CHAPTER
1
THERMODYNAMICS–I
1.1 INTRODUCTION
The energy can be transformed from one form to another, but it can neither be created nor destroyed.
This is the fundamental law of nature known as the law of conservation of energy. This accounts
accurately for the energy changes that accompany chemical and physical transformations of matter.
Thermodynamics is that branch of science, which describes in detail the principles that govern energy
and energy changes. The study of thermodynamics is based essentially on three fundamental laws
which correlate the inter conversion of different forms of energy. These three laws of thermodynamics
can be stated in mathematical forms. Historically, the science of thermodynamics was developed to
provide a better understanding of heat engines, with special reference to the conversion of heat into
useful work. Two forms of energy, generally associated with the process occurring in the universe, are
heat and work. Thermodynamics, in general is the study of heat and work. The processes subjected to
thermodynamic studies include not only the natural phenomenon but also control chemical reactions,
the performance of engines, and even hypothetical processes or chemical reactions that do not occur
but can be imagined. Thermodynamics can successfully provide answers to the questions such as
(i) when two or more substances are mixed together, whether the reaction will occur or not, (ii) if the
reaction takes place, then what will be the final concentrations of the reactants and products at equilibrium
and (iii) how much heat will be associated with these chemical changes. However, thermodynamics
cannot tell us how fast the reaction or process can occur and what will be the mechanism of that
reaction or process. Thermodynamics provide useful informations about how energy is stored, released,
transferred, used and the nature and consequences of the transfer of energy.
1.2 THERMODYNAMIC TERMS AND CONCEPTS
There are certain terms frequently used in thermodynamics whose concept must be defined carefully
before we go deeper into this subject. These are:
(a) System, boundary and surroundings
Any part of the universe chosen for thermodynamic study is called a system. The system is confined at
a definite place in space by the boundary.
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PHYSICAL CHEMISTRY–II
The boundary separates the system from the rest of the universe. This rest of the universe which
is not under investigation is called surroundings. This can exchange matter or energy with the system.
A system may either be homogeneous or heterogeneous. A system is said to be homogeneous
when it is completely uniform throughout. A homogeneous system contains only one phase.
A system is said to be heterogeneous if it is not uniform throughout. A heterogeneous system
consists of two or more phases.
Types of systems
Based on exchange of energy, or matter or both with the surroundings, three types of systems occur.
These are:
(i) An open system: A system is said to be an open system when energy and matter both can pass
across the boundary to the surroundings so that there is change in mass (Dm) and energy (DE). Thus,
we can write
Dm ¹ 0 and DE ¹ 0.
(ii) A closed system: A closed system is one that exchange energy (DE ¹ 0) but not the matter
(Dm = 0) with the surroundings.
(iii) An isolated system: A system is an isolated system when the boundary prevents any type of
interactions or exchange with the surroundings i.e., it does not exchange matter or energy with the
surroundings (Dm = 0 and DE = 0).
An isolated system does not produce any effect or disturbance in its surroundings.
(b) Properties of a system
The properties of a system may either be (i) intensive or (ii) extensive in nature.
(i) Intensive properties
Intensive properties do not depend on the quantity of matter present in the system but depends on the
nature of the substance. Temperature, pressure, specific heat, chemical potential, boiling point, melting
point, freezing point, molar entropy, molar enthalpy, dielectric constant, viscosity, surface tension,
refractive index, density, molar volume, molarity and mole fraction etc. are all intensive properties.
(ii) Extensive properties
Extensive properties depend on the quantity of the matter of a system which is being kept under
investigation. Volume, energy and mass are familiar examples of extensive properties.
(c) State of a system, state functions and path functions
In order to study and discuss the changes which occur in a system, it is necessary to define its properties
accurately before and after the change has occurred. This is done by specifying the state of a system.
This means that temperature, pressure and composition of the system are specified in such a way that
all other properties of the system are automatically fixed. Thus, a thermodynamic system is said to be
in a definite state when each of its properties has a definite value.
Pressure (P), temperature (T), volume (V), mass and composition are the fundamental properties
of a system which specify the state of a thermodynamic system. Any change in the value of these
properties brings change in the state of the system, therefore, the quantities P, V, T etc. are called state
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THERMODYNAMICS–I
functions or state variables (thermodynamic variables). Variables that depend only on the state of the
system are called state functions. The state functions determine the physical state of a given system. In
a particular state, the values of the state functions do not depend on the previous history of the sample.
For example, the volume of 1 mole of water at 25°C and 1 atm pressure does not depend on what was
its temperature and pressure sometimes before in the past. Further a change in one of these variables
depends on the difference between initial conditions and final conditions, but it does not depend how
the change of state has been done. The change in state functions (P, V, T etc.) from one state to another
state is measured only as the difference between the final state and the initial state.
→ State II (P2, V2, T2)
State I (P1, V1, T1) 
On the other hand those variables, that depend on how a change takes place in a system, are called
path functions. For example, amount of work done in a chemical process depends on how the change
has taken place. Therefore, it is called path function. The path of the change in state is defined by
giving the initial state, the sequence of intermediate states traveled by the system and then the final
state. The three state variables P, V, and T in a pure gas are inter related with each other by a relationship
known as the equation of state. For one mole of a pure gas, the equation of state is written as
PV = RT (where R is gas constant)
In this case, if out of the three variables, say only P and T are specified, then the value of the third
state variable V gets fixed automatically and is obtained with the help of equation of state. The state
variables P and T which were specified to define the state of a system are called as independent state
variables while the state variable V which depends on the values of P and T is known as dependent state
variable.
(d) Thermodynamic equilibrium
When the state variables such as temperature, pressure and concentration of all the components of a
system are same at all the parts in the system and do not change with time, then the system is said to be
in a state of thermodynamic equilibrium. However, a system is said to be in a state of non-equilibrium
when the state variables have different values at different parts of the system.
(e) Thermodynamic processes
A process is the method of operation, which occurs in a system when any sort of change or transformation
(chemical or physical) takes place. During the process, the system goes from an initial state to a final
state through a series of intermediate states. This series of states is called the path of the process. In a
thermodynamic system this change is effected in two ways.
1. Reversible process
If the change in the system and its surroundings is brought in such a way that the system can regain its
original state then the process is known as reversible process.
2. Irreversible process
If the change in the state of system is effected by a process in such a way that the system cannot be
brought back to its original state without the help of some external force then that process is called an
irreversible process. In this process, the change is so rapid that the system is unable to attain equilibrium.
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PHYSICAL CHEMISTRY–II
All natural processes are irreversible and therefore, occur spontaneously. Irreversible process has a
marked influence on the properties of the surroundings. Some examples of irreversible processes are:
sudden expansion of a gas in vacuum, water flowing from a hill, flow of heat from a hot body to a cold
body, etc.
Thermodynamic processes occur under different conditions, hence are given different names.
Some of the processes are:
(i) Isothermal process: In this process the temperature of the system remains constant. The
system can exchange heat from the surroundings and the change in temperature, DT = 0.
(ii) Adiabatic process: A process in which no heat is exchanged between the system and
surroundings at any stage is said to be an adiabatic process (i.e., q = 0) and the system is completely
isolated from the surroundings. In this process heat evolved increases the temperature of the system
and heat absorbed decreases the temperature.
(iii) Isobaric process: In this process, the pressure of the system remains constant through out the
period of change. For an isobaric process, the change in pressure is zero (DP = 0). For example,
reaction taking place in an open system at one atmospheric pressure is isobaric.
(iv) Isochoric process: When the volume of the system remains constant during various operations
of the change in state of the system, then the process is said to be an isochoric. For such changes
DV= 0. Reaction taking place in a sealed container is an example of an isochoric process.
(v) Cyclic process: The process which brings back a system to its original state after a series of
changes in the system, is called a cyclic process and the path of the process is known as a cycle. In a
cyclic process the change in internal energy and enthalpy are zero i.e., DE = 0 and DH = 0.
(f) Work, heat and energy
Before we discuss the first law of thermodynamics we must understand the basic concept of work, heat
and energy.
Work: Work is a quantity of fundamental significance in a thermodynamic change in the state of
a system.
The mechanical work is given by
W = Force × Displacement
W = F . dx
or
where F is the force and dx is the displacement.
It can be shown that work has units of energy. The unit of work in SI system is a joule which is the
work done when a force of one Newton (N) acts through a distance of one meter (1 N = 1 kg ms–2). Work
can also be expressed in terms of calorie and erg. All three units are related as
7
1 × 10 ergs = 1 joule
and
4.18 joules = 1 calorie
In chemical thermodynamics, the general concept of work done is in the expansion or compression
of gases which is frequently referred to pressure-volume (P-V) work.
The pressure-volume work (W) is given by the relation
or
W = – P . DV
…(1.1)
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THERMODYNAMICS–I
where P is the pressure and DV is the change in volume. When the system expands, the work done by
the system on the surroundings is given a negative sign. On the other hand if work is done on the
system by the surroundings then the sign of work is positive.
Heat: Heat is another significant quantity in thermodynamic processes and is defined as a quantity
which describes the transfer of the thermal energy from one substance to another. The quantity of heat
q, is a product of heat capacity (C) of the substance and difference in temperature.
q = Heat capacity × Temperature difference
= C × (T2 – T1)
…(1.2)
where T1 and T2 are initial and final temperatures respectively. By convention ‘q’ is positive when it
is added to the system from the surroundings.
Energy: Energy is defined as the capacity to do work. The energy can be converted into work or
one can get energy when the work is done. The energy of the system increases if work is done on the
system. On the other hand energy of the system is decreased when work is done by the system on its
surroundings.
The total energy of a system is obtained when the external and internal energies are added
together. The external energy is the sum of kinetic and potential energies. In thermodynamics, actually
we are concerned with the study of internal energy and its changes for the system which is denoted by
E or sometimes by U.
Internal energy is a state function which depends only on the state variables P, V and T. The
change in internal energy of a system, when the system goes from an initial state I (P1, V1, T1) with
energy E1 to a final state II (P2, V2, T2) having energy E2 is given by
DE = Efinal – Einitial
= E2 (P2, V2, T2) – E1 (P1, V1, T1)
or simply as
DE = E2 – E1
…(1.3)
Internal energy is a property of a system and its absolute value cannot be determined.
In a closed system, the internal energy is a function of two variables T and V. Thus, one can write
that
E = f (T, V)
…(1.4)
Therefore, the total change in energy with small changes in the variables T and V is given by
 ∂E 
 ∂E 
dE =   dT + 
 dV
 ∂T  V
 ∂V  T
…(1.5)
1.3 ZEROTH LAW OR LAW OF THERMAL EQUILIBRIUM
This law was proposed by Fowler. This law states that “Two systems in thermal equilibrium with a third
system separately are also in thermal equilibrium with each other.”
Thus, if a system A is in thermal equilibrium with the system B, and the system B maintains
thermal equilibrium with system C, then if A and C are brought into contact with each other, it is
observed that systems A and C are also in thermal equilibrium.
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PHYSICAL CHEMISTRY–II
1.4 FIRST LAW OF THERMODYNAMICS
This law deals with the fundamental concept of energy, work and heat and the relation between them.
The first law of thermodynamics simply says ‘Energy is conserved’. Therefore, it is also known as the
law of conservation of energy. It has been stated in various forms; One of the statements of the first law
is that “Energy can neither be created nor destroyed, however, it can be transformed from one form to
other.”
Since energy is related to mass by the relation E = mc2, this law can also be stated as “The total
mass and energy of an isolated system remains unchanged.”
Another definition of the first law can be given as “The total energy of a system and its surroundings
must remain constant; however, it can change from one form to another.”
The mathematical statement of the first law of thermodynamics may be given in the following
manner:
Suppose a system which possesses an internal energy E1 in its initial state is given a heat ‘q’. The
heat taken up by the system can be used to increase either its internal energy, or perform work, or may
be used for both the purpose. If the energy of the system increases and attains a value E2 in the final
state, then E2 – E1 = DE is the increase in the internal energy. Further if W is the amount of work done
by the system, then the law of conservation of energy requires.
Heat absorbed by the system = Increase in internal energy of the system + Work done by the
system
i.e.,
q = DE + (–W) = DE – W
…(1.6)
or
DE = q + W
…(1.7)
The minus sign of work (–W) appears as work is done by the system on the surroundings so the
internal energy of the system decreases. Eq.(1.7) also suggests that when ‘q’ amount of heat is given to
the system and at the same time W amount of work is done on the system, then the internal energy of
the system is increased by DE.
For infinitesimal small changes, Eq. (1.7) can be written as
dE = dq + dW
…(1.8)
where dq is the amount of heat given to the system, dE is infinitesimal changes in the internal energy
and dW is the work done by the system.
If only expansion work (pressure-volume work), the work arising from a change in volume is
considered, then one can write
dW = –P dV
…(1.9)
So, Eq. (1.8) can be written as
dE = dq – P dV
…(1.10)
If the volume of the system is kept constant i.e., no work is done, then dV = 0 and Eq. (1.10)
reduces to
…(1.11)
dE = [dq]V
or
DE = [q]V
…(1.12)
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THERMODYNAMICS–I
This means that heat absorbed by a closed system (at constant volume, the system becomes a
closed one) increases the internal energy of the system.
Further, for a cyclic process, DE = 0, therefore, Eq. (1.7) becomes
q+W =0
or
q = (–W) = Work done by the system
…(1.13)
According to this equation the total heat given to the system is utilized in doing work by the
system.
Heat change at constant volume, [q]V
From the definition of first law of thermodynamics, we can write
q = DE – W
and in terms of work of expansion
q = DE + P DV
At constant volume, DV = 0, so we can write
[q]V = DE
…(1.14)
Thus, at constant volume, the heat absorbed is equal to the increase in energy of the system
during the process.
Heat change at constant pressure, [q]P — Enthalpy (H)
Several chemical and physical processes generally occur at constant pressure rather than at constant
volume since most of the reactions are carried out in an open system instead of in a closed system.
Thus, at constant pressure, the total heat given to the system is used in increasing its internal
energy and doing the work of expansion. The first law enables us to write
[q]P = DE + PDV
or
[q]P = E2 – E1 + P [V2 – V1]
or
[q]P = [E2 + PV2] – [E1 + PV1]
…(1.15)
…(1.16)
The terms (E2 + PV2) and (E1 + PV1) are measures of heat content of the system in the final and
in the initial states respectively. At this stage, a new thermodynamic function ‘H’ called the enthalpy or
heat content of the system must be introduced which expresses the thermal changes at constant pressure
and is defined as
H = E + PV
…(1.17)
where E is the internal energy and P and V are pressure and volume of the system. Since E, P and V are
all state functions, therefore, the term (E + PV) and hence the enthalpy (H) will also be a state function
or state variable (The word enthalpy comes from a German word ‘Enthalen’ which means to contain.
In Greek literature ‘en’ means ‘in’ and ‘thalpos’ means ‘heat’).
If H1 and H2 be the enthalpies of the initial and final states, then one can write
H1 = E1 + PV1
…(1.18)
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PHYSICAL CHEMISTRY–II
and
H2 = E2 + PV2
or
DH = H2 – H1 = (E2 + PV2) – (E1 + PV1)
The magnitude of DH depends only on the enthalpies of the initial and final states of the system
DH = Hfinal – Hinitial
i.e.,
…(1.19)
Thus, from Eq. (1.16) one can write
[q]P = H2 – H1
[q]P = DH
or
…(1.20)
According to Eq. (1.20) we can say that the heat absorbed by the system at constant pressure is
equal to the increase in the heat content or enthalpy of the system. If the work done is
pressure-volume work only then on substituting DH for [q]P in Eq. (1.15) we get
DH = DE + P DV
…(1.21)
For infinitesimal changes, we can write Eq. (1.17) as
dH = dE + P dV + V dP
or
dH = dq + V dP
(3 dq = dE + P dV)
…(1.22)
If the pressure is kept constant, then dP = 0, therefore, Eq. (1.22) is written as
dH = [dq]P
…(1.23)
This suggests that the enthalpy of the system is equal to the heat absorbed or evolved at constant
pressure.
Heat capacity
The heat capacity of a system is defined as the amount of energy required to raise the temperature of a
given quantity of substance by 1°C. However if the amount of the substance is 1 g, it is said specific
heat capacity or only specific heat. The unit of specific heat is J g–1K–1. If the amount of substance is
1 mole then it is called molar heat capacity. Its unit is J mol–1 K–1.
If a very small amount of heat, dq is given to the system and the temperature of the system is
raised by dT °C, then the heat capacity, is given by
C=
dq
dT
…(1.24)
According to the first law of thermodynamics, (Eq. 1.10),
dq = dE + P dV
Therefore,
or
C =
dE + P dV
dT
C =
dE
dV
+ P
dT
dT
…(1.25)
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THERMODYNAMICS–I
At constant volume, since dV = 0, therefore, P dV = 0, as such the heat capacity at constant
volume, CV is given by
 dE 
CV = 

