V dT
or (ƏE/ƏV)T = - (ƏE/ƏT)V (ƏT/ƏV)E
By definition,
(ƏE/ƏT)V = CV → E = f(T)
჻ (ƏE/ƏT)V = CV (ƏT/ƏV)E
For an isothermal process dT = 0
჻ (ƏT/ƏV)E = 0
჻ (ƏE/ƏV)T = 0 → E = f(V)

28. Heat Capacity
⚫ “The quantity of heat required to raise the temperature of a
subatance (solid, liquid or gas) by one degree” is called its “heat
capacity”.
⚫ If one gram of the substance is considered, the heat capacity is
known as “specific heat capacity” or “specific heat”.
⚫ For one mole of the substance, it is called “molar heat capacity”.
⚫ If δq is a very small quantity of heat absorbed by a substance and
dT is its rise in temperature, then the heat capacity is defined as
⚫ C = δq/dT
⚫ According to the First law of Thermodynamics,
⚫ δq = dE +PdV
⚫ ჻ C = (dE +PdV)/dT
⚫ If this substance is a gas, there are two types of heat capacities

29. i) Heat capacity at constant volume (CV)
⚫ C = (dE +PdV)/dT (By definition)
⚫ If the absorption of heat takes place at constant volume, dV = 0. Then, the
equation reduces as
⚫ C = dE/dT
⚫ or CV = (ƏE/ƏT)V
⚫ Thus, the heat capacity at constant volume is defined as “ the internal energy
change per degree rise of temperature”.
⚫ ii) Heat capacity at constant pressure (CP)
⚫ C = (dE +PdV)/dT (By definition)
⚫ If the gas absorbs heat at constant pressure, dP = 0. Then,
⚫ dH = dE+PdV
⚫ ჻ C = dH/dT
⚫ or CP = (ƏH/ƏT)P
⚫ Thus, the heat capacity at constant pressure is defined as “ the enthalpy change
per degree rise of temperature”.

30. Thermodynamic derivation of the CP – CV relation
⚫ By definition,
⚫ CP = (ƏH/ƏT)P
⚫ But H = E + PV
⚫ ჻ CP = [Ə(E + PV)/ƏT]p
⚫ or CP = (ƏE/ƏT)p + P(ƏV/ƏT)P
⚫ E is a state function and dE is a complete differential.
⚫ E = f(T, V)
⚫ ჻ dE = (ƏE/ƏT)V dT + (ƏE/ƏV)T dV
⚫ Dividing both sides by dT at constant pressure.
⚫ (ƏE/ƏT)P = (ƏE/ƏT)V + (ƏE/ƏV)T (ƏV/ƏT)P
⚫ Substituting the value of (ƏE/ƏT)P in the Eqn.
⚫ CP = (ƏE/ƏT)V + (ƏE/ƏV)T (ƏV/ƏT)P + P(ƏV/ƏT)P
⚫ By definition,
⚫ CV = (ƏE/ƏT)V
⚫ ჻ CP = CV + (ƏE/ƏV)T (ƏV/ƏT)P + P(ƏV/ƏT)P
or CP - CV = (ƏE/ƏV)T (ƏV/ƏT)P + P(ƏV/ƏT)P
⚫ = (ƏV/ƏT)P [P + (ƏE/ƏV)T]
⚫ For an ideal gas (ƏE/ƏV)T = 0
჻ CP - CV = P(ƏV/ƏT)P
⚫ For one mole of an ideal gas
⚫ PV = RT
⚫ Differentiating the equation w.r.to T at constant pressure
⚫ P(ƏV/ƏT)P = R
⚫ Comparing both we get
⚫ CP – CV = R

31. Joule-Thomson Effect
⚫ When a compressed gas is allowed to expand by
passing through a porous plug under adiabatic
condition, the gas gets cooled. This phenomenon is
called Joule–Thomson effect.

32. Cause for Joule-Thomson Effect
When a gas expands from the state P1V1 to state P2V2, the
work done by the gas is given as
w = P2V2- P1V1
q = ΔE + w
For adiabatic processes, q = 0
჻ ΔE = - w
Thus, the gas performs work at the cost of its internal
energy. That is, internal energy and hence the temperature of
the gas falls.
Importance of Joule-Thomson Effect
i) It forms the basis for liquefaction of gases.
ii) For an ideal gas (ƏE/ƏV)T = 0 and so it does not involve any
temperature change when subjected to Joule-Thomson
effect.
iii) Hydrogen and helium experience a heating effect during
adiabatic expansion at room temperature. This is because of
their very low “inversion temperature”.

