THERMODYNAMICS–I 23

ã 1– ã
 T2   P1 
or   =   …(1.100)
 T1   P2 

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
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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)

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 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
aA + bB 
→ cC + dD ….(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.
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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
PV1 = n1RT 
and  …(1.109)
PV2 = n2 RT 

where n1 and n2 are the number of moles of the reactants and products respectively. V1 is the volume of

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 THERMODYNAMICS–I 25

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
RT
= (n 2 − n1 )
P

RT
or DV = Än …(1.110)
P
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,
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H2 (g) + Cl2 (g) 

→ 2HCl (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

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 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

CH4 (g) + 2O2 (g) 

→ CO2 (g) + 2H2 O (l); DH (298 K) = –890.3 kJ mol–1

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 + O2 
→ 2H2 O; DH = –571.6 kJ mol–1

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 O2(g) 
→ H2O (l); DH = –285.8 kJ mol–1.

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:
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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.

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 THERMODYNAMICS–I 27

S (s) + O2 (g) 
→ SO2 (g); 0 –1
DH (298 K) = –297.5 kJ mol

But when SO2 (g) is decomposed, exactly the same amount of heat as mentioned above is needed
and the thermochemical equation is written as
SO2 (g) 
→ S (s) + O2 (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.

CO2 (g) + 2H2O (l) 
→ CH4 (g) + 2O2 (g); 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.

,H
A P
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,H1
,H5
B E
,H2
,H 4

,H3
C D

Fig. 1.3: The representation of direct and indirect paths

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 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.
A  → P; DH = q joules
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:
A  → B; DH1 = q1
B 
→ C; DH2 = q2
C 
→ D; DH3 = q3

D 
→ E; DH4 = q4
E 
→ P; 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
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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

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 THERMODYNAMICS–I 29

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. C (s) + O2 (g) 
→ CO2 (g); DH1 = –393.7 kJ mol–1

Step 2. CO2 (g) + NH3 (g) 

→ Urea (s); DH2 = –813.58 kJ mol–1

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:

0 H2 (g) +½O2 (g)

∆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):
H2O (l) 
→ H2 (g) + –1
1
2 O2 (g); DH1 = +285.8 kJ mol

H2 (g) + ½ O2 (g) 
→ H2O (g); DH2 = –242.3 kJ mol–1
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The total amount of enthalpy change is

–1
DH1 + DH2 = +43.5 kJ mol

However, when the reaction H2O (l) 

→ H2O (g) takes place directly then the value of
–1
DH = +43.5 kJ mol
Thus, DH = DH1 + DH2 holds true.

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 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

C (graphite) + ½ O2 (g) 
→ CO (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) C (graphite) + O2 (g) 

→ CO2 (g); DH1 = –393.5 kJ mol–1

(ii) CO (g) + 1
2 O2 (g) 
→ CO2 (g); DH2 = –283.0 kJ mol–1
On subtracting (ii) from (i), one gets

C (graphite) + 1
2 O2 (g) 
→ CO (g); DHf = –110.5 kJ mol–1

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) → 2CO2 (g) + 3H2O (l); DH = –1380.7 kJ mol–1

C2H5OH (l) + 3O2 (g) 
(ii) C (s) + O2 (g) 
→ CO2 (g); DH = –393.5 kJ mol–1

(iii) 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 O2 (g) 
→ 2CO2 (g) + 3H2O (l); DH = – 1446.8 kJ

Equation (i) is then subtracted from equation (iv) and then the final result is written as
2C (s) + 3H2 (g) + 1
2 O2 (g) 
→ C2H5OH (l); DH = –66.1 kJ
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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

C2H4 (g) + H2 (g) 

→ C2H6 (g)

at 298 K can be obtained by using the heat of combustion values of ethylene (C2H4), hydrogen (H2)

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 THERMODYNAMICS–I 31

and ethane (C2H6). The thermochemical equations for the heats of combustion of C2H4, H2 and C2H6
are written as
(i) C2H4 (g) + 3O2 (g) 
→ 2CO2 (g) + 2H2O (l); DH0 = –1410.8 kJ mol–1