 dT V
…(1.26)
Similarly the heat capacity at constant pressure, CP is written as
 dq 
CP =  dT 
 P
…(1.27)
Since according to Eq. (1.22)
dq = dH – V dP
Therefore, at constant pressure
(dq)p = dH
As such Eq. (1.27) may be written as
 dH 
CP = 

 dT  P
…(1.28)
Eqs. (1.26) and (1.28) suggest that CV is the rate of change of internal energy with temperature
at constant volume while CP is the rate of change of enthalpy with temperature at constant pressure.
The ratio of the two heat capacities of a gas is represented by g i.e.,
CP
= g
CV
…(1.29)
where for monoatomic gas the value of g is about 1.67, for diatomic gas g is 1.40 and for triatomic gas
g is 1.3.
Difference between CP and CV for an ideal gas
At constant pressure when heat is added to a system, then a part of it is used in doing work of expansion
besides increasing its internal energy. However, at constant volume all the heat added increases only the
internal energy of the system. Therefore, CP is generally larger than CV. The difference between these
two heat capacities can be derived thermodynamically in the following manner.
We know as per definition
 ∂H 
CP = 

 ∂T  P
Therefore,
and
 ∂E 
CV =  
 ∂T  V
 ∂H  –  ∂E 
CP − CV = 
 T

 ∂ V
 ∂T P
…(1.30)
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PHYSICAL CHEMISTRY–II
We also know that the enthalpy H is given by
H = E + PV
On differentiating it with respect to temperature T, at constant pressure, we get
 ∂H 
 ∂E 
 ∂V 
 T  =  T  + P T 
 ∂ P
 ∂ P
 ∂ P
...(1.31)
On substituting the value of (¶H/¶T)P from Eq. (1.31) into Eq. (1.30), we can write
 ∂E 
 ∂V 
 ∂E 
CP − CV =   + P 
−  

 ∂T P
 ∂T P  ∂T V
…(1.32)
If we consider the variation of E with temperature, T and volume, V then internal energy can be
written below as a function of T and V as
E = f (T, V)
whose total differential will become
 ∂E 
 ∂E 
dV +   dT
dE = 

 ∂V T
 ∂T  V
…(1.33)
On dividing Eq. (1.33) by dT on both the sides and keeping the pressure constant, we get
 ∂E 
 ∂E   ∂V 
 ∂E 
+  
 T = 



 ∂ P
 ∂V T  ∂T  P  ∂T  V
…(1.34)
The value of (¶E/¶T)P is then substituted from Eq. (1.34) into Eq. (1.32) and then Eq. (1.32)
becomes
 ∂E  ∂V ∂E
∂V ∂E
+
+ P
CP – CV = 




 −

 ∂ V T  ∂ T P  ∂ T  V
 ∂ T P  ∂ T  V
or
 ∂E 

CP – CV = 
+ P

 ∂V T

 ∂V 
 ∂T 

P
…(1.35)
The term (¶E/¶V)T is known as the internal pressure. In solids and liquids this is large because of
strong cohesive forces. In case of gases, this term is small as compared to P. Also for an ideal gas
 ∂E 
 V = 0
 ∂ T
as DE = 0
for an isothermal process
Therefore, Eq. (1.35) may further be simplified as
 ∂V 
CP – CV = P 

 ∂T P
…(1.36)
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THERMODYNAMICS–I
For one mole of an ideal gas
PV = RT
or
P dV = R dT
 ∂V 
P
 =R
 ∂T P
or
…(1.37)
On comparing, Eqs. (1.36) and (1.37), one can write
CP – CV = R.
…(1.38)
Variation of internal energy with temperature and volume
For a closed system of constant composition, the internal energy E is a function of volume and
temperature. When the volume changes from V to V + dV and temperature T changes to T + dT, then
the change in internal energy is given by the relation
 ∂E 
 ∂E 
dV +   dT
dE = 

 ∂V T
 ∂T V
…(1.39)
The term (¶E/¶T)V is the heat capacity CV at constant volume. Therefore, Eq. (1.39) is written as
 ∂E 
dE = 
 dV + C V dT
 ∂V T
…(1.40)
The coefficient (¶E/¶T)T is known to be a measure of the internal energy of a substance when its
volume is changed at constant temperature. This is known as the internal pressure (p).
i.e.,
 ∂E 
(p) T = 

 ∂V  T
…(1.41)
The internal pressure is a measure of the strength of the cohesive forces in the substance.
Eq. (1.40) may then be written as
dE = (π) T dV + C V dT.
…(1.42)
1.5 JOULE’S LAW
Joule’s law deals with the study of changes in the internal energy of a gas with volume when the
temperature is kept constant. This law may be defined as “The change in the internal energy of an ideal
gas with volume, at constant temperature, is equal to zero”. Mathematically, it may be represented as
 ∂E 

 =0
 ∂V  T
…(1.43)
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PHYSICAL CHEMISTRY–II
The above statement is given on the basis of
Stirrer
Thermometer
the results of Joule’s experiment. In this experiment
two metal containers A and B, connected together
by a tube containing a stopcock, were taken and
immersed in a water bath maintained at a constant
temperature T (Fig. 1.1).
One container A was filled with a gas at a
certain pressure and the other, B was evacuated. Both
the containers A and B were allowed to come in
thermal equilibrium with water. The stopcock is
Fig. 1.1: Joule’s expansion experiment
opened and the gas was allowed to expand freely
from the filled container into the evacuated one.
The system is then again brought into thermal equilibrium. The temperature of water is recorded and
no change in temperature was observed.
As the gas has freely expanded in the vacuum, P = 0. Hence work done during expansion,
W = – P dV = 0. According to the first law of thermodynamics, dE = dq + W, hence dE = dq (since W
= 0). Since no change in temperature took place in this experiment, it means that dq = 0. Thus, the
change in internal energy, dE = 0.
We know that E is a function of volume and temperature and the total change in internal energy
is given by the relation;
 ∂E 
 ∂E 
dE = 
 dV +   dV
V
∂
 ∂T  V
T

But in the above experiment, dE = 0 and dT = 0, therefore, the above relation reduces to
 ∂E 

 dV = 0
 ∂V  T
…(1.44)
Also in this experiment dV ¹ 0, this means that
 ∂E 

 =0
 ∂V  T
This is Joule’s law as stated above. Thus, it may be concluded that energy is a function of
temperature only. Mathematically, it is written as
E = f (T)
However, this law is applicable only for ideal gases.
Variation of enthalpy with temperature and pressure
The enthalpy H is a function of pressure P and temperature T. Mathematically it is written as
H = f (T, P)
13
THERMODYNAMICS–I
Therefore, the total change in enthalpy due to the changes in T and P is given by the relation
 ∂H 
 ∂H 
dH = 
dT + 

 dP
 ∂T P
 ∂P T
…(1.45)
 ∂H 
 = C P , therefore, Eq. (1.45) can be written as
 ∂T  P
Since 
 ∂H 
dH = C P dT + 
 dT
 ∂P  T
…(1.46)
CP can be determined directly by the calorimetric methods and the value of (¶H/¶P)T can be
evaluated.
We know that H = E + PV, so
dH = dE + P dV + V dP
…(1.47)
On substituting the values of dH and dE from Eqs. (1.46) and (1.40) into Eq. (1.47), one gets
 ∂H 
 ∂E 
CP dT + 
dP = CV dT + 

 dV + V dP + P dV
 ∂P T
 ∂V T
…(1.48)
At constant temperature, dT = 0, so Eq. (1.48) reduces to
 ∂E 

 ∂H 
 (dP )T = 

 + P  (dV)T + V(dP)T
 ∂P  T
 ∂V  T

…(1.49)
On dividing both sides of Eq. (1.49) by dP, we get
 ∂E 
  ∂V 
 ∂H 
 ∂P  =  ∂V  + P   ∂P  + V

 T 
T
T
 
…(1.50)
The differential coefficient (∂V ∂P)T appearing in Eq. (1.50) can be recognized as isothermal
compressibility, kT which is defined as
κT = −
1  ∂V 
V  ∂P T
…(1.51)
For solid and liquid substances the compressibility is very small, therefore (¶V/¶P)T will be very
small (approximately zero). As such the value of the first term on the right hand side of Eq. (1.50) can
easily be neglected in comparison to the volume V. Therefore, Eq. (1.50) becomes
 ∂H 