33. Joule Thomson Co-efficient
⚫ The drop in temperature that gas experiences during adiabatic expansion is
proportional to the pressure difference maintained
⚫ ΔT α ΔP
⚫ “The rate of change of temperature with pressure at constant enthalpy” is
known as the “Joule-Thomson co-efficient” and is denoted by the symbol μJ.T.
⚫ μJ.T = (ƏT/ƏP)H
⚫ Expression for μJ.T
⚫ H = f(T, P)
⚫ dH = (ƏH/ƏT)V dT + (ƏH/ƏP)T dP
⚫ By definition,
(ƏH/ƏT)P = Cp
⚫ In Joule-Thomson experiment, enthalpy remains constant,
⚫ dH = 0
⚫ ჻ 0 = Cp dT + (ƏH/ƏP)T dP
⚫ or (ƏT/ƏP)H = -1/Cp (ƏH/ƏP)T
⚫ or μJ.T = -1/Cp (ƏH/ƏP)T

34. μJ.T for an Ideal gas
⚫ H = E+PV
Differentiating w.r. to P at constant T
(ƏH/ƏP)T = (ƏE/ƏP)T + (ƏPV/ƏP)T
μJ.T = -1/Cp (ƏH/ƏP)T
Substituting the value of (ƏH/ƏP)T
μJ.T = -1/Cp [(ƏE/ƏP)T + (ƏPV/ƏP)T]
=-1/Cp[(ƏE/ƏV)T(ƏV/ƏP)T+ (ƏPV/ƏP)T]
⚫ For ideal gases, (ƏE/ƏV)T = 0
⚫ ჻ μJ.T = -1/Cp (ƏPV/ƏP)T
⚫ For an ideal gas,
⚫ PV = Constant
⚫ Differentiating the above equation w.r.to P at constant temperature
(ƏPV/ƏP)T = 0
⚫ Hence,
⚫ μJ.T = 0

35. μJ.T for Real gases
⚫ For a van der Waals gas
⚫ [P+(a/V2)](V-b) = RT
⚫ or PV+ (a/V)-Pb-(ab/V2) = RT
⚫ Since a and b are small, the term (ab/V2) can be neglected.
⚫ PV+ (a/V)-Pb = RT
⚫ or PV= RT- (a/V)+Pb
⚫ or V= (RT/P)- (a/PV)+b
⚫ = (RT/P)- (a/RT)+b [sincePV=RT]
⚫ Differentiating the above equation w.r.to T at constant pressure.
⚫ (ƏV/ƏT)P = (R/P)+(a/RT2)
⚫ Rearranging eqn
⚫ V-b = (RT/P)-(a/RT)
⚫ Dividing both sides by T
⚫ (V-b)/T = (R/P) - (a/RT2)
⚫ or R/p = (V-b)/T + (a/RT2)
⚫ Substituting the value of R/P
⚫ (ƏV/ƏT)P = (V-b)/T + (a/RT2) + (a/RT2)
⚫ = (V-b)/T + (2a/RT2)

36. Multiplying both sides by T
T (ƏV/ƏT)P = (V-b) + (2a/RT)
or T (ƏV/ƏT)P - V = (2a/RT) – b
According to Second law of Thermodynamics
T (ƏV/ƏT)P – V = -(ƏH/ƏP)T
On comparing both equations
-(ƏH/ƏP)T = (2a/RT) – b
We know,
μJ.T = -(1/CP) (ƏH/ƏP)T
Substituting the value of (ƏH/ƏP)T
μJ.T = (1/CP) [(2a/RT) – b]

37. Inversion Temperature
⚫ For ideal gases, (ƏH/ƏP)T is zero
⚫ so that μJ.T = 0
⚫ This implies that adiabatic expansion of an ideal does not involve any temperature
change.
⚫ For a real gas,
⚫ μJ.T = (1/CP) [(2a/RT) – b]
⚫ μJ.T may be positive, negative or zero depending upon the values of (2a/RT) and b
⚫ i) When 2a/RT > b, μJ.T is positive (cooling)
⚫ ii) When 2a/RT = 0, μJ.T is zero (no change)
⚫ iii) When 2a/RT < b, μJ.T is negative (heating)
⚫ Since a, b and R are constants, the sign of μJ.T depends on the temperature at which
the gas is allowed to expand under adiabatic conditions.
⚫ The temperature at which the Joule Thomson coefficient changes its sign from
positive to negative or vice versa is called the “inversion temperature”. Evidently, at this
temperature μJ.T = 0. That is, 2a/RTi = b
⚫ or Ti = 2a/Rb
⚫ where Ti represents the inversion temperature.

