CHEMISTRY160

THERMODYNAMICS

It is the only physical theory of universal content concerning

which I am convinced that, within the framework of the

applicability of its basic concepts, it will never be overthrown.

Albert Einstein

After studying this Unit, you will

be able to

••

• explain the terms : system and

surroundings;

••

• discriminate between close,

open and isolated systems;

••

• explain internal energy, work

and heat;

••

• state first law of

thermodynamics and express

it mathematically;

••

• calculate energy changes as

work and heat contributions

in chemical systems;

••

• explain state functions: U, H.

••

• correlate ∆U and ∆H;

••

• measure experimentally ∆U

and ∆H;

••

• define standard states for ∆H;

••

• calculate enthalpy changes for

various types of reactions;

••

• state and apply Hess’s law of

constant heat summation;

••

• differentiate between extensive

and intensive properties;

••

• define spontaneous and non-

spontaneous processes;

••

• explain entropy as a

thermodynamic state function

and apply it for spontaneity;

••

• explain Gibbs energy change

(∆G); and

••

• establish relationship between

∆G and spontaneity, ∆G and

equilibrium constant.

Chemical energy stored by molecules can be released as heat

during chemical reactions when a fuel like methane, cooking

gas or coal burns in air. The chemical energy may also be

used to do mechanical work when a fuel burns in an engine

or to provide electrical energy through a galvanic cell like

dry cell. Thus, various forms of energy are interrelated and

under certain conditions, these may be transformed from

one form into another. The study of these energy

transformations forms the subject matter of thermodynamics.

The laws of thermodynamics deal with energy changes of

macroscopic systems involving a large number of molecules

rather than microscopic systems containing a few molecules.

Thermodynamics is not concerned about how and at what

rate these energy transformations are carried out, but is

based on initial and final states of a system undergoing the

change. Laws of thermodynamics apply only when a system

is in equilibrium or moves from one equilibrium state to

another equilibrium state. Macroscopic properties like

pressure and temperature do not change with time for a

system in equilibrium state. In this unit, we would like to

answer some of the important questions through

thermodynamics, like:

How do we determine the energy changes involved in a

chemical reaction/process? Will it occur or not?

What drives a chemical reaction/process?

To what extent do the chemical reactions proceed?

UNIT 6

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THERMODYNAMICS 161

6.1 THERMODYNAMIC TERMS

We are interested in chemical reactions and the

energy changes accompanying them. For this

we need to know certain thermodynamic

terms. These are discussed below.

6.1.1 The System and the Surroundings

A system in thermodynamics refers to that

part of universe in which observations are

made and remaining universe constitutes the

surroundings. The surroundings include

everything other than the system. System and

the surroundings together constitute the

universe .

The universe = The system + The surroundings

However, the entire universe other than

the system is not affected by the changes

taking place in the system. Therefore, for

all practical purposes, the surroundings

are that portion of the remaining universe

which can interact with the system.

Usually, the region of space in the

neighbourhood of the system constitutes

its surroundings.

For example, if we are studying the

reaction between two substances A and B

kept in a beaker, the beaker containing the

reaction mixture is the system and the room

where the beaker is kept is the surroundings

(Fig. 6.1).

be real or imaginary. The wall that separates

the system from the surroundings is called

boundary. This is designed to allow us to

control and keep track of all movements of

matter and energy in or out of the system.

6.1.2 Types of the System

We, further classify the systems according to

the movements of matter and energy in or out

of the system.

1. Open System

In an open system, there is exchange of energy

and matter between system and surroundings

[Fig. 6.2 (a)]. The presence of reactants in an

open beaker is an example of an open system

Here the boundary is an imaginary surface

enclosing the beaker and reactants.

2. Closed System

In a closed system, there is no exchange of

matter, but exchange of energy is possible

between system and the surroundings

[Fig. 6.2 (b)]. The presence of reactants in a

closed vessel made of conducting material e.g.,

copper or steel is an example of a closed

system.

Fig. 6.2 Open, closed and isolated systems.

Fig. 6.1 System and the surroundings

Note that the system may be defined by

physical boundaries, like beaker or test tube,

or the system may simply be defined by a set

of Cartesian coordinates specifying a

particular volume in space. It is necessary to

think of the system as separated from the

surroundings by some sort of wall which may

* We could have chosen only the reactants as system then walls of the beakers will act as boundary.