 =V
 ∂P  T
…(1.52)
14
PHYSICAL CHEMISTRY–II
For ideal gases, we know that PV = nRT. Thus one can write
P dV + V dP = nR dT
At constant temperature, dT = 0, so that
P (dV)T + V (dP)T = 0
or
 ∂V 
P
 +V=0
 ∂P  T
…(1.53)
Also the Joule’s law suggests that
 ∂E 

 =0
 ∂V  T
…(1.54)
Therefore, with the help of Eqs. (1.53) and (1.54) one can simplify Eq. (1.50) for an ideal gas as
 ∂H 

 =0
 ∂P  T
…(1.55)
The value of (¶H/¶P)T for real gases is very small and can be measured with the help of JouleThomson’s experiment.
1.6 JOULE-THOMSON’S EFFECT (ADIABATIC EXPANSION OF A REAL GAS)
When a real gas is allowed to pass adiabatically from a high pressure to a low pressure region through
a porous plug or a nozzle, expansion takes place. As a result of this, cooling occurs. This phenomenon
is known as Joule-Thomson’s effect.
Just as Joule’s experiment directly gives information about the dependence of energy on volume,
Joule-Thomson’s experiment give information about the dependence of enthalpy on pressure. The
Joule-Thomson’s experiment involves the expansion of a gas from one fixed pressure to another pressure
under adiabatic conditions. The experimental set up is shown in Fig.1.2. It consists of a cylindrical
tube, divided into two parts by a porous plug in its center, or a needle valve. A fixed quantity of gas is
contained in the cylinder by frictionless pistons placed on both sides of the porous plug. The entire
apparatus is insulated to prevent the transfer of heat to the surroundings.
Porous plug
P 1 V1
Thermal
insulation
P 2 V2
P1
P2
Fig. 1.2: Joule-Thomson’s experiment
15
THERMODYNAMICS–I
The pressure of the gas was kept P1 on the left side of the porous plug and P2 on the right side
where P1 > P2 · V1 volume of the gas is then allowed to pass through the porous plug from left to the
right side of the cylinder in such a way that its pressure on the left side of the plug is maintained
constant at P1 by moving the piston very slowly towards the porous plug. Simultaneously the pressure
on the right side of the plug is maintained constant at P2 by moving the piston away from the porous
plug. V2 is the change in volume of the gas on the right side of the porous plug. Temperatures T1 and
T2 on the two sides of the porous plug then measured accurately. It was found that temperature of the
gas falls in the low pressure side of the cylinder. This cooling was noticed with a number of gases
except hydrogen and helium when the experiments were carried out at ordinary temperatures.
Now as mentioned above, volume V1 of gas has passed through the porous plug. Since, on the left
side of the porous plug, the work is done on the system, whereas, on the right side, the work is done by
the system, we then have
Change in volume on the left side = – V1
Work done on the system on the left side = P1V1
Change in volume on the right side = V2
Work done by the system on the right side = – P2V2
\ Net work involved in the system, W = – P2V2 + P1V1
Because the process is carried out adiabatically i.e., q = 0, therefore, by the first law of
thermodynamics, we have
DE = W
or
DE = – P2V2 + P1V1
…(1.56)
where DE is the net change in energy of the gas as it passes from left to right side of the porous plug.
But since DE = E2 – E1, as such one can write
E2 – E1 = – P2V2 + P1V1
or
E2 + P2V2 = E1 + P1V1
…(1.57)
But by definition H = E + PV, thus Eq. (1.57) is written as
H 2 = H1
or
H2 – H1 = 0
or
DH = 0
…(1.58)
where H1 and H2 are the enthalpies of the gas on the left and right side of the porous plug.
This leads to the conclusion that the enthalpy of the gas which has moved across the porous plug
remains constant when the process of expansion of the gas is carried out adiabatically. Thus, JouleThomson’s expansion if isenthalpic process. In this experiment the measured change in temperature
–DT and the measured change in pressure –DP are combined in the ratio
 −ÄT 
 ÄT 

 =

 −ÄP  H  ÄP  H
…(1.59)
16
PHYSICAL CHEMISTRY–II
1.7 JOULE-THOMSON COEFFICIENT (mJT)
The thermodynamic quantity (¶T/¶P)H measured under constant enthalpy conditions is known as the
Joule-Thomson coefficient denoted by mJT. It is defined as the change in temperature per unit change in
pressure when a gas expands under adiabatic conditions at constant enthalpy. Thus
 ∂T 
 ÄT 
ì JT =   ≅  
 ∂P H  ÄP  H
…(1.60)
Physical Significance of mJT
From Eq. (1.60) the following significant points are noted:
(i) For cooling, mJT will be positive because dT and dP both will be negative.
(ii) For heating, mJT will be negative because dT will then be positive while dP is negative.
(iii) If mJT = 0, then the gas will neither be heated nor cooled on adiabatic expansion because
mJT = 0 only if dT = 0 for any value of dP.
Thus, at any pressure every gas has a definite temperature at which mJT = 0. Below this temperature
mJT has a positive value and above this temperature the value of mJT is negative. The temperature at
which mJT = 0, i.e., neither cooling nor heating of the gas takes place on adiabatic expansion, when the
gas is passed through a porous plug, is called the inversion temperature.
Relationship of mJT with other thermodynamic quantities
We know that enthalpy is a function of temperature and pressure. Mathematically, it is written as
H = f (T, P)
Total differential equation is written as
 ∂H 
 ∂H 
dH = 
 dT + 
 dP
∂
T

P
 ∂P  T
Since
Therefore,
…(1.61)
 ∂H 

 = CP
 ∂T  P
 ∂H 
dH = C P dT + 
 dP
 ∂P T
…(1.62)
In Joule-Thomson’s effect, the enthalpy of the system, dH = 0. Putting this value in Eq. (1.62),
we get
 ∂H 

 (dP )H + C P (dT )H = 0
 ∂P  T
or
 ∂H 
 ∂T 

 = − CP  
 ∂P  T
 ∂P  H
…(1.63)
17
THERMODYNAMICS–I
 ∂T 
Since   = ì JT (Joule-Thomson’s coefficient)
 ∂P  H
Therefore
 ∂H 
 P  = − ì JT CP
 ∂ T
ì JT = −
or
…(1.64)
1  ∂H 
C P  ∂P T
…(1.65)
Eq. (1.65) suggests that if CP and mJT are known then (¶H/¶P)T for real gases can be calculated.
1.8 SOME USEFUL RELATIONS FOR IDEAL GASES
Calculations of thermodynamic quantities W, P, DE and DH for the expansion of an ideal gas
under isothermal and adiabatic conditions.
First law of thermodynamics provides a base for the calculation of various thermodynamic quantities
such as work done (W), Heat (q), energy change (DE) and enthalpy change (DH) for different types of
processes under different conditions. The process may be carried out either reversibly or irreversibly.
However presently we shall be discussing only the reversible ones.
1. Isothermal reversible expansion of an ideal gas
The temperature of the system is kept constant in isothermal expansion of a gas by providing heat to
the system from the surroundings. The isothermal expansion of a gas can be done reversibly by placing
the container of the gas in a thermostat (constant temperature bath).
Let us consider ‘n’ moles if an ideal gas kept in a cylindrical container with a frictionless piston.
This gas expands reversibly from its initial volume V1 to a final volume V2 at constant temperature.
The pressure of the gas then reduces from an initial value P1 to P2.
(a) Expression for work of expansion: The work done by the gas for very small volume change
on expansion is given by
dW = – P dV (as per definition)
…(1.66)
The total work done will then be the sum of all the continuous series of small P dV terms from
the initial state to the final state. This result can mathematically be expressed by means of an integral
W=∫
Since for an ideal gas
so that
Therefore,
V2
V1
dW = − ∫
V2
V1
P dV
…(1.67)
PV = nRT
P=
nRT
V
W = −
∫
V2
V1
nRT
dV
V
...(1.68)
18
PHYSICAL CHEMISTRY–II
V2
dV
V
…(1.69)
V 
W = − nRT ln  2 
 V1 
…(1.70)
W = − nRT
or
∫V
1
or
For an ideal gas, P1V1 = P2V2 therefore, one can write Eq. (1.70) as
V 
W = − nRT ln  1 
 V2 
…(1.71)
Eqs. (1.70) and (1.71) represents the work done during isothermal expansion. The work done
during isothermal compression has the same value but with the changed sign.
(b) Expression for change in internal energy (DE): The internal energy of an ideal gas remains
unchanged during isothermal expansion because as per Joule’s law the internal energy of a gas does not
depend upon its volume. It is a function of temperature only. Thus
E = f (T)
and at constant temperature dT = 0.
Hence
DE = E2 – E1 = 0
…(1.72)
(c) Expression for q (heat absorbed) during isothermal expansion: The first law of
thermodynamics is mathematically written as
DE = q + W
Since for an isothermal expansion of an ideal gas
 V2 

W = – nRT ln 
 V1 
or
and
 V2
W = – 2.303 nRT log 
 V1



DE = 0
Therefore, the heat absorbed (q) is given by
V 
P 
q = 2.303 nRT log  2  = 2.303 nRT log  1 
 V1 
 P2 
…(1.73)
This means that heat absorbed in an isothermal reversible expansion of an ideal gas is equal to the
work of expansion.
(d) Expression for enthalpy change (DH): We know that the enthalpy change
DH = H2 – H1
= (E2 + P2V2) – (E1 + P1V1)
19
THERMODYNAMICS–I
= (E2 – E1) + P2V2 – P1V1
= DE + D(PV)
= DE + D(nRT)
= DE + nRDT
For an isothermal expansion of an ideal gas DE = 0, and DT = 0, therefore
DH = 0 + 0 = 0
…(1.74)
So the enthalpy change is also zero in an isothermal reversible expansion of an ideal gas.
2. Reversible adiabatic expansion of an ideal gas
An adiabatic process is defined as that process in which no heat is allowed to enter or leave the system
at any stage. Thus, q = 0. The first law of thermodynamics under this condition becomes;
DE = W
…(1.75)
This equation indicates that DE and W both have same sign. Thus, if work is done by the system
(W is negative), DE will be negative which means internal energy of the system decreases and the
temperature of the system (gas) falls. On the other hand, if W is positive i.e., work is done on the
system, DE will be positive suggesting internal energy to increase and hence the temperature of the gas
would increase.
(a) Expression for work of expansion and change in internal energy in an adiabatic change:
Let us take ‘n’ moles of an ideal gas kept in such a vessel which does not allow heat to be exchanged
between the system and surroundings. The gas is permitted to expand or compress adiabatically under
reversible conditions. Let P1, T1 and V1 represent the initial state of the gas while P2, T2 and V2 its
final state. The work of expansion can be calculated provided DE is known. The value of DE is
calculated in the following manner;
We know that for an ideal gas
dE = CV dT
…(1.76)
In order to obtain DE, let us integrate Eq. (1.76) as
E2
∫
T2
dE = CV ∫ dT
…(1.77)
E2 – E1 = CV (T2 – T1)
…(1.78)
DE = CV (T2 – T1)
…(1.79)
E1
T1
This yields
Since according to Eq. (1.75)
DE = W
Therefore, the work of expansion in an adiabatic process is given by
W = CV (T2 – T1)
…(1.80)
20
PHYSICAL CHEMISTRY–II
(b) Expression for DH: Similarly the enthalpy change in an adiabatic process can also be obtained.
We know that
DH = H2 – H1
= (E2 + P2V2) – (E1 + P1V1)
= (E2 – E1) + (P2V2 – P1V1)
= DE + nR (T2 – T1)
DH = CV (T2 – T1) + nR (T2 – T1)
From Eq. (1.80)
or
DH = (CV + nR) (T2 – T1)
or
DH = CP (T2 – T1)
(Since CP – CV = nR)
…(1.81)
Relationship between temperature, volume and pressure in reversible adiabatic
expansion
The exact relationship between volume and temperature, pressure and temperature and volume and
pressure in a reversible adiabatic expansion of an ideal gas may be obtained in the following manner:
(i) Relationship between volume and temperature
In case of an ideal gas, the energy change is related to the heat capacity by the relation
dE = CV dT
and the work done during expansion of small volume dV is given as
dW = – P dV
Since q = 0 during adiabatic process, therefore, the first law of thermodynamics permits us to
write
dE = dW
or
CV dT = – P dV
or
CV dT = −
or
CV
nRT
dV
V
(as PV = nRT)
dT
 dV 
= − nR 