38. Physical significance of Inversion Temperature
⚫ i) Gases except hydrogen and helium experience cooling effect hen allowed
to expand adiabatically at room temperature. This is due to the high
inversion temperature of the gases.
⚫ When Ti > T
⚫ 2a/Rb > T [Since Ti = 2a/Rb]
⚫ or 2a/RT > b
⚫ ჻ μJ.T is +ve (cooling effect)
⚫ ii) In the case of H2 and He the inversion temperatures are so far below
the room temperature. Hence, they have μJ.T = negative and experience
heating effect during adiabatic expansion at room temperature.
⚫ When Ti < T
⚫ 2a/Rb < T
⚫ or 2a/RT < b
⚫ ჻ μJ.T is -ve (heating effect)

39. THERMOCHEMISTRY
⚫ Chemical reactions are accompanied by energy changes
mainly in the form of heat. The branch of science which
deals with heat changes during chemical reactions is
called “thermochemistry”.
⚫Heat of Reaction:
⚫ The quantity of heat evolved or absorbed during
chemical reactions is called the “heat of reaction”. When a
chemical reaction proceeds at constant volume , the heat
of reaction is equal to the “internal energy change” ΔE. If
the chemical reaction is carried out at constant pressure,
the heat of reaction is equal to the “enthalpy change” ΔH.
Since chemical reactions usually occur at constant
pressure, the heat of reaction is called “enthalpy of
reaction”.

40. Relation between ΔE and ΔH of a chemical
reaction
⚫ ΔH = ΔE +PΔV
⚫ PΔV = P(V2-V1)
⚫ = PV2-PV1
⚫ For an ideal gas,
⚫PV = nRT
⚫჻ PV2-PV1 =(n2-n1)RT
⚫or PΔV = ΔnRT
⚫Substituting the value of PΔV
⚫ΔH = ΔE +ΔnRT

41. Exothermic and Endothermic Reactions
⚫ΔH = Hproducts – Hreactants
⚫ = HSurroundings – Hsystem
⚫When the reaction proceeds with absorption of heat
from the surroundings
⚫ Hsur > Hsys
⚫or HP > HR
⚫჻ ΔH is +ve................. “Endothermic Reaction”
⚫If ΔH is -ve................. “Exothermic Reaction”

42. Enthalpy change of Reactions
1. Enthalpy of Solution: When a substance is dissolved in a solvent,
heat may either be liberated or absrbed. The heat (enthalpy) change
associated with the dissolution of a substance in a solvent is called
the “heat (enthalpy) of solution”. It is not a constant quantity but
varies with the concentration of the solution. With reference to the
concentration of the solution, enthalpy of solution is expressed in
two different ways.
i) Integral Enthalpy of Solution
“ The enthalpy change per mole of solute dissolved in a specified
quantity of solvent” is called the “Integral heat of solution”. If ΔH is
the total enthalpy change when n2 moles of a solute are dissolved in a
definite quantity of a solvent(e.g.n1 mole of water or 1000 gs of water)
then ΔH/n2 is the integral enthalpy of solution at the given
concentration.
ii) Differential Enthalpy of solution
It is defined as the “enthalpy change caused when 1 mole of the
solute is dissolved in such a large volume of solution that there is no
appreciable change in its concentration”. It is expressed in
d(ΔH)/dn2.

43. 2. Enthalpy of Dilution: Dilution of a solution from one
concentration to another is accompanied by enthalpy change called
“enthalpy of dilution”. It is expressed in two different ways:
i) Integral enthalpy of dilution
With respect to the integral heat, consider a process in which a
certain amount of solution diluted from an initial concentration to a
final concentration. The enthalpy change in this process, normalized
by the mole number of solute, is evaluated as the molar integral heat
of dilution.
ii) Differential enthalpy of dilution
It is defined as “the enthalpy change associated when 1 mole of the
solvent is added to such a large volume of solution that its
concentration does not change appreciably. This is given by the slope
of the curve obtained by plotting ΔH of dilution at various
concentrations against the ratio n1/n2.

44. Variation of Enthalpy of reaction with
temperature (Kirchoff’s Equation)
⚫ For a given reaction, ΔH varies with temperature. The variation of enthalpy of reaction (ΔH)
with temperature is given by “Kirchoff’s equation” which may be derived as follows:
⚫ If H1 and H2 are the enthalpy of reactants and products respectively
⚫ ΔH = H1+H2
⚫ Differentiating the equation w.r.to T at constant pressure
⚫ [ə(ΔH)/əT]P = [əH2/əT]P - [əH1/əT]P
By definition,[əH/əT]P = CP
⚫ Now, [ə(ΔH)/əT]P = (CP)2-(CP)1 = ΔCP
⚫ or d(ΔH)/dT = ΔCP
⚫ ჻ d(ΔH) = ΔCP.dT
⚫ On integration,
⚫ ΔH1 ʃ ΔH2 d(ΔH) = ΔCP T1ʃ T2 dT
⚫ ΔH2 – ΔH1 = ΔCP (T2 – T1)
⚫ where, ΔH1 = Enthalpy of reaction at T1
⚫ ΔH2 = Enthalpy of reaction at T2
⚫ Application
⚫ Usually, the enthalpy of a reaction is determined at 298 K and 1 atm. pressure. The Kirchoff’s
equation is used to calculate the enthalpy of a reaction at any desired temperature.