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CHEMISTRY162

3. Isolated System

In an isolated system, there is no exchange of

energy or matter between the system and the

surroundings [Fig. 6.2 (c)]. The presence of

reactants in a thermos flask or any other closed

insulated vessel is an example of an isolated

system.

6.1.3 The State of the System

The system must be described in order to make

any useful calculations by specifying

quantitatively each of the properties such as

its pressure (p), volume (V), and temperature

(T ) as well as the composition of the system.

We need to describe the system by specifying

it before and after the change. You would recall

from your Physics course that the state of a

system in mechanics is completely specified at

a given instant of time, by the position and

velocity of each mass point of the system. In

thermodynamics, a different and much simpler

concept of the state of a system is introduced.

It does not need detailed knowledge of motion

of each particle because, we deal with average

measurable properties of the system. We specify

the state of the system by state functions or

state variables.

The state of a thermodynamic system is

described by its measurable or macroscopic

(bulk) properties. We can describe the state of

a gas by quoting its pressure (p), volume (V),

temperature (T ), amount (n) etc. Variables like

p, V, T are called state variables or state

functions because their values depend only

on the state of the system and not on how it is

reached. In order to completely define the state

of a system it is not necessary to define all the

properties of the system; as only a certain

number of properties can be varied

independently. This number depends on the

nature of the system. Once these minimum

number of macroscopic properties are fixed,

others automatically have definite values.

The state of the surroundings can never

be completely specified; fortunately it is not

necessary to do so.

6.1.4 The Internal Energy as a State

Function

When we talk about our chemical system

losing or gaining energy, we need to introduce

a quantity which represents the total energy

of the system. It may be chemical, electrical,

mechanical or any other type of energy you

may think of, the sum of all these is the energy

of the system. In thermodynamics, we call it

the internal energy, U of the system, which may

change, when

•

• heat passes into or out of the system,

•

• work is done on or by the system,

••

• matter enters or leaves the system.

These systems are classified accordingly as

you have already studied in section 6.1.2.

(a) Work

Let us first examine a change in internal

energy by doing work. We take a system

containing some quantity of water in a thermos

flask or in an insulated beaker. This would not

allow exchange of heat between the system

and surroundings through its boundary and

we call this type of system as adiabatic. The

manner in which the state of such a system

may be changed will be called adiabatic

process. Adiabatic process is a process in

which there is no transfer of heat between the

system and surroundings. Here, the wall

separating the system and the surroundings

is called the adiabatic wall (Fig 6.3).

Let us bring the change in the internal

energy of the system by doing some work on

Fig. 6.3 An adiabatic system which does not

permit the transfer of heat through its

boundary.

it. Let us call the initial state of the system as

state A and its temperature as T

. Let the

internal energy of the system in state A be

called U

. We can change the state of the system

in two different ways.

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THERMODYNAMICS 163

One way: We do some mechanical work, say

1 kJ, by rotating a set of small paddles and

thereby churning water. Let the new state be

called B state and its temperature, as T

. It is

found that T

> T

and the change in

temperature, ∆T = T

–T

. Let the internal

energy of the system in state B be U

and the

change in internal energy, ∆U =U

– U

Second way: We now do an equal amount (i.e.,

1kJ) electrical work with the help of an

immersion rod and note down the temperature

change. We find that the change in temperature

is same as in the earlier case, say, T

– T

In fact, the experiments in the above

manner were done by J. P. Joule between

1840–50 and he was able to show that a given

amount of work done on the system, no matter

how it was done (irrespective of path) produced

the same change of state, as measured by the

change in the temperature of the system.

So, it seems appropriate to define a

quantity, the internal energy U, whose value

is characteristic of the state of a system,

whereby the adiabatic work, w

required to

bring about a change of state is equal to the

difference between the value of U in one state

and that in another state, ∆U i.e.,

2 1 ad

= = w

∆ −

U U U

Therefore, internal energy, U, of the system

is a state function.

By conventions of IUPAC in chemical

thermodynamics. The positive sign expresses

that w

is positive when work is done on the

system and the internal energy of system

increases. Similarly, if the work is done by the

system,w

will be negative because internal

energy of the system decreases.