T
 V 
…(1.82)
…(1.83)
or, one can write Eq. (1.83) as
CV d(ln T) = – nR d(ln V)
…(1.84)
On integrating Eq. (1.84) between the limits T1 and T2 and V1 and V2, we obtain
V 
V 
T 
CV ln  2  = − nR ln  2  = nR ln  1 
 V1 
 V2 
 T1 
…(1.85)
21
THERMODYNAMICS–I
For one mole of an ideal gas CP – CV = R, so that one can write Eq. (1.85) as
T 
V
C V ln  2  = (C P − C V ) ln  1
 T1 
 V2



or
C

V 
T 
ln  2  =  P − 1  ln  1 
 CV

 V2 
 T1 
or
V 
T 
ln  2  = ( γ − 1) ln  1 
 V2 
 T1 
or
T 
V 
ln 2  = ln  1 
 T1 
 V2 
γ −1
…(1.86)
where the symbol g represents the ratio of CP/CV and for monoatomic gas, its value is 1.67. The
Eq. (1.86) can be written in the following form by taking antilogarithm.
 T2

 T1

V
 =  1
 V2




ã –1
V 
T 2 = T1  1 
 V2 
or
…(1.87)
ã –1
T1V1γ − 1 = T2 V2γ − 1
or
or
simply as TV
g–1
…(1.88)
= constant.
(ii) Relation between pressure and volume
For many thermodynamic applications, it is needed to develop relations between volume and pressure
in a reversible adiabatic process. Let us write the equation of state PV = nRT as
P dV + V dP = nR dT
So that
dT =
=
or
dT =
…(1.89)
P dV + V dP
nR
P
V
dV +
dP
nR
nR
nRT dV nRT dP
+
nR V
nR P
= T d(ln V) + T d(ln P)
…(1.90)
22
PHYSICAL CHEMISTRY–II
On substituting the value of dT in the Eq. (1.82), we get
CV [(T d(ln V) + T d(ln P)] = – nRT d(ln V)
or
CV d(ln P) = – (CV + nR) d(ln V)
…(1.91)
For one mole of an ideal gas, Eq. (1.91) becomes
or
CV d(ln P) = – (CV + R) d(ln V)
…(1.92)
CV d(ln P) = – CP d(ln V)
…(1.93)
d(ln P) = – (CP/CV) d(ln V) = – g d(ln V)
or
…(1.94)
Integrating Eq. (1.94) between the limits P1 and P2 and V1 and V2, one gets
V 
P 
ln  2  = γ ln  1 
 V2 
 P1 
…(1.95)
On taking antilog on both the sides of Eq. (1.95), one gets
P2
 V1 


P1 =  V2 
g
ã
…(1.96)
g
P2 V2 = P1V1
or
…(1.97)
g
PV = constant.
so that
…(1.98)
(iii) Relation between pressure and temperature
Since an ideal gas is under consideration, we can easily write
P1V1 = RT1 and P2V2 = RT2
V1 =
i.e.,
RT2
RT1
and V2 =
P2
P1
V1
RT1 / P1 T1P2
=
=
V2
RT2 / P2 T2 P1
So that
On putting the value of
V1
in Eq. (1.88), one gets
V2
TP 
T2
=  1 2
T1
 T2 P1 
or
…(1.99)
 T2 
 
 T1 
ã
 P2 
=  
 P1 
ã −1
ã –1
T2  T2 
× 
or
T1  T1 
ã −1
P 
= 2 
 P1 
ã −1
23
THERMODYNAMICS–I
or
 T2 
 
 T1 
ã
1– ã
 P1 
=  
 P2 
…(1.100)
On taking the gth root on the both sides of Eq. (1.100), one obtains
(1 – ã ) / ã
T2
 P1 
T1 =  P 
 2
or
T1P1(l − γ ) / γ = T2 P2(l − γ ) / γ
…(1.101)
or
TP(l − γ) / γ = constant.
…(1.102)
1.9 THERMOCHEMISTRY
Thermochemistry deals with the study of heats of chemical reactions. When a chemical reaction takes
place, energy in various forms may either be emitted or absorbed. During many reactions the temperature
rises indicating that heat is being evolved. In others, when temperature falls, absorption of heat is
indicated. Thus, the study of the heat produced or required by chemical reaction forms the basis of
thermochemistry. Reactions in which heat is produced are called exothermic reactions, whereas, those
in which heat is absorbed are known as endothermic reactions. The release of heat suggests a decrease
in the enthalpy a system when the pressure is constant. It may therefore, be concluded that in a exothermic
process at constant pressure, the enthalpy change is negative i.e., DH < 0. On the other hand, absorption
of heat leads to an increase in enthalpy. Thus, endothermic process at constant pressure has positive
value of enthalpy change i.e., DH > 0.
1.10 HEAT OF REACTION
The heat of reaction is defined as the amount of heat liberated or absorbed at a given temperature when
the reactants are converted into products as represented by the balanced chemical equation. If q is the
heat of reaction, W the work done by the reactants during reaction, then according to the first law of
thermodynamics, one can write
q = DE – W
…(1.103)
DE is the energy difference between the internal energy of products and that of reactants as given
below
DE = å E (products) – å E (reactants)
…(1.104)
If the chemical reaction occurs at constant volume (bomb calorimeter) then W = 0 and in that
case
(q)V = DE
…(1.105)
Thus, the heat of reaction at constant volume (q)V is the change in internal energy of the system
during a chemical reaction. However, if the chemical reaction is carried out at constant pressure in an
open calorimeter then W = – PDV and in that case
(q)V = DE – (– P DV) = DE + PDV = DH
…(1.106)
24
PHYSICAL CHEMISTRY–II
In this case the amount of heat evolved, or absorbed during a chemical reaction occurring at a
constant pressure and at a particular temperature, is referred to as enthalpy of reaction. Let us consider
a general reaction represented as
→ cC + dD
aA + bB 
….(1.107)
The change in enthalpy of the system when the reactants are converted into products is given by
the difference between the total heat content of the products and that of the reactants at constant
temperature and pressure. This quantity is called the heat of reaction.
or
DH = å H (products) – å H (reactants)
or
DH = (cHC + dHD) – (aHA + bHB)
…(1.108)
where HA, HB, HC and HD are enthalpies of A, B, C and D respectively and a, b, c and d are their
respective number of moles taking part in the reaction.
The change in volume DV is very small for those reactions involving solids and liquids. Thus,
P DV is very small and can be neglected, so DH is equal to DE. However, if the reactants and products
are gases then volume changes cannot be ignored and one must specify the manner in which the
reaction is performed. Mostly gaseous reactions are performed at constant pressure in which change in
enthalpy (DH) is measured.
Exothermic and endothermic reactions
The value of enthalpy change (DH) may be zero, positive or negative. When DH = 0, the enthalpies of
the reactants and products are equal and no heat is absorbed or evolved. When DH < 0 (i.e., negative),
the total enthalpy of the products å H (product) is less than the total enthalpy of the reactants
å H (reactants). In such reactions the heat is liberated and given to the surroundings. Thus, reactions
associated with negative DH values are known as exothermic reactions. When DH > 0 (i.e., positive)
the total enthalpy of the products are greater than the total enthalpy of the reactants. This means an
equivalent amount of heat must be absorbed by the system from the surroundings. Therefore, those
reactions which have positive DH values are called endothermic reactions and involve absorption of
heat from the surroundings.
The sign convention
If heat is absorbed by the system, the reaction is endothermic and DE, or DH value has a positive sign.
If heat is liberated by the system, the reaction is exothermic and DE, or DH value bears a negative
sign.
Relationship between heat of reaction at constant pressure and heat of reaction at
constant volume
In case of gaseous reactants and products and assuming them to be ideal, we can write at constant
temperature and pressure
and
PV1 = n1RT 