(ii) H2 (g) + 1
2 O2 (g) 
→ H2O (l); DH0 = –286.2 kJ mol–1

(iii) C2H6 (g) + 3 12 O2 (g) 

→ 2CO2 (g) + 3H2O (l); DH0 = –1560.6 kJ mol–1

The heat of reaction of

C2H4 (g) + H2 (g) 
→ C2H6 (g)

can be obtained by adding equations (i) and (ii) which yields

C2H4 (g) + H2 (g) + 3 12 O2 (g) 

→ 2CO2 (g) + 3H2O (l); DH0 = –1697.0 kJ mol–1

From which the equation (iii) is subtracted to have the desired result for the standard heat of
formation of ethane.

C2H4 (g) + H2 (g) 

→ C2H6 (g); DH = –136.4 kJ mol–1.

(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) C (diamond) + O2 (g) 
→ CO2 (g); DH0 = –395.4 kJ mol–1
(ii) C (graphite) + O2 (g) 
→ CO2 (g); DH0 = –393.5 kJ mol–1
By subtracting equation (ii) from (i), we get
C (diamond) 
→ C (graphite); DH0tran = – 1.9 kJ mol–1.

(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
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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.
(i) Na (s) is vapourized into Na (g); Na (s) 
→ Na (g); –1
DH1 = 317.5 kJ mol

(ii) Na (s) is then ionized; → Na+ (g) + e– ; DH2= 495.8 kJ mol–1

Na (g) 

Cl2 (g) 
→ Cl (g); –1
(iii) Dissociation of Cl2; 1
2 DH3 = ½ × 241.84 kJ mol

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 32 PHYSICAL CHEMISTRY–II

Cl (g) + e 
→ Cl– (g);
– – –1
(iv) Dissociation ofCl (g); DH4 = –365.26 kJ mol

(v) Combination of Na+ (g) and Cl– (g) to form NaCl (s)

Na (g) + Cl (g) 
→ NaCl(s); DH5 = ?
+ –

On adding steps (1) to (v), the net change is written as

Na (s) + 1
2 Cl2 (g) 
→ NaCl (s); DH = –410.87 kJ mol–1

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 + 1
2 × 241.84 – 365.26 + DH5
–1
or DH5 = –979.9 kJ mol
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
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coefficient of elements is common in formation reactions.

Thus, the reactions
–1
H2 (g) + 1
2 O2 (g) 
→ H2O (l); ∆H 0f = –285.8 kJ mol
–1
C (s) + 2H2 (g) + 1
2 O2 (g) 
→ CH3OH (l); ∆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.

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 THERMODYNAMICS–I 33

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]
i.e., ÄH0f =[ ÄH0f (products) − ÄH0f (reactants)]
Therefore, for the general reaction aA + bB 
→ cC + dD, the standard heat of reaction is
obtained as
ÄH 0 = [{c × ÄHf (C) + d × ÄHf (D)} − { a × ÄHf (A) + b × ÄHf (B)}]
0 0 0 0

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) H2 (g) + ½O2 (g)

C (s) + 2H 2 (g)
0
2C (g) + H2 (g)
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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

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 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.
C2H5OH (l) + 3O2 (g) 
→ 2CO2 (g) + 3H2O (l); –1
∆H0C = –1367 kJ mol
The other examples of combustion reactions are
CH4 (g) + 2O2 (g) 
→ CO2 (g) + 2H2O (l); –1
∆H0C = – 890.3 kJ mol

C (diamond) + O2 (g) 
→ CO2 (g); –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
C2H5OH (l) + 3O2 (g) 
→ 2CO2 (g) + 3H2O (l); –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)
–1
or x = –277.5 kJ mol
Therefore, the heat of formation of liquid ethanol from its elements in the standard states is
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obtained very easily and is written as

O2 (g) 
→ C2H5OH (l); –1
2C (s) + 3H2 (g) + 1
2 ∆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

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 THERMODYNAMICS–I 35

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 (g) + nH2O (l) 
→ HCl (nH2O)

where ‘n’ represents the number of moles of the solvent (H2O). Thus,
HCl (g) + 5H2O (l)  → HCl · 5H2O; –1
DH = –63.99 kJ mol
HCl (g) + 25H2O (l) 
→ HCl · 25 H2O; DH = –72.17 kJ mol–1
HCl (g) + 2H2O(l) 
→ HCl (aq); –1
and DH = –75.14 kJ mol
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
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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.