Can you name some other familiar state

functions? Some of other familiar state

functions are V, p, and T. For example, if we

bring a change in temperature of the system

from 25°C to 35°C, the change in temperature

is 35°C–25°C = +10°C, whether we go straight

up to 35°C or we cool the system for a few

degrees, then take the system to the final

temperature. Thus, T is a state function and

the change in temperature is independent of

the route taken. Volume of water in a pond,

for example, is a state function, because

change in volume of its water is independent

of the route by which water is filled in the

pond, either by rain or by tubewell or by both.

(b) Heat

We can also change the internal energy of a

system by transfer of heat from the

surroundings to the system or vice-versa

without expenditure of work. This exchange

of energy, which is a result of temperature

difference is called heat, q. Let us consider

bringing about the same change in temperature

(the same initial and final states as before in

section 6.1.4 (a) by transfer of heat through

thermally conducting walls instead of

adiabatic walls (Fig. 6.4).

We take water at temperature, T

in a

container having thermally conducting walls,

say made up of copper and enclose it in a huge

heat reservoir at temperature, T

. The heat

absorbed by the system (water), q can be

measured in terms of temperature difference ,

– T

. In this case change in internal energy,

∆U= q, when no work is done at constant

volume.

By conventions of IUPAC in chemical

thermodynamics. The q is positive, when

heat is transferred from the surroundings to

the system and the internal energy of the

system increases and q is negative when

heat is transferred from system to the

surroundings resulting in decrease of the

internal energy of the system..

Earlier negative sign was assigned when the work is done on the system and positive sign when the work is done by the

system. This is still followed in physics books, although IUPAC has recommended the use of new sign convention.

Fig. 6.4 A system which allows heat transfer

through its boundary.

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Let us consider the general case in which a

change of state is brought about both by

doing work and by transfer of heat. We write

change in internal energy for this case as:

∆U = q + w (6.1)

For a given change in state, q and w can

vary depending on how the change is carried

out. However, q +w = ∆U will depend only on

initial and final state. It will be independent of

the way the change is carried out. If there is

no transfer of energy as heat or as work

(isolated system) i.e., if w = 0 and q = 0, then

∆ U = 0.

The equation 6.1 i.e., ∆U = q + w is

mathematical statement of the first law of

thermodynamics

, which states that

The energy of an isolated system is

constant.

It is commonly stated as the law of

conservation of energy i.e., energy can neither

be created nor be destroyed.

Note: There is considerable difference between

the character of the thermodynamic property

energy and that of a mechanical property such

as volume. We can specify an unambiguous

(absolute) value for volume of a system in a

particular state, but not the absolute value of

the internal energy. However, we can measure

only the changes in the internal energy, ∆U of

the system.

Problem 6.1

Express the change in internal energy of

a system when

(i) No heat is absorbed by the system

from the surroundings, but work (w)

is done on the system. What type of

wall does the system have ?

(ii) No work is done on the system, but

q amount of heat is taken out from

the system and given to the

surroundings. What type of wall does

the system have?

(iii) w amount of work is done by the

system and q amount of heat is

supplied to the system. What type of

system would it be?

Solution

(i) ∆ U = w

, wall is adiabatic

(ii) ∆ U = – q, thermally conducting walls

(iii) ∆ U = q – w, closed system.

6.2 APPLICATIONS

Many chemical reactions involve the generation

of gases capable of doing mechanical work or

the generation of heat. It is important for us to

quantify these changes and relate them to the

changes in the internal energy. Let us see how!

6.2.1 Work

First of all, let us concentrate on the nature of

work a system can do. We will consider only

mechanical work i.e., pressure-volume work.

For understanding pressure-volume

work, let us consider a cylinder which

contains one mole of an ideal gas fitted with a

frictionless piston. Total volume of the gas is

and pressure of the gas inside is p. If

external pressure is p

which is greater than

p, piston is moved inward till the pressure

inside becomes equal to p

. Let this change

Fig. 6.5(a) Work done on an ideal gas in a

cylinder when it is compressed by a

constant external pressure, p

(in single step) is equal to the shaded

area.