PV2 = n2 RT 
…(1.109)
where n1 and n2 are the number of moles of the reactants and products respectively. V1 is the volume of
25
THERMODYNAMICS–I
gaseous reactants while V2 is the volume of gaseous products. The change in volume, DV is therefore,
given by
DV = V2 – V1 = n2 RT − n1RT
P
P
= (n 2 − n1 )
or
DV = Än
RT
P
RT
P
…(1.110)
where Dn is the change in number of moles. In case of the general reaction expressed by Eq. (1.107),
the change in number of moles Dn is equal to
Dn = (c + d) – (a + b)
Further at constant pressure the change in enthalpy and change in internal energy are related to
each other by the relation
DH = DE + P DV
…(1.111)
On substituting the value of P DV from Eq. (1.110) into Eq. (1.111), one gets
DH = DE + Dn RT
…(1.112)
This relation is used to calculate the enthalpy of reaction from the energy of reaction at a given
temperature. The value of gas constant R is considered in calories or joules (R = 1.987 calories, or
8.314 joules). Further, since DH = [q]P and DE = [q]V, therefore, one can write
[q]P = [q]V + Dn RT
…(1.113)
where ‘q’ denotes the heat of reaction.
Conditions under which [q]P = [q]V, or DH = DE
1. When the reaction takes place in a closed vessel the volume remains constant i.e., DV = 0.
2. When only solids, liquids or solutions are involved in the reaction (no gaseous reactant or
product), the volume changes are negligible during a chemical reaction.
3. When the number of moles of gaseous reactants and products are equal in a reaction,
Dn = 0 (i.e., nproduct = nreactant). For example,
→ 2HCl (g).
H2 (g) + Cl2 (g) 
1.11 THERMOCHEMICAL EQUATIONS
Any chemical equation which is properly balanced and expresses the amount of heat change (liberated
or absorbed) in the reaction alongwith the physical states of the reactant and products is known as
thermochemical equation.
Firstly the balanced chemical equation representing the actual reaction and the physical states of
the reactants and products are written. The heat change associated with this chemical reaction is then
26
PHYSICAL CHEMISTRY–II
expressed by writing DH or DE with proper sign (positive or negative) to the right side of the equation.
The physical states are indicated by the symbols (s), (l), (g) or (aq) for solid, liquid, gas and aqueous
states respectively. Thus, the thermochemical equation for the combustion of methane is written as
→ CO2 (g) + 2H2 O (l); DH (298 K) = –890.3 kJ mol–1
CH4 (g) + 2O2 (g) 
This equation indicates that when 1 mole of methane burns in presence of 2 moles of oxygen,
1 mole of CO2 gas and 2 moles of water are formed and 890.3 kJ heat is liberated at constant pressure
and 298 K temperature. However, writing the thermochemical equation simply by the following equation.
→ 2H2 O; DH = –571.6 kJ mol–1
2H2 + O2 
is not correct and complete because it does not indicate the physical states of the reactants and products;
whether water is formed in the form of steam, or as liquid water. Hence the correct and complete
thermochemical equation is written as
H2 (g) +
1
2
→ H2O (l); DH = –285.8 kJ mol–1.
O2(g) 
Standard states
It must be noted that change in enthalpy of a reaction can vary with temperature and pressure. So
changes in enthalpy are generally measured for processes taking place under a set of specific conditions.
This set of specific conditions is called as standard states. The conventional standard state of a
substance is its pure form at 298.15 K and 1 atm pressure.
The standard enthalpy change DH0, or energy change DE0 is the change in enthalpy, or energy
for a process in which the initial and final substances i.e., the reactants and products are in their
standard states. Thus, the standard enthalpy change for a chemical reaction or a physical process is the
difference between the total enthalpy of the products and the total enthalpy of the reactants in their
standard states at a particular temperature. Thus
ÄH 0 =
∑ H 0products − ∑ H 0reactants
The standard molar enthalpy of every element (i.e., at 1 atm pressure and 298.15 K) in the most
stable state is taken as zero.
1.12 THERMOCHEMICAL LAWS
The principle of conservation of energy provides a basis for two very important laws of thermochemistry:
1. First law of thermochemistry
The first law of thermochemistry is due to the experimental observations of A.L. Lavoisier and
P.S. Laplace (1780). The law suggests that the quantity of heat which is given to decompose a compound
into its elements is equal to the heat liberated when that compound is formed from its elements. In
simpler word one can say that when a compound decomposes, its heat of decomposition is equal in
magnitude to its heat of formation but with opposite sign. The first law of thermochemistry can be
applied to those reactions which involve a compound and its constituent elements. For example, in the
formation of SO2 (g) from its elements S and O2, 297.5 kJ heat is released.
27
THERMODYNAMICS–I
→ SO2 (g);
S (s) + O2 (g) 
0
DH (298 K) = –297.5 kJ mol
–1
But when SO2 (g) is decomposed, exactly the same amount of heat as mentioned above is needed
and the thermochemical equation is written as
→ S (s) + O2 (g);
SO2 (g) 
0
–1
DH (298 K) = +297.5 kJ mol
As a consequence of this law, thermochemical equations can be reversed by changing the sign of
DH. For example, the following thermochemical equation:
→ CO2 (g) + 2H2O (l); DH0 (298 K) = –890.36 kJ mol–1
CH4 (g) + 2O2 (g) 
can be reversed with changed sign of DH.
→ CH4 (g) + 2O2 (g);
CO2 (g) + 2H2O (l) 
0
–1
DH (298 K) = 890.4 kJ mol .
2. Hess’s law—The law of constant heat summation
G.H. Hess (1840) experimentally gave the second law of the thermochemistry which has important
applications and is known as the law of constant heat summation. The following statements can be
given on the basis of this law.
This law may be stated as “The net heat change in a chemical reaction, either at constant pressure,
or at constant volume, has the same value whether it occurs in one step or in several steps”. Thus,
according to this law, the resultant heat change in the reaction depends only on the initial and final
states and not on the intermediate steps through which the final state is achieved. Another important
significance of this law is that the thermochemical equations, like algebraic equations, can be added, or
substracted. This result enables one to give one more statement of this law as “If two or more
thermochemical equations are added to give another thermochemical equation then the corresponding
enthalpies of reactions must be added.” This is a statement of energy conservation as applied to chemical
reaction.
Hess’s law is simply a consequence of the first law of thermodynamics. In order to prove this
statement, let us consider a system in which A and P represents the reactants and products respectively.
It is further assumed that there are two ways by which the system can go to the final state, P from the
initial state, A.
A
,H
P
,H1
,H5
E
B
,H2
,H 4
C
,H3
D
Fig. 1.3: The representation of direct and indirect paths
28
PHYSICAL CHEMISTRY–II
In the first way, A is converted directly into P and the enthalpy change in this process is DH. Let
DH be equal to ‘q’ amount of heat evolved in the direct change.
→ P; DH = q joules
A 
In the second way which is an indirect path; first A is converted into B; B into C; C into D; D into
E and finally E is converted to P in the following manner:
→ B; DH1 = q1
A 
→ C;
B 
DH2 = q2
→ D; DH3 = q3
C 
→ E; DH4 = q4
D 
→ P;
E 
DH5 = q5
where DH1, DH2, DH3, DH4 and DH5 are enthalpy changes in these intermediate steps. So the total heat
involved in these steps is given by
DH1 + DH2 + DH3 + DH4 + DH5 = q1 + q2 + q3 + q4 + q5 = q' joules (say)
The Hess’s law requires that we must have
DH = DH1 + DH2 + DH3 + DH4 + DH5
i.e.,
q = q'
The Hess’s law also requires that
DH1 + DH2 + DH3 + DH4 + DH5 + (–DH) = 0
or
q' + (–q) = 0
But suppose the Hess’s law is not correct and the change in heat in these two ways of performing
the reactions is not the same and let q' be greater than q (i.e., q' > q). Then on going from A to P
through different intermediate steps and then returning back directly to A, an energy (q' – q) joules
should be produced. But this is against the statement of first law of thermodynamics of energy
conservation. Hence q' must be equal to q i.e., the Hess’s law is correct which is in accordance with
the first law. This law has been experimentally found to be true. The law of constant heat summation
can be illustrated with the help of the following examples:
(i)
The formation of urea from carbon, oxygen and NH3 either takes place directly in
one step, or in two steps as shown below
C + O2 + NH3
C + O2
,H1 = –393.7 kJ
CO2 + NH3
,H = –1207.1 kJ
,H2 = –813.58 kJ
Urea
Urea
Fig. 1.4: Formation of urea
29
THERMODYNAMICS–I
The enthalpy change when the reaction takes place directly is –1207.1 kJ mol–1 and the enthalpy
change when the reaction takes place in two steps is equal to DH1 + DH2.
Step 1.
Step 2.
→ CO2 (g); DH1 = –393.7 kJ mol–1
C (s) + O2 (g) 
→ Urea (s); DH2 = –813.58 kJ mol–1
CO2 (g) + NH3 (g) 
Thus, DH1 + DH2 = –1207.28 kJ mol–1 which is equal to DH value of one step process. Hence,
DH = DH1 + DH2.
(ii) Conversion of 1 mole of H2O (l) to 1 mol of H2O (g) at constant temperature and
pressure
The enthalpy diagram for the reaction is shown below:
H2 (g) +½O2 (g)
0
∆H2 = –242.3 kJ
(Energy released)
∆H1 = +285.8 kJ
(Energy absorbed)
H2 O (g)
∆H
∆H = +43.5 kJ
H2 O (l)
Fig. 1.5: Conversion of H2O (l) to H2O (g)
Following two step processes are involved in the conversion of H2O (l) to H2O (g):
→ H2 (g) +
H2O (l) 
1
2
O2 (g); DH1 = +285.8 kJ mol
–1
→ H2O (g); DH2 = –242.3 kJ mol–1
H2 (g) + ½ O2 (g) 
The total amount of enthalpy change is
DH1 + DH2 = +43.5 kJ mol
–1
→ H2O (g) takes place directly then the value of
However, when the reaction H2O (l) 
DH = +43.5 kJ mol
Thus, DH = DH1 + DH2 holds true.
–1
30
PHYSICAL CHEMISTRY–II
Applications of Hess’s law
Hess’s law has several important practical applications for determining heats of formation, heats of
reaction and heats of transition of substances which cannot be measured directly by experiments.
(i) Determination of heat of formation of substances: The law of constant heat summation is
conveniently used as an alternative procedure to calculate the heat of formation (DHf) of those substances
for which heat of formation cannot be determined by direct experiments. For example, the value of
DHf for the reaction
→ CO (g)
C (graphite) + ½ O2 (g) 
is difficult to determine experimentally. But it can be estimated from the following two reactions
whose DHf values can be obtained from experimental measurements:
(i)
(ii)
→ CO2 (g); DH1 = –393.5 kJ mol–1
C (graphite) + O2 (g) 
CO (g) +
1
2
→ CO2 (g); DH2 = –283.0 kJ mol–1
O2 (g) 
On subtracting (ii) from (i), one gets
C (graphite) +
1
2
→ CO (g); DHf = –110.5 kJ mol–1
O2 (g) 
Consequently, DHf = DH1 – DH2 can easily be calculated.
Example: Calculation of standard heat of formation of liquid ethanol.
It is done with the help of three thermochemical equations written as
(i)
(ii)
(iii)
→ 2CO2 (g) + 3H2O (l); DH = –1380.7 kJ mol–1
C2H5OH (l) + 3O2 (g) 
→ CO2 (g); DH = –393.5 kJ mol–1
C (s) + O2 (g) 
H2 (g) +
1
2
→ H2O (l); DH = –286.6 kJ mol–1
O2 (g) 
First multiply equation (ii) by 2 and equation (iii) by 3 and then add them. This will give
(iv) 2C (s) + 3H2 (g) +
7
2
→ 2CO2 (g) + 3H2O (l); DH = – 1446.8 kJ
O2 (g) 
Equation (i) is then subtracted from equation (iv) and then the final result is written as
2C (s) + 3H2 (g) +
1
2
→ C2H5OH (l); DH = –66.1 kJ
O2 (g) 
The standard heat of formation DHf of ethanol is, therefore, –66.1 kJ mol –1 at 298 K.
(ii) Determination of heat of reaction: The heat of reaction of various reactions can be obtained
from making use of Hess’s law. This may be illustrated with the help of several examples.
Example: The standard heat of reaction for the reaction
→ C2H6 (g)
C2H4 (g) + H2 (g) 
at 298 K can be obtained by using the heat of combustion values of ethylene (C2H4), hydrogen (H2)
31
THERMODYNAMICS–I
and ethane (C2H6). The thermochemical equations for the heats of combustion of C2H4, H2 and C2H6
are written as
(i)
→ 2CO2 (g) + 2H2O (l); DH0 = –1410.8 kJ mol–1
C2H4 (g) + 3O2 (g) 
(ii)
H2 (g) +
(iii)
1
2
→ H2O (l); DH0 = –286.2 kJ mol–1
O2 (g) 
→ 2CO2 (g) + 3H2O (l); DH0 = –1560.6 kJ mol–1
C2H6 (g) + 3 12 O2 (g) 
The heat of reaction of
→ C2H6 (g)
C2H4 (g) + H2 (g) 
can be obtained by adding equations (i) and (ii) which yields
→ 2CO2 (g) + 3H2O (l); DH0 = –1697.0 kJ mol–1
C2H4 (g) + H2 (g) + 3 12 O2 (g) 
From which the equation (iii) is subtracted to have the desired result for the standard heat of
formation of ethane.
→ C2H6 (g); DH = –136.4 kJ mol–1.
C2H4 (g) + H2 (g) 
(iii) Determination of heat of transition: The heat of transition from one crystalline form of a
substance to the other can be calculated by heats of combustion data of the two allotropic forms of the
substance. For example, the heat of transition of carbon (diamond) to carbon (graphite) can be obtained
from the heats of combustion data of diamond and graphite respectively. The thermochemical equations
for the combustion of diamond and graphite are written as:
(i)
→ CO2 (g); DH0 = –395.4 kJ mol–1
C (diamond) + O2 (g) 
(ii)
→ CO2 (g); DH0 = –393.5 kJ mol–1
C (graphite) + O2 (g) 
By subtracting equation (ii) from (i), we get
→ C (graphite); DH0tran = – 1.9 kJ mol–1.
C (diamond) 
(iv) Determination of lattice energy of a crystal: Hess’s law can be used to determine the
lattice energy of a crystal from the Born-Haber cycle. The lattice energy is defined as the energy
required to separate completely one mole of a solid ionic compound into gaseous ions. The more is the