HCl (g) + 5H2O (l) 

→ HCl · 5H2O; –1
(i) DH1 = –63.99 kJ mol

(ii) HCl (g) + 25H2O (l) 

→ HCl · 25H2O; DH2 = –72.17 kJ mol–1
so that

(iii) HCl · 5H2O + 20 H2O (l) 

→ HCl · 25H2O; DH3 = DH2 – DH1 = –8.18 kJ mol–1

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 36 PHYSICAL CHEMISTRY–II

Thus, DH3 = DHdilution

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 (s) + 5H2O (l) 
→ CuSO4 · 5H2O (s)

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:
(i) CuSO4 (s) + 2H2O(l)  → CuSO4 (aq); DH = –66.5 kJ mol–1

and (ii) CuSO4 (s) . 5H2O (s) + 2H2O(l) 

→ CuSO4 (aq); DH = 11.7 kJ mol–1
Now Eq. (i) can be expressed in two steps as
(iii) CuSO4 (s) + 5H2O (l)  → Cu SO4 · 5H2O (s); DH = q1 kJ mol–1 (say)
–1
and (iv) CuSO4 · 5H2O + 2H2O(l) 
→ CuSO4 (aq); DH = q2 kJ mol (say)
According to Hess’s law these equations (iii) and (iv), on addition, yields
(v) CuSO4 (s) + 2H2O(l) 
→ CuSO4 (aq); DH = (q1 + q2) kJ mol–1

Since equations (i) and (v) are same, therefore, one can write
q1 + q2 = –66.5 kJ mol–1
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Further, since equations (ii) and (iv) are same, therefore,

–1
q2 = +11.7 kJ mol
–1
Hence q1 = –78.2 kJ mol
Thus, Eq. (iii) is now written as
CuSO4 (s) + 5H2O (l) 
→ CuSO4 · 5H2O(s); DH = –78.2 kJ mol–1

Thus, the heat of hydration of CuSO4 is equal to –78.2 kJ mol–1.

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 THERMODYNAMICS–I 37

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

HCl (aq) + NaOH (aq) 

→ NaCl (aq) + H2O (l); DH = –57.1 kJ mol–1

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) + OH (aq) + H (aq) + Cl (aq) 

→ Na+(aq) + Cl– (aq) + H2O (l);
+ – + –

–1
DH = –57.1 kJ mol

H (aq) + OH (aq) 
→ H2O (l); DH = –57.1 kJ mol–1
+ –
or

Thus, the neutralization process is simply a reaction between H+ ions (furnished by the acid) and
OH– 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.
(i) CH3COOH CH3COO– + H+
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(ii) NaOH 
→ Na+ + OH–
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.

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 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 (s) 
→ H2O (l); DHfusion = +6.0 kJ mol–1
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 (l) 
→ H2O (g); DHvap = +40.7 kJ mol–1

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 (s) 
→ C (g); DHsublimation = +715.1 kJ mol–1.
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(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 (diamond) 
→ C (graphite); –1
DHtransition = –1.9 kJ mol
S (rhombic) 
→ S (monoclinic); DHtransition = –0.3 kJ mol–1.

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 THERMODYNAMICS–I 39

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) 
C6H6 (g) + 3H2 (g) 
→ C6H12 (g); DH = –208.4 kJ mol–1.