45. Thermochemical laws
1. Lavoisier and Laplace law:
⚫ The law states that “the quantity of heat required to
decompose a compound into its element is equal to the
heat evolved when the compound is formed from its
elements”. It may also be stated in another form as “the
heat change accompanying a chemical reaction in one
direction is exactly equal in magnitude but opposite in
sign to that associated with the same reaction in the
opposite direction”. Thus, the law admits that
thermochemical equations can be reversed, provided the
sign of ΔH is changed.e.g.
⚫CH4(g)+2O2(g)→CO2(g)+2H2O(l) ΔH = -212.8 K cal
⚫CO2(g)+2H2O(l) → CH4(g)+2O2(g) ΔH = 212.8 K cal

46. 2. Hess’s law of constant heat summation
The law states that “the total enthalpy change for a
chemical reaction is the same whether the reaction takes place
in one step or in several steps”. It is another way of stating the
first law of thermodynamics which implies that the enthalpy of
a reaction depends only on the initial and final states of the
system and is independent of the path connecting them.
As an illustration
i) Formation of CO2 in a single step
C(s)+O2 → CO2(g) ΔH = -Q1 cals
ii) Formation of CO2 in two steps
C(s)+ ½ O2 → CO(g) ΔH = -Q2 cals
CO(g)+ ½ O2 → CO2(g) ΔH = -Q3 cals
It is found that Q1 = Q2 + Q3

47. Applications
i)Calculation of enthalpy of formation
It is found that there is no suitable method
to measure directly the enthalpy of formation of
carbon monoxide. However, it can be calculated
from thermochemical data by applying Hess’s
law.
a) C(s)+ ½ O2 → CO(g) ΔH = x cals
b) CO(g)+ ½ O2 → CO2(g) ΔH = Q1 cals
c) C(s)+O2 → CO2(g) ΔH = Q2 cals
Add eqns (a) and (b) and equate to (c)
x + Q1 = Q2
჻ x = Q2 - Q1

48. ii) Calculation of enthalpy of transition
⚫ The transition of rhombic sulphur to monoclinic sulphur
is so slow that direct measurement of enthalpy change is
not possible for such a transition. However, it may be
calculated using Hess’s law.
⚫ a) SR + O2(g) → SO2 (g) ΔH = Q1 cals
⚫ b) SM + O2(g) → SO2(g) ΔH = Q2 cals
⚫ c) SR → SM ΔH = x cals
⚫ Subtract eqn (b) from (a) and equate to (c)
⚫ Q1 - Q2 = x

49. Bond Energy and Bond dissociation Energy
⚫ Whenever a chemical bond is formed, energy is released while bond breaking occurs
with absorption of energy. “The energy required to break a bond between two atoms in
a molecule in the gaseous state” is called the “bond dissociation energy”. For example,
04 K cal of heat is necessary to break the H-H bond in H2 molecule
⚫ H2(g) → 2H(g) ΔH = 104 K cal
⚫ In a polyatomic molecule which contains more than one bond of the same type, the
bond dissociation energy is not the same for all the bonds. For example, the bond
dissociation energies of the C-H bond in methane are
⚫ CH4(g) → CH3(g) + H(g) ΔH = 435 K J
⚫ CH3(g) → CH2(g) + H(g) ΔH = 370 K J
⚫ CH2(g) → CH (g) + H(g) ΔH = 385 K J
⚫ CH (g) → C (g) + H(g) ΔH = 338 K J
⚫ The average of the bond dissociation energies in a polyatomic molecule in the
gaseous state is known as “bond energy”. For example , the bond energy of the C-H
bond is
⚫ EC-H = (435+370+385+338)/4
⚫ = 1528/4
⚫ = 382 KJ mol-1
⚫ Thus, bond energy is defined as “ the average energy per mole required to break a
particular bond in a molecule into its constituent atoms in the gaseous state”.

50. Calculation of Enthalpy of reaction from Bond energy
data
⚫ If the energies of the various bonds in the reactant and
product molecules are known. It is possible to calculate
the enthalpy of reaction. However, the values so estimated
are only approximate. For example, the enthalpy of
hydrogenation of ethylene is calculated from the bond
energy data.
H H
I I
H- C= C- H + H- H → H- C- C- H
I I I I
H H H H

51. a) Heat absorbed when the various bonds of the reactant
molecules are broken = Ha
4 C-H bonds = 4a
One C=C bond = b
One H-H bond = c
჻ Ha = 4a + b + c
b) Heat liberated when the various bonds of the product
molecules are formed = Hf
6 C-H bonds = 6a
One C-C bond = d
ΔH = Ha - Hf
= 4a + b + c - 6a – d
= b + c - 2a – d K cal mole-1