lattice energy, the more stable will be the ionic compound (the ions will be more tigthly held). But the
lattice energy is not measured directly. It can be measured from Born- Haber cycle. For example, in the
formation of NaCl crystals from Na (s) and Cl2 (g), the following steps are written.
→ Na (g);
(i) Na (s) is vapourized into Na (g); Na (s) 
–1
→ Na+ (g) + e– ; DH2= 495.8 kJ mol–1
Na (g) 
(ii) Na (s) is then ionized;
(iii) Dissociation of Cl2;
DH1 = 317.5 kJ mol
1
2
→ Cl (g);
Cl2 (g) 
–1
DH3 = ½ × 241.84 kJ mol
32
PHYSICAL CHEMISTRY–II
–
(iv) Dissociation ofCl (g);
–
→ Cl– (g);
Cl (g) + e 
DH4 = –365.26 kJ mol
–1
(v) Combination of Na+ (g) and Cl– (g) to form NaCl (s)
+
–
→ NaCl(s); DH5 = ?
Na (g) + Cl (g) 
On adding steps (1) to (v), the net change is written as
Na (s) +
1
2
→ NaCl (s); DH = –410.87 kJ mol–1
Cl2 (g) 
In the above sequential steps DH1, DH2, DH3, DH4, DH5 and DH are the enthalpy changes in the
reactions as respectively mentioned respectively above. Now according to Hess’s law, one may write
that
DH = DH1 + DH2 + DH3 + DH4 + DH5
Since in all these reactions, the changes of enthalpy (except DH5) can be experimentally determined,
therefore, the value of DH5 can very easily be evaluated and the required lattice energy be found out.
Now putting the experimental values in the above expression, we have
–410.87 = 317.57 + 495.80 +
DH5 = –979.9 kJ mol
or
1
2
× 241.84 – 365.26 + DH5
–1
Thus, the lattice energy of NaCl is negative and is equal to –979.9 kJ mol–1.
1.13 DIFFERENT TYPES OF HEATS OF REACTION
Various types of heat or enthalpy changes occur in the chemical reactions depending upon the nature of
the reaction.
1. Heat (enthalpy) of formation
The change in enthalpy involved in the formation of one mole of a compound from the constituent
elements in their standard states is known as enthalpy of formation of the compound. The formation
reactions exhibit the following features:
(i) There is a single product, with a stoichiometric coefficient of one.
(ii) All the starting materials are elements and in their standard states at T = 298.15 K and
P = 1 atm.
(iii) Since the reaction must generate exactly 1 mole of product therefore, fractional stoichiometric
coefficient of elements is common in formation reactions.
Thus, the reactions
–1
H2 (g) +
1
2
→ H2O (l);
O2 (g) 
∆H 0f = –285.8 kJ mol
C (s) + 2H2 (g) +
1
2
→ CH3OH (l);
O2 (g) 
–1
∆H 0f = –238.66 kJ mol
suggest that the heat of formation of 1 mol of water is –285.8 kJ mol–1 and that of 1 mole of methanol
is –238.66 kJ mol–1.
33
THERMODYNAMICS–I
Standard enthalpy of formation, ∆H0f
The standard enthalpy of formation of a substance is defined as the change in enthalpy of a reaction in
which one mole of the compound is formed from its elements, all substances being in their standard
states (298.15 K and 1 atm pressure). It is denoted as ∆H 0f . The superscript ‘0’ in ∆H 0f indicates under
standard conditions of 298.15 K and 1 atm pressure and the subscript ‘f’ means ‘formation of 1 mole’.
Every chemical substance has a characteristic ∆H 0f expressed in kilojoules per mole.
It is a convention that the formation reaction for one element in its standard state is considered
to involve no change at all. Thus, ∆H 0f for an element in its standard state is zero. But if an element is
not in its standard state then ∆H 0f is not zero.
Standard heat of reaction (DH0) from standard heat of formation ( ∆H0f )
In general, the standard heat of reaction is equal to the difference in the standard heats of formation of
products and reactants.
DH 0 = [Total standard heat of formation of products]
– [Total standard heat of formation of reactants]
ÄH0f
i.e.,
=[
ÄH0f (products)
− ÄH0f (reactants)]
→ cC + dD, the standard heat of reaction is
Therefore, for the general reaction aA + bB 
obtained as
0
0
0
0
ÄH 0 = [{c × ÄHf (C) + d × ÄHf (D)} − { a × ÄHf (A) + b × ÄHf (B)}]
The enthalpy changes for the formation of some compounds from their respective elements are
given in Fig. 1.6.
C2 H2 (g) [+226.7 kJ]
200
C (s) + O2 (g)
0
C (s) + 2H 2 (g)
H2 (g) + ½O2 (g)
2C (g) + H2 (g)
CH4 (g) [–74.8 kJ]
– 200
H2O (g) [–241.8 kJ]
– 400
CO2 (g) [–393.5 kJ]
Fig. 1.6: Enthalpy changes for the formation of some compounds at 298 K and 1 atm pressure
34
PHYSICAL CHEMISTRY–II
2. Heat of combustion
In case of organic compounds, a very important reaction is combustion. The heat of combustion is
therefore, defined as “the change in enthalpy, or heat content when one mole of a compound is burnt
completely in oxygen.”
The combustion is always carried out in excess of oxygen so that the products are only CO2 and
H2O and not CO. If the substances concerned are all in their standard states then the symbol ∆H0C is
used. For example, the heat of combustion of ethanol is ∆H0C = –1367 kJ mol–1.
→ 2CO2 (g) + 3H2O (l);
C2H5OH (l) + 3O2 (g) 
–1
∆H0C = –1367 kJ mol
The other examples of combustion reactions are
→ CO2 (g) + 2H2O (l);
CH4 (g) + 2O2 (g) 
→ CO2 (g);
C (diamond) + O2 (g) 
–1
∆H0C = – 890.3 kJ mol
–1
∆H0C = –395.4 kJ mol
The combustion reactions are always exothermic hence the value of ∆H0C is negative at all times.
There are several important applications of heat of combustion values.
(i) In the calculation of heat of formation
The heat of formation of a compound can easily be calculated from its heat of combustion values.
Example: Let us calculate the heat of formation of liquid ethanol when the standard heat of
combustion value is –1366.9 kJ mol–1 at 25°C. The thermochemical equation for combustion of ethanol
is written as
→ 2CO2 (g) + 3H2O (l);
C2H5OH (l) + 3O2 (g) 
–1
∆H0C = –1366.9 kJ mol .
Suppose the heat of formation of C2H5OH (l) is x kJ while those of CO2 (g), H2O (l) and O2 (g)
are –393.5 kJ mol–1, –285.8 kJ mol–1 and 0 kJ respectively. The heat of combustion is the change in
heats of reaction of the products and reactants. Thus,
ÄH 0C =
∑ ÄH 0f (products) − ∑ ÄH0f (reactants)
–1366.9 = (2 × –393.5) + (3 × –285.8) – (x + 0)
or
x = –277.5 kJ mol
–1
Therefore, the heat of formation of liquid ethanol from its elements in the standard states is
obtained very easily and is written as
2C (s) + 3H2 (g) +
1
2
→ C2H5OH (l);
O2 (g) 
–1
∆H0C = –277.5 kJ mol .
(ii) In the determination of calorific values of fuels and food
The calorific value is defined as the amount of energy produced as heat in calories (or joules) when one
gram of substance is completely burnt. Its unit is kJ g–1. By comparing the calorific values of substances,
one can know which will be used as better fuel. For example, the heat of combustion of one mole of
methane is –890.3 kJ and that of one mole of ethane is –1559.7 kJ. Therefore, the calorific value of
35
THERMODYNAMICS–I
methane will be ∆H0C / gram = + (890.3/16) kJ g–1 = 55.64 kJ g–1 and the calorific value of ethane
= (–1559.30/30) kJ g–1 = –51.90 kJ g–1. Hence, methane has a better fuel efficiency.
3. Heats of solution and dilution
When a substance is dissolved in a solvent, heat may either be liberated or absorbed. This thermal
change is known as the heat of solution. The heat change when one mole of solute gets dissolved is not
constant; it generally varies with the concentration of the solution. Thus, during the course of the
solution process, the heat of solution per mole at any time changes with the concentration of the
solution. This quantity of heat change is known as the ‘differential heat of solution’. However, the total
heat change per mole of solute when the solution process is complete is called the ‘integral heat of
solution’. Therefore, the integral heat of solution of any substance may be defined as the enthalpy
change when one mole solute is dissolved in a definite quantity of a pure solvent to form a solution of
the desired concentration under conditions of constant temperature and pressure. The solution process
of HCl in general may be represented by chemical equation
→ HCl (nH2O)
HCl (g) + nH2O (l) 
where ‘n’ represents the number of moles of the solvent (H2O). Thus,
–1
→ HCl · 5H2O;
HCl (g) + 5H2O (l) 
DH = –63.99 kJ mol
→ HCl · 25 H2O; DH = –72.17 kJ mol–1
HCl (g) + 25H2O (l) 
→ HCl (aq);
HCl (g) + 2H2O(l) 
and
DH = –75.14 kJ mol
–1
suggest that when one mole of hydrogen chloride gas is dissolved in 5 mole of water 63.99 kJ of heat
is evolved; when dissolved in 25 mole of water 72.17 kJ heat is evolved but if 1 mole HCl is dissolved
in a very large quantity of water then 75.14 kJ of heat is liberated at 298.15 K. The term HCl (aq)
represents an aqueous solution of HCl which is so dilute that on further dilution no heat change takes
place.
(i) Integral heat of dilution
It has been mentioned above that the heat of solution of a substance varies with its concentration. This
fact suggests that there must be heat change when a solution is diluted further by adding more solvent.
This heat change is known as heat of dilution which can be defined as the change in heat content when
a solution containing 1 mole of solute is diluted further from one known concentration (C1) to another
concentration (C2). This quantity is the difference between the integral heats of solution at the two
concentrations. For example, as mentioned above for the case of HCl, one easily see that when 20
moles of water are added to a solution of HCl · 5H2O, then a solution HCl · 25H2O is obtained. The
change in enthalpy during this process is actually called as the integral heat of dilution.
(i)
(ii)
→ HCl · 5H2O;
HCl (g) + 5H2O (l) 
–1
DH1 = –63.99 kJ mol
→ HCl · 25H2O; DH2 = –72.17 kJ mol–1
HCl (g) + 25H2O (l) 
so that
→ HCl · 25H2O; DH3 = DH2 – DH1 = –8.18 kJ mol–1
(iii) HCl · 5H2O + 20 H2O (l) 
36
PHYSICAL CHEMISTRY–II
DH3 = DHdilution
Thus,
Hence,
DHdilution = –8.18 kJ mol–1.
(ii) Differential heats of solution and dilution
The differential heat of solution may be defined as the change in enthalpy, or heat content when one
mole of a substance (solute) is added (dissolved) to a large amount of a solution in such a way that the
concentration of the solution remains unchanged.
Whereas, the differential heat of dilution is defined as the heat change when one mole of a solvent
is added to a large volume of the solution of known concentration in such a way that the concentration
of the solution is not altered. The heat change associated with this process is called the differential heat
of dilution.
4. Heat of hydration
Heat of hydration of a particular anhydrous or partially hydrated salt is defined as the heat change
when it combines with the specific amount of water to form a stable hydrated salt. For example, the
hydration of anhydrous CuSO4 can be shown by the following expression:
→ CuSO4 · 5H2O (s)
CuSO4 (s) + 5H2O (l) 
The DH value for such a reaction is negative. The experimental determination of heat of hydration
is very difficult (almost impossible). But, with the help of Hess’s law, the value of heat of hydration
can easily be calculated from integral heats of solution of the hydrated and anhydrous salts in the
following manner:
→ CuSO4 (aq); DH = –66.5 kJ mol–1
(i)
CuSO4 (s) + 2H2O(l) 
→ CuSO4 (aq); DH = 11.7 kJ mol–1
and (ii) CuSO4 (s) . 5H2O (s) + 2H2O(l) 
Now Eq. (i) can be expressed in two steps as
→ Cu SO4 · 5H2O (s); DH = q1 kJ mol–1 (say)
(iii)
CuSO4 (s) + 5H2O (l) 
–1
→ CuSO4 (aq); DH = q2 kJ mol
and (iv) CuSO4 · 5H2O + 2H2O(l) 
(say)
According to Hess’s law these equations (iii) and (iv), on addition, yields
→ CuSO4 (aq); DH = (q1 + q2) kJ mol–1
CuSO4 (s) + 2H2O(l) 
(v)
Since equations (i) and (v) are same, therefore, one can write
q1 + q2 = –66.5 kJ mol–1
Further, since equations (ii) and (iv) are same, therefore,
q2 = +11.7 kJ mol
Hence
q1 = –78.2 kJ mol
–1
–1
Thus, Eq. (iii) is now written as
→ CuSO4 · 5H2O(s); DH = –78.2 kJ mol–1
CuSO4 (s) + 5H2O (l) 
Thus, the heat of hydration of CuSO4 is equal to –78.2 kJ mol–1.
37
THERMODYNAMICS–I
5. Heat of neutralization
The heat of neutralization of an acid by a base may be defined as the enthalpy change when one gram
equivalent of the acid is completely neutralized by one gram equivalent of the base in dilute aqueous
solution at a definite temperature. The solution must be dilute, so on mixing the acid and base no heat
change occurs due to dilution. The heat of neutralization of a base by an acid can also be defined in a
similar manner. For example, in the neutralization of one gram equivalent of strong acid HCl by a
strong base NaOH, or one gram equivalent NaOH by HCl, when both solutions are dilute and aqueous,
57.1 kJ of heat is produced. Thus, one can write
→ NaCl (aq) + H2O (l); DH = –57.1 kJ mol–1
HCl (aq) + NaOH (aq) 
Hence the enthalpy or heat of neutralization of HCl with NaOH or NaOH with HCl is 57.1 kJ mol–1.
The value of DH for the neutralization of any strong acid (such as HCl, HNO3 or H2SO4) by a
strong base (such as NaOH, KOH or LiOH) or vice versa, is always the same i.e., –57.1 kJ mol–1. The
reason for this is due to the fact that all the strong acids, strong bases and the salts which they form are
completely ionized in dilute aqueous solutions. Therefore, the reaction between them e.g., in the above
mentioned case of HCl and NaOH is written as
+
–
+
–
→ Na+(aq) + Cl– (aq) + H2O (l);
Na (aq) + OH (aq) + H (aq) + Cl (aq) 
DH = –57.1 kJ mol
–1
+
–
→ H2O (l); DH = –57.1 kJ mol–1
H (aq) + OH (aq) 
or
Thus, the neutralization process is simply a reaction between H+ ions (furnished by the acid) and
ions (furnished by the base) to form one mole of H2O. Since strong acids and strong bases ionize
completely in dilute aqueous solution, the number of H+ and OH– ions produced by one gram equivalent
of strong acid and strong base is always the same. Hence the value of heat of neutralization between a
strong acid and a strong base remains always constant.