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THERMODYNAMICS 165

be achieved in a single step and the final

volume be V

. During this compression,

suppose piston moves a distance, l and is

cross-sectional area of the piston is A

[Fig. 6.5(a)].

then, volume change = l × A = ∆V = (V

– V

)

We also know,

force

pressure

area

Therefore, force on the piston = p

. A

If w is the work done on the system by

movement of the piston then

w = force × distance = p

. A .l

= p

. (–∆

V) = – p

∆

V = – p

– V

) (6.2)

The negative sign of this expression is

required to obtain conventional sign for w,

which will be positive. It indicates that in case

of compression work is done on the system.

Here (V

– V

) will be negative and negative

multiplied by negative will be positive. Hence

the sign obtained for the work will be positive.

If the pressure is not constant at every

stage of compression, but changes in number

of finite steps, work done on the gas will be

summed over all the steps and will be equal

p V

− ∑ ∆

[Fig. 6.5 (b)]

If the pressure is not constant but changes

during the process such that it is always

infinitesimally greater than the pressure of the

gas, then, at each stage of compression, the

volume decreases by an infinitesimal amount,

dV. In such a case we can calculate the work

done on the gas by the relation

w = −

∫

p dV

( 6.3)

Here, p

at each stage is equal to (p

+ dp) in

case of compression [Fig. 6.5(c)]. In an

expansion process under similar conditions,

the external pressure is always less than the

pressure of the system i.e., p

= (p

– dp). In

general case we can write, p

= (p

+ dp). Such

processes are called reversible processes.

A process or change is said to be

reversible, if a change is brought out in

such a way that the process could, at any

moment, be reversed by an infinitesimal

change. A reversible process proceeds

infinitely slowly by a series of equilibrium

states such that system and the

surroundings are always in near

equilibrium with each other. Processes

Fig. 6.5 (c) pV-plot when pressure is not constant

and changes in infinite steps

(reversible conditions) during

compression from initial volume, V

final volume, V

. Work done on the gas

is represented by the shaded area.

Fig. 6.5 (b) pV-plot when pressure is not constant

and changes in finite steps during

compression from initial volume, V

final volume, V

. Work done on the gas

is represented by the shaded area.

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other than reversible processes are known

as irreversible processes.

In chemistry, we face problems that can

be solved if we relate the work term to the

internal pressure of the system. We can

relate work to internal pressure of the system

under reversible conditions by writing

equation 6.3 as follows:

rev ex

p dV p dp dV

= − = − ±

∫ ∫

( )

Since dp × dV is very small we can write

rev

= −

∫

p V

(6.4)

Now, the pressure of the gas (p

which we

can write as p now) can be expressed in terms

of its volume through gas equation. For n mol

of an ideal gas i.e., pV =nRT

⇒ =p

Therefore, at constant temperature (isothermal

process),

rev

= − = −

∫

n nR RT

= – 2.303 nRT log

(6.5)

Free expansion: Expansion of a gas in

vacuum (p

= 0) is called free expansion. No

work is done during free expansion of an ideal

gas whether the process is reversible or

irreversible (equation 6.2 and 6.3).

Now, we can write equation 6.1 in number

of ways depending on the type of processes.

Let us substitute w = – p

∆V (eq. 6.2) in

equation 6.1, and we get

∆ ∆U q p V

= −

If a process is carried out at constant volume

(∆V = 0), then

∆U = q

the subscript V in q

denotes that heat is

supplied at constant volume.

Isothermal and free expansion of an

ideal gas

For isothermal (T = constant) expansion of an

ideal gas into vacuum ; w = 0 since p

= 0.

Also, Joule determined experimentally that

q = 0; therefore, ∆U = 0

Equation 6.1,

∆ = +

U q

can be

expressed for isothermal irreversible and

reversible changes as follows:

1. For isothermal irreversible change

q = – w = p

– V

)

2. For isothermal reversible change

q = – w = nRT ln

= 2.303 nRT log

3. For adiabatic change, q = 0,

∆U = w

Problem 6.2

Two litres of an ideal gas at a pressure of

10 atm expands isothermally at 25 °C into

a vacuum until its total volume is 10 litres.

How much heat is absorbed and how much

work is done in the expansion ?

Solution

We have q = – w = p

(10 – 2) = 0(8) = 0

No work is done; no heat is absorbed.