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

H (aq) + OH (aq) 
→ H2O (l); DH0 = –57.1 kJ mol–1
+ –

On reversing, we get

H2O (l) 
→ H+ (aq) + OH– (aq); 0 –1
(i) DH = +57.1 kJ mol

and (ii) H2 (g) + 1

2 O2 (g) 
→ H2O (l); DH0 = –285.8 kJ mol–1

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
H2 (g) 
→ H+ (aq) + e–; DH0 = 0

Therefore, the heat of formation of OH– (aq) ion is obtained from equation (iii).
Thus, DH0 of OH– (aq) = – 228.7 kJ mol–1.
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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
Cl2 (g) + (aq) 
→ H+ (aq) + Cl– (aq); DH0 = –168.0 kJ mol–1

Further the standard heat of formation of H+ ions is taken as zero.

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 40 PHYSICAL CHEMISTRY–II

(ii) 1
2 H2 (g) 
→ H+ (aq) + e– ; DH0 = 0

On subtracting Eq. (ii) from Eq. (i), one gets

Cl2 (g) + (aq) + e 
→ Cl– (aq);
– 0 –1
1
2 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
∈ C—H = = 1663.4 kJ mol –1/4 = 415. 85 kJ mol –1
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:
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(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
C (g) + 4H (g) 
→ CH4 (g).

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 THERMODYNAMICS–I 41

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 energy Bond energy
Bond Bond
(kJ mol–1) (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
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dissociation energies
ÄH
CH4 (g) 
→ C (g) + 4 H (g); ÎC—H =
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) C (graphite) + 2H2 (g) 
→ CH4 (g); DH0 = –74.9 kJ

This DH value is reversed when one mole of methane is split into separate atoms of carbon and
hydrogen.

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 42 PHYSICAL CHEMISTRY–II

(ii) CH4 (g) 

→ C (graphite) + 2H2 (g); DH0 = +74.9 kJ

Further, it is given that

(iii) C (graphite) 
→ C (g); DH = +716.68 kJ

(iv) 2H2 (g) 

→ 4H (g); DH = 2 × 435.9 = 871.8 kJ

On adding equations (ii), (iii) and (iv), one gets

CH4 (g) 
→ C (g) + 4H (g); DH = 1663.39 kJ

ÄH 1663.39
Thus, C—H bond energy, ÎC—H = = = 415.85 kJ mol–1.
4 4
2. 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:
(i) 2C (graphite) + 3H2 (g) 
→ C2H6 (g); DH = –87.9 kJ

On reversing this equation and then combining with other given equations, one can write

(ii) C2H6 (g) 

→ 2C (graphite) + 3H2 (g); DH = +87.9 kJ
Further, we know
(iii) 2C (graphite) 
→ 2C (g); DH = 716.68 × 2 = 1433.36 kJ
(iv) 3H2 (g) 
→ 6H (g); DH = 435.9 × 3 =1307.7 kJ

On adding equations (ii), (iii) and (iv), one gets

C2H6 
→ 2C (g) + 6H (g); DH = 2828.9 kJ

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)
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–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) 2C (graphite) + 2H2 (g) 
→ C2H4 (g); DH = +52.6 kJ

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 THERMODYNAMICS–I 43

This on reversing may be written as

(ii) C2H4 (g) 
→ 2C (graphite) + 2H2 (g); DH = –52.6 kJ

(iii) 2C (graphite) 
→ 2C (g); DH = 1433.36 kJ

(iv) 2H2 (g) 

→ 4H (g); DH = 871.8 kJ
On adding equations (ii), (iii) and (iv), one may write
C2H4 (g) 
→ 2C (g) + 4H (g); DH = 2252.56 kJ

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) 2C (graphite) + H2 (g) 
→ C2H2 (g); DH = 226.9 kJ

On reversing this, we get

(ii) C2H2 (g) 

→ 2C (graphite) + H2 (g); DH = –226.9 kJ

(iii) 2C (graphite) 
→ 2C (g); DH = 1433.36 kJ

(iv) H2 (g) 
→ 2H (g); DH = 435.9 kJ

On addition of equations (ii), (iii) and (iv),we can write

C2H2 (g) 
→ 2C (g) + 2H (g); DH = 1642.36 kJ

Thus, the C=C bond energy is given as

ÎC= C = 1642.36 – 2ÎC—H
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= (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

C2H4 (g) + 2H (g) 

→ C2H6 (g); DH = ?