But in those cases when either the acid, or the base, or both are weak, the heat of neutralization is
generally found to be less than 57.1 kJ mol–1. The reason for this may be explained by considering the
neutralization of a weak acid such as acetic acid with a strong base like sodium hydroxide (heat of
neutralization = 55.2 kJ mol–1). Acetic acid ionizes to a small extent whereas ionization of sodium
hydroxide is complete.
OH–
(i) CH3COOH
(ii)
CH3COO– + H+
→ Na+ + OH–
NaOH 
When H+ ions furnished by the weak acetic acid combine with the OH– ions produced by the
strong base NaOH, the equilibrium of reaction step (i) shifts to the right i.e., more of acetic acid
dissociates. A part of the heat produced during the combination of H+ and OH– ions is used for the
complete dissociation of acetic acid. This heat is called the heat of ionization or heat of dissociation.
For acetic acid this value is +1.9 kJ mol–1. Therefore, the net heat evolved in the above reaction is
(57.1 – 1.9) kJ or 55.2 kJ mol–1.
38
PHYSICAL CHEMISTRY–II
Similarly 5.6 kJ heat is used for the dissociation of the weak base NH4OH in the neutralization of
NH4OH with strong acid HCl. Therefore, the heat of neutralization in this case will be
= 57.1 – 5.6 = 51.5 kJ mol–1.
6. Heat of phase change
Phase transformations are accompanied by heat changes. Thus, heat of phase change is defined as the
enthalpy change when one mole of substance is converted from one phase to another at a particular
temperature and pressure.
(a) Heat of fusion
The heat change associated with the conversion of one mole of a solid substance into the liquid state is
defined as the heat of fusion. For example,
→ H2O (l); DHfusion = +6.0 kJ mol–1
H2O (s) 
Ice
Water
when one mole of ice melts at 0°C it absorbs +6.0 kJ mol–1 of heat. The greater is the magnitude of
intermolecular forces the larger will be the heat of fusion.
(b) Heat of vapourization
When one mole of a liquid is converted into the vapour phase at a given temperature and pressure then
the enthalpy change is known as the heat of vaporization. For example,
→ H2O (g); DHvap = +40.7 kJ mol–1
H2O (l) 
Conversion of 1 mole water into steam at 100°C is associated with absorption of +40.7 kJ mol–1.
(c) Heat of sublimation
It is defined as the enthalpy change when 1 g mole of a solid substance changes directly into gaseous
state without changing into liquid state. This process takes place below the melting point of the solid.
For example, the heat of sublimation of 1 g atom of graphite to monoatomic carbon vapour is
+715.1 kJ mol–1.
→ C (g);
C (s) 
DHsublimation = +715.1 kJ mol–1.
(d) Heat of transition in solid phase
The change in enthalpy, or change in heat content when one mole of a solid changes from its one
allotrope to another, is defined as the heat of transition. For example, the transitions of diamond into
graphite and sulphur (rhombic) into sulphur (monoclinic) are given below:
→ C (graphite);
C (diamond) 
–1
DHtransition = –1.9 kJ mol
→ S (monoclinic); DHtransition = –0.3 kJ mol–1.
S (rhombic) 
39
THERMODYNAMICS–I
7. Heat of hydrogenation
When one gram mole of an unsaturated hydrocarbon is hydrogenated into saturated compound by
gaseous hydrogen, then the change in heat content is known as the heat of hydrogenation. For example,
ethylene and benzene are hydrogenated as shown below:
→ C2H6 (g); DH = –135.6 kJ mol–1
C2H4 (g) + H2 (g) 
→ C6H12 (g); DH = –208.4 kJ mol–1.
C6H6 (g) + 3H2 (g) 
8. Heat of formation of ions in solution
In aqueous solution the formation of H+ and OH– ions are involved in one way or the other. With the
heat of formation of these two ions, the heat of formation of any other ion present in the aqueous
solution of various electrolytes can be found. Let us first calculate the standard heats of formation of
H+ and OH– ions. We know that the heat of formation of one mole of water from H+ and OH– ions is
equal to the heat of neutralization of strong acid by a strong base (–57.1 kJ mol–1). The standard heat
of formation of water from its elements H2 and O2 is –285.8 kJ mol–1. Thus, we may write
+
–
→ H2O (l); DH0 = –57.1 kJ mol–1
H (aq) + OH (aq) 
On reversing, we get
→ H+ (aq) + OH– (aq);
H2O (l) 
(i)
and
(ii)
H2 (g) +
1
2
0
–1
DH = +57.1 kJ mol
→ H2O (l); DH0 = –285.8 kJ mol–1
O2 (g) 
On adding the above two equations, we get
(iii)
H2 (g) +
1
2
→ H+ (aq) + OH– (aq); DH0 = –228.7 kJ mol–1
O2 (g) 
Thus, the sum of heat of formation of both the ions H+ and OH– is equal to –228.7 kJ mol–1. It
is a convention that the standard heat of formation of H+ ions in aqueous solution is taken to be
zero i.e.,
1
2
→ H+ (aq) + e–; DH0 = 0
H2 (g) 
Therefore, the heat of formation of OH– (aq) ion is obtained from equation (iii).
Thus,
DH0 of OH– (aq) = – 228.7 kJ mol–1.
Knowing the enthalpies of formation of these two ions, the standard heat of formation of other
ions in solution can be calculated.
Problem 1 : Obtain the standard heat of formation of chloride ions when the standard heat of formation
of HCl in water at 25°C is –168.0 kJ mol–1.
Solution : The heat of formation of HCl is written as:
(i)
1
2
H2 (g) +
1
2
→ H+ (aq) + Cl– (aq); DH0 = –168.0 kJ mol–1
Cl2 (g) + (aq) 
Further the standard heat of formation of H+ ions is taken as zero.
40
PHYSICAL CHEMISTRY–II
1
2
(ii)
→ H+ (aq) + e– ; DH0 = 0
H2 (g) 
On subtracting Eq. (ii) from Eq. (i), one gets
1
2
–
→ Cl– (aq);
Cl2 (g) + (aq) + e 
0
–1
DH = –168.0 kJ mol .
1.14 BOND ENERGIES OR BOND ENTHALPIES
During a chemical reaction some bonds are broken and some new bonds are formed. When a bond is
formed between two atoms, there is release of energy from the system. The same amount of energy is
absorbed when the bond is broken.
The bond energy is, therefore, defined as the average amount of energy required to break all bonds
of a particular type in one mole of the substance. It is expressed in kJ mol–1 or kcal mol–1. For
example, the bond energy of H—H bond is ~435.9 kJ mol–1 or 104.18 kcal mol–1. Generally the
experiments are performed at constant pressure, so the bond energy is mostly called the bond enthalpy.
The bond energy of a particular bond depends on
(i) the specific molecule in which it occurs, and
(ii) its exact position in the molecule.
The concept of bond energy in diatomic and polyatomic molecules is quite different. Therefore,
it is very essential to distinguish between the bond energy and the bond dissociation energy of a given
linkage. The bond dissociation energy is the energy required to dissociate a given bond of some specific
compound, whereas the bond energy is the average value of the dissociation energies of the said bond
in a series of different dissociating species.
For diatomic molecules such as H2, N2, O2, Cl2 and HCl etc., the bond energies are equal to their
bond dissociation energies. But in case of polyatomic molecules the bond energy of a particular bond
is not the same when present in different types of compounds. For example, bond energy of C—Cl is
not same in CH3Cl, CH2Cl2, CHCl3 and CCl4. In fact, even in the same compound (such as methane,
CH4) the bond energy of C—H bond is not the same. In CH4, there are four C—H bonds but the bond
energy for first, second, third and fourth C—H bonds are not equal. In such cases, average value is
taken. Thus,
ÄH1 + ÄH 2 + ÄH 3 + ÄH 4
= 1663.4 kJ mol –1/4 = 415. 85 kJ mol –1
∈ C—H =
4
The bond energy is a measure of strength of the bond. In other words, bond energy is the force
with which the atoms are bonded together. It depends upon:
(i) size of the atom,
(ii) electronegativity, and
(iii) bond length.
A knowledge of bond energy is useful for calculating heat of reactions for gaseous reactions for
which no thermal data is available and which involve substances having covalent bonds. For example,
if we want to know the bond energy of C—H bond in methane then we should know the energy change
for the reaction
→ CH4 (g).
C (g) + 4H (g) 
41
THERMODYNAMICS–I
The energy change for the above reaction can be obtained by combining the heat of formation of
methane from C (s) + H2 (g) with the heat of sublimation of carbon i.e., C (s) ® C (g) and the heat of
dissociation of hydrogen into atoms i.e., H2 (g) ® 2H (g), which have been determined by spectroscopic
methods. The value for the energy change, thus obtained is ~1663.39 kJ mol–1. This represents the
bond energy for four C—H bonds. Since all the bonds in methane are identical as such the bond energy
of C—H bond is one-fourth of the total change in energy i.e., 1663.39 kJ mol–1/4 = 415.85 kJ mol–1.
In a similar way the bond energies of other bonds can also be calculated. Further, it should be noted
that when a bond is broken, the bond energy is given a positive sign because heat is absorbed during the
process, but when a bond is formed the bond energy is represented with a minus sign as heat is evolved
during this process.
The bond energies of some common bonds are given in Table 1.
TABLE 1.1: Bond energy values of some common bonds
Bond
Bond energy
(kJ mol–1)
Bond
Bond energy
(kJ mol–1)
H– H
435.9
C–H
415.85
H– F
565
O–H
462.6
H– l
431
N–H
389
H–Br
364
C– C
336.8
H– I
297
C=C
615
F– F
155
CºC
812
Cl–Cl
242
C–Cl
328
Br– Br
190
C–O
335
I– I
149
C=O
707
O=O
494
C–N
293
NºN
941
Bond energy calculations
Bond energies can easily be calculated by the knowledge of heat of formation of molecules from the
atom and heat of combustion etc. Bond energy calculations in few cases are discussed below:
1.
C—H bond energy in methane
We have four C—H bonds in methane so the C—H bond energy will be the average of the bond
dissociation energies
ÄH
→ C (g) + 4 H (g); ÎC—H =
CH4 (g) 
4
The values of heat of formation of methane, the heat of sublimation of C (graphite) and the heat
of dissociation of H2 (g) molecules are used to calculate the ÎC—H in methane.
(i)
→ CH4 (g); DH0 = –74.9 kJ
C (graphite) + 2H2 (g) 
This DH value is reversed when one mole of methane is split into separate atoms of carbon and
hydrogen.
42
PHYSICAL CHEMISTRY–II
→ C (graphite) + 2H2 (g); DH0 = +74.9 kJ
CH4 (g) 
(ii)
Further, it is given that
(iii)
→ C (g); DH = +716.68 kJ
C (graphite) 
→ 4H (g); DH = 2 × 435.9 = 871.8 kJ
2H2 (g) 
(iv)
On adding equations (ii), (iii) and (iv), one gets
→ C (g) + 4H (g); DH = 1663.39 kJ
CH4 (g) 
Thus, C—H bond energy, ÎC—H =
2.
ÄH 1663.39
=
= 415.85 kJ mol–1.
4
4
C—C bond energy in ethane
In ethane there are 6C—H and 1C—C bonds. The heat of formation of ethane is given by the following
equation:
→ C2H6 (g); DH = –87.9 kJ
(i) 2C (graphite) + 3H2 (g) 
On reversing this equation and then combining with other given equations, one can write
→ 2C (graphite) + 3H2 (g);
C2H6 (g) 
(ii)
DH = +87.9 kJ
Further, we know
(iii)
→ 2C (g);
2C (graphite) 
(iv)
→ 6H (g);
3H2 (g) 
DH = 716.68 × 2 = 1433.36 kJ
DH = 435.9 × 3 =1307.7 kJ
On adding equations (ii), (iii) and (iv), one gets
→ 2C (g) + 6H (g); DH = 2828.9 kJ
C2H6 
This suggests 2828.9 kJ energy is needed to break 6C—H bonds and one C—C bond in ethane.
So the C—C bond energy is given as
Î C—C = DH – 6 ÎC—H
= 2828.9 – 6 (415.85)
–1
= 338.8 kJ mol .
3. C=C bond energy in ethylene
In ethylene there are 4C—H bonds and one C = C bond. The heat of formation of ethylene is given by
the reaction
(i)
→ C2H4 (g); DH = +52.6 kJ
2C (graphite) + 2H2 (g) 
43
THERMODYNAMICS–I
This on reversing may be written as
(ii)
(iii)
(iv)
→ 2C (graphite) + 2H2 (g); DH = –52.6 kJ
C2H4 (g) 
→ 2C (g); DH = 1433.36 kJ
2C (graphite) 
→ 4H (g);
2H2 (g) 
DH = 871.8 kJ
On adding equations (ii), (iii) and (iv), one may write
→ 2C (g) + 4H (g); DH = 2252.56 kJ
C2H4 (g) 
2252.56 kJ energy is needed to break 4C—H and 1 C=C bond in ethylene. So the C=C bond
energy is given as
ÎC= C = 2252.56 – 4ÎC—H
= (2252.56 – 415.85 × 4)
= (2252.56 – 1663.4)
–1
= 589.16 kJ mol .
4. C=C bond energy in acetylene
In acetylene there are 2C—H bonds and one C=C bond. The heat of formation of acetylene is given by
the equation
(i)
→ C2H2 (g); DH = 226.9 kJ
2C (graphite) + H2 (g) 
On reversing this, we get
(ii)
(iii)
(iv)
→ 2C (graphite) + H2 (g); DH = –226.9 kJ
C2H2 (g) 
→ 2C (g); DH = 1433.36 kJ
2C (graphite) 
→ 2H (g); DH = 435.9 kJ
H2 (g) 
On addition of equations (ii), (iii) and (iv),we can write
→ 2C (g) + 2H (g); DH = 1642.36 kJ
C2H2 (g) 
Thus, the C=C bond energy is given as
ÎC= C = 1642.36 – 2ÎC—H
= (1642.36 – 831.7)
–1
= 810.66 kJ mol .
5. Estimation of heat of reaction from bond energy data
It is possible to calculate heats of gaseous reactions from bond energy data. For example, let us calculate
the heat change for the reaction
→ C2H6 (g);
C2H4 (g) + 2H (g) 
DH = ?
44
PHYSICAL CHEMISTRY–II
This reaction is represented diagrammatically as
H
H H
H
C
H
2C (g)
(g) + H — H (g)
C
H
Break
bonds
+
∆rH
H
C C
H
H H
Break
bonds
4H (g) + 2H (g)
Make bond
H
H
H C
C
H
H
H
Here we can imagine that firstly ethylene and hydrogen dissociates into atoms which absorbs
energy equivalent to the energy required in breaking four C—H bonds, one C=C bond and one
H—H bond thus, the heat required to dissociate C2H4 (g) and H2 (g) into the gaseous atoms is
610.0 kJ mol–1
4 × 415.85 kJ mol–1
435.9 kJ mol–1
—————————
Total heat =
2709.3 kJ mol–1
—————————
Now if ethane molecule is formed from separate gaseous atoms, then 1C — C bond and 6C — H
bonds are to be formed and equivalent amount of energy needed for the formation of these bonds will
have to be released. Thus, the energy released for the formation of C2H6 (g) will be
For making
1 C—C bond
338.8 kJ mol–1
For making
6 C—H bond
6 × 415.85 kJ mol–1
—————————
Total heat =
2833.9 kJ mol–1
—————————
The heat of reaction, DH is thus given by
DH = Bond enthalpy in – Bond enthalpy out
For breaking
For breaking
For breaking
1 C=C bond
4 C—H bond
1 H—H bond
 Energy required for   Energy required in 

−

 breaking the bonds   the formation bonds 
= 2709.3 – 2833.9
= –124.6 kJ mol–1
Thus heat of reaction for the reaction
→ C2H6 (g)
C2 H4 (g) + 2H (g) 
Comes out to be –124.6 kJ mol
–1
.
6. Evaluation of heat of formation from bond energy data
Heats of formation of several compounds can also be obtained from bond energy data. For example, heat
of formation of ethane can be calculated with the following data: Î H—H = Q 1 kJ mol –1 ,
45
THERMODYNAMICS–I
ÎC—H = Q2 kJ mol–1, ÎC—C = Q3 kJ mol–1 and heat of vaporization of 1 g atom of carbon = Q4 kJ mol–1.
The formation of ethane may be represented as
→
2C (s) + 3 H—H (g) 
H
H
H
C
C
H
H
H
This reaction involves vaporization of two gram atoms of solid carbon and breaking of three
moles of H—H bonds which requires energy. Further, one C—C bond and 6C—H bonds are formed in
which energy will be released. Thus, the heat of formation of ethane is calculated as
DH = Bond enthalpy in – Bond enthalpy out
= (Total energy required for – (Total energy released in
breaking the bonds)
bond formation)
= [2ÎC(s)—C(g) + 3(ÎH—H] – [6 (ÎC—H) – 1 (ÎC—C)]
–1
= (2Q4 + 3Q1 – 6Q2 – Q3) kJ mol .
Effect of temperature on the heat of reaction: The Kirchhoff’s relation
It is possible to calculate the heat of reaction at standard temperature (298.15 K) and pressure (1 atm)
by using the data of standard heats of formation from table. But several times, the heats of reaction at
other temperature and pressures are also required. The variation of heat of reaction with temperature at
constant pressure was given by Kirchhoff who derived an expression which explains the temperature
dependence of the heat of reaction. This expression can be derived in the following manner:
Let us consider a general reaction of the type
A

→
(Reactant)
B
(Product)
If this reaction takes place at constant volume, then the heat of reaction will be equal to change in
internal energy of the system. Thus
DE = Eproduct – Ereactant = EB – EA
…(1.114)
where EA and EB are internal energies of the reactant and product respectively. If we want to study the
change in internal energy with respect to temperature, then one needs to differentiate Eq. (1.114) with
respect to T at constant volume. On doing so, we get
 ∂E 
 ∂(ÄE ) =  ∂E B 

 −  A
 ∂T 
T

V
 ∂ V
 ∂T  V
But since
 ∂E 
  = CV
 ∂T  V
…(1.115)
46
PHYSICAL CHEMISTRY–II
We have
 ∂(ÄE) =
(CV )B − (CV )A
 ∂T 

V
d (DE) = DCV dT
or
…(1.116)
…(1.117)
where DCV = å CV (product) – å CV (reactant)
where (CV)A and (CV)B are heat capacities of the reactant and product at constant volume. Further,
integrating Eq. (1.117) within the limits T1 and T2, we have
2
∫
2
∂(∆E) =
1
∫ (∆CV ) dT
1
or
DE2 – DE1 = DCV (T2 – T1)
or
DE2 (T2) – DE1(T1) = DCV (T2 – T1).
…(1.118)
However, if the reaction is occurring at constant pressure, then the heat of reaction for the above
mentioned general reaction is equal to change in enthalpy, DH and one can write
DH = Hproduct – Hreactant = HB – HA
…(1.119)
where HA and HB are the enthalpies of the reactants and products respectively. In order to study the
effect of temperature on the heat of reaction at constant pressure, Eq. (1.119) is differentiated with
respect to the temperature, at constant pressure. Thus,
 ∂HB 
 ∂H A 
 ∂(ÄH)
 ∂T  =  ∂T  −  ∂T 

P

P

P
…(1.120)
As we know that the heat capacity at constant pressure,
 ∂H 
CP = 

 ∂T  P
Therefore,
 ∂(ÄH)
 ∂T  = (C P )B − (CP )A = ∆C P

P
d(DH) = DCP dT
or
(1.121)
where DCP = å CP (products) – å CP (reactants).
where (CP)A and (CP)B are the heat capacities of the reactants and products respectively at constant
pressure. If the heat of reaction DH1 is known at one temperature T1, then the heat of reaction DH2 at
temperature T2 can be obtained by integrating Eq. (1.121) within the temperature limits T1 and T2.
Thus,
T2
∫
T1
T2
d (∆H) =
∫
T1
(∆CP ) dT
47
THERMODYNAMICS–I
or
DH2 – DH1 = DCP (T2 – T1)
or
DH2 (T2) – DH1(T1) = DCP (T2 – T1).
…(1.122)
Eqs. (1.118) and (1.122) are known as Kirchhoff’s equation which enable one to calculate heat
of reaction at a particular temperature when other quantities like heat of reaction at other temperature
and molar heat capacities of the reactants and products are given.
Problem 2: The heat of dissociation of gaseous water per mole at 18°C and 1 atm pressure is 241.75
kJ. Calculate its value at 70°C when the following data are available:
–1
–1
CP (H2O) = 33.56, CP (H2) = 28.83, CP (O2) = 29.12 JK mol .
Solution: We can write the dissociation reaction of water as
→ H2 (g) +
H2O (g) 
For this reaction,
1
2
O2 (g);
0
DH (291.15 K) = 241.75 kJ
DCP = å CP (product) – å CP (reactant)
= CP (H2) +
1
2
CP (O2) – CP (H2O)
= (28.33) +
1
2
29.12 – 33.56 JK mol
–1
–1
–1
–1
= 9.83 JK mol .
Therefore,
DCP + (T2 – T1) = 9.83 (70 – 18) = (9.83 × 52)
= 511.16 J mol
–1
–1
= 0.51116 kJ mol
Hence,
0
0
DH (343.15 K) = DH (291.15 K) + DCP (DT)
–1
= (241.75 + 0.51116) kJ mol
= 242.251 kJ mol–1.
Effect of pressure on the heat of reaction
Very often heats of reaction at other pressure are needed. So it is necessary to discuss the effect of
pressure on the heats of reaction. For this purpose, the thermodynamic equation of state is used. This
equation is (Eq. 2.300 of chapter 2)
 ∂H 
 ∂V 
 = V − T


P
∂

T
 ∂T P
…(1.123)
Eq. (1.123) can also be written as
 ∂ ( ∆V) 
 ∂( ∆H) 
 ∂P  = ∆V − T  ∂T 

T
P
where DV is the volume change of the reaction at pressure P.
…(1.124)
48
PHYSICAL CHEMISTRY–II
ÄV =
∑ V (products) − ∑ V (reactants)
On differentiating with respect to T at constant P, we get
 ∂(∆V) 
 ∂T  =

P
∑
 ∂V (products) 
 ∂V (reactants) 
−∑ 



∂T
∂T

P

P
…(1.125)
Further the coefficient of thermal expansion a is defined as
á =
1  ∂V 


V  ∂T  P
Therefore, Eq. (1.125) becomes
 ∂(∆V) 
 ∂T  =

P
∑ áV (products) − ∑ áV (reactants)
= D (aV)
…(1.126)
So Eq. (1.124) is then written as
 ∂(∆H) 
 ∂T  = ∆V − T [∆(á V)]

P
…(1.127)
Eq. (1.127) can be integrated between the pressure limits P1 and P2 as
P2
(ÄH)P2 − (ÄH)P1
=
∫ [ÄV − T Ä (α V)]
dP
…(1.128)
P1
P2
(ÄH)P2
or
=
(ÄH)P1 + ∫ [ÄV − T ∆(αV)] dP
…(1.129)
P1
This equation describes the pressure dependence of the heat of reaction. In case of ideal gases the
heat of reaction is independent of pressure.
For ideal gases,
So that,
and thus,
V = nRT/P
ÄV = Än
RT
P
R
 ∂(∆V) 
 ∂T  = Än P

P
On substituting these values in Eq. (1.124), one gets
RT
R
 ∂(∆H) 
 ∂P  = Än P − T ∆n P = 0

T
Thus, heat of reaction is thus independent of the pressure.
…(1.130)
49
THERMODYNAMICS–I
QUESTIONS
1. Explain the following:
(a) System, boundary and surroundings
(b) Thermodynamic processes
(c) Work, heat and energy
2. State first law of thermodynamics. Derive a relationship between E, q and W.
3. Define CP and CV.
4. For an ideal gas, show that CP – CV = R.
5. What is Joule-Thomson effect? Derive and expression for Joule-Thomson coefficient of ideal
gas and a van der Wall’s gas.
6. Prove the following:
(i) TV
(ii)
PVg
g–1
= Constant
= Constant
(iii) (T2/T1) g = (P1/P2)1 – g.
7. Discuss different laws of thermochemistry.
8. Discuss Hess’s law.
9. Define the following:
(i) Heat of formation
(ii) Heat of reaction
(iii) Heat of transition
(iv) Lattice energy of a crystal.
10. What is standard enthalpy of formation?
11. Discuss the following:
(i) Heat of combustion
(ii) Heat of solution
(iii) Heat of hydration
(iv) Heat of neutralization.
12. Discuss the following:
(i) Heat of phase change
(ii) Heat of hydrogenation
(iii) Heat of formation of ions in solution.
13. Distinguish in the following pairs:
(i) Exothermic and endothermic reactions.
(ii) Enthalpy of a reaction at constant volume and constant pressure.
(iii) Integral and differential heat of solution.
(iv) Enthalpy of dissociation and enthalpy of ionization.
50
PHYSICAL CHEMISTRY–II
14. Explain the following giving reasons:
(i) For reactions involving condensed phases, i.e., solids and liquids,
DH = DE
(ii) Heat of neutralization of a strong monobasic acid and strong base is always –57.32 kJ mol–1.
(iii) The Hess’s law of constant heat summation is a direct consequence of the first law of
thermodynamics.
(iv) The energy required to break a OH bond in water is 498 kJ mol–1 while in hydryl radical it
is 430 kJ mol–1.
15. Given the following reactions:
→ C 
→D
(i) A + B 
→ P 
→D
(ii) A+ B 
How the enthalpy changes for reactions (i) and (ii) are related to each other?
16. Given the following informations:
(i) H2(g) + ½ O2(g) = H2O(l); DH0 = – 285.84 kJ
(ii) C(s) + O2(g) = CO2(g); DH0 = – 393.51 kJ
(iii) CH4(g) + 2O2(g) = CO2(g) + 2H2O(l); DH0 = – 890.35 kJ
Calculate, at 298 K the enthalpy change for the reaction
CH4(g) + O2(g) = C(s) + 2H2O(l)
17. Estimate heat of reaction and heat of formation from bond energy data.
18. (i)
Deduce the following relationships:
DE2(T2) – DE1(T1) = DCv (T2 – T1)
DH2(T2) – DH1(T1) = DCp (T2 – T1)
(ii) Derive the following relationship:
( ∆H) P2 = ( ∆H) P1 +
P2
∫ [∆V − T∆(αV)] dP
P1