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 44 PHYSICAL CHEMISTRY–II

This reaction is represented diagrammatically as

H H H
H
∆rH H C C H
C C (g) + H — H (g)
H H H H
Break Break
bonds bonds Make bond
H H

2C (g) + 4H (g) + 2H (g) 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
For breaking 610.0 kJ mol–1
1 C=C bond
For breaking 4 × 415.85 kJ mol–1
4 C—H bond
For breaking 435.9 kJ mol–1
1 H—H bond
—————————
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
 Energy required for   Energy required in 
 − 
 breaking the bonds   the formation bonds 
= 2709.3 – 2833.9
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= –124.6 kJ mol–1
Thus heat of reaction for the reaction
C2 H4 (g) + 2H (g) 
→ C2H6 (g)
–1
Comes out to be –124.6 kJ mol .

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 ,

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 THERMODYNAMICS–I 45

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
H H

H C C H
2C (s) + 3 H—H (g) 
→
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 
→ B
(Reactant) (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
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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

 ∂(ÄE ) =  ∂E B   ∂E  …(1.115)
 ∂T    −  A
 V  ∂ V
T  ∂T  V

But since  ∂E 
  = CV
 ∂T  V

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 46 PHYSICAL CHEMISTRY–II

 ∂(ÄE) =
We have
 ∂T  (CV )B − (CV )A …(1.116)
 V

or d (DE) = DCV dT …(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) = ∫ (∆CV ) dT
1 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,
 ∂(ÄH)  ∂HB   ∂H A 
 ∂T  =  ∂T  −  ∂T  …(1.120)
 P  P  P

As we know that the heat capacity at constant pressure,

 ∂H 
CP =  
 ∂T  P

 ∂(ÄH)
Therefore,  ∂T  = (C P )B − (CP )A = ∆C P
 P
or d(DH) = DCP dT (1.121)
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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 T2

∫ d (∆H) = ∫ (∆CP ) dT
T1 T1

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 THERMODYNAMICS–I 47

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

H2O (g) 
→ H2 (g) + 0
1
2 O2 (g); DH (291.15 K) = 241.75 kJ
For this reaction, DCP = å CP (product) – å CP (reactant)
1
= CP (H2) + 2 CP (O2) – CP (H2O)
1 –1 –1
= (28.33) + 2 29.12 – 33.56 JK mol
–1 –1
= 9.83 JK mol .
Therefore,
DCP + (T2 – T1) = 9.83 (70 – 18) = (9.83 × 52)
–1
= 511.16 J mol
–1
= 0.51116 kJ mol
0 0
Hence, 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)
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 ∂H   ∂V 
  = V − T  …(1.123)
 ∂P T  ∂T P

Eq. (1.123) can also be written as

 ∂( ∆H)   ∂ ( ∆V) 
 ∂P  = ∆V − T  ∂T  …(1.124)
 T P

where DV is the volume change of the reaction at pressure P.

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 48 PHYSICAL CHEMISTRY–II

ÄV = ∑ V (products) − ∑ V (reactants)
On differentiating with respect to T at constant P, we get
 ∂(∆V)   ∂V (products)   ∂V (reactants) 
 ∂T  =
 P
∑ 
 ∂T 
P
−∑ 
 ∂T 
P
…(1.125)

Further the coefficient of thermal expansion a is defined as

1  ∂V 
á =  
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)] …(1.127)
 P

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

or (ÄH)P2 = (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, V = nRT/P
RT
So that, ÄV = Än
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P
 ∂(∆V)  R
and thus,  ∂T  = Än P
 P

On substituting these values in Eq. (1.124), one gets

 ∂(∆H)  RT R
 ∂P  = Än P − T ∆n P = 0 …(1.130)
 T

Thus, heat of reaction is thus independent of the pressure.

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 THERMODYNAMICS–I 49

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 g–1 = Constant
(ii) PVg = 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.
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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.

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 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:

(i) A + B 
→ C 
→D

(ii) A+ B 
→ P 
→D
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:
P2
( ∆H) P2 = ( ∆H) P1 + ∫ [∆V − T∆(αV)] dP
P1
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