Problem 6.3

Consider the same expansion, but this

time against a constant external pressure

of 1 atm.

Solution

We have q = – w = p

(8) = 8 litre-atm

Problem 6.4

Consider the expansion given in problem

6.2, for 1 mol of an ideal gas conducted

reversibly.

Solution

We have q = – w = 2.303 nRT

log

= 2.303 × 1 × 0.8206 × 298 × log

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THERMODYNAMICS 167

= 2.303 x 0.8206 x 298 x log 5

= 2.303 x 0.8206 x 298 x 0.6990

= 393.66 L atm

Remember ∆H = q

, heat absorbed by the

system at constant pressure.

∆∆

∆H is negative for exothermic reactions

which evolve heat during the reaction and

∆

∆∆

∆H is positive for endothermic reactions

which absorb heat from the surroundings.

At constant volume (∆V = 0), ∆U = q

therefore equation 6.8 becomes

∆H = ∆U = q

The difference between ∆H and ∆U is not

usually significant for systems consisting of

only solids and / or liquids. Solids and liquids

do not suffer any significant volume changes

upon heating. The difference, however,

becomes significant when gases are involved.

Let us consider a reaction involving gases. If

is the total volume of the gaseous reactants,

is the total volume of the gaseous products,

is the number of moles of gaseous reactants

and n

is the number of moles of gaseous

products, all at constant pressure and

temperature, then using the ideal gas law, we

write,

= n

and pV

= n

Thus, pV

– pV

= n

RT – n

RT = (n

–n

)RT

or p (V

– V

) = (n

– n

) RT

or p ∆V = ∆n

RT (6.9)

Here, ∆n

refers to the number of moles of

gaseous products minus the number of moles

of gaseous reactants.

Substituting the value of p∆V from

equation 6.9 in equation 6.8, we get

∆H = ∆U + ∆n

RT (6.10)

The equation 6.10 is useful for calculating

∆H from ∆U and vice versa.

Problem 6.5

If water vapour is assumed to be a perfect

gas, molar enthalpy change for

vapourisation of 1 mol of water at 1bar

and 100°C is 41kJ mol

–1

. Calculate the

internal energy change, when

6.2.2 Enthalpy, H

(a) A Useful New State Function

We know that the heat absorbed at constant

volume is equal to change in the internal

energy i.e., ∆U = q

. But most of chemical

reactions are carried out not at constant

volume, but in flasks or test tubes under

constant atmospheric pressure. We need to

define another state function which may be

suitable under these conditions.

We may write equation (6.1) as

∆

U = q

– p

∆

V at constant pressure, where q

is heat absorbed by the system and –p∆V

represent expansion work done by the system.

Let us represent the initial state by

subscript 1 and final state by 2

We can rewrite the above equation as

–U

= q

– p (V

– V

)

On rearranging, we get

= (U

+ pV

) – (U

+ pV

) (6.6)

Now we can define another thermodynamic

function, the enthalpy H [Greek word

enthalpien, to warm or heat content] as :

H = U + pV (6.7)

so, equation (6.6) becomes

= H

– H

= ∆H

Although q is a path dependent function,

H is a state function because it depends on U,

p and V, all of which are state functions.

Therefore, ∆H is independent of path. Hence,

is also independent of path.

For finite changes at constant pressure, we

can write equation 6.7 as

∆H = ∆U + ∆pV

Since p is constant, we can write

∆H = ∆U + p∆V (6.8)

It is important to note that when heat is

absorbed by the system at constant pressure,

we are actually measuring changes in the

enthalpy.

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1 mol of water is vapourised at 1 bar

pressure and 100°C.

Solution

(i) The change

(

)

(

)

→

2 2

H O H O g

∆ = ∆ + ∆

H U n T

or ∆U = ∆H –

∆n T

, substituting the

values, we get

∆U = −

× ×

−

− −

41 00 1

8 3 373

1 1

kJ mol

J mol K K

= 41.00 kJ mol

-1

– 3.096 kJ mol

-1

= 37.904 kJ mol

–1

(b) Extensive and Intensive Properties

In thermodynamics, a distinction is made

between extensive properties and intensive

properties. An extensive property is a

property whose value depends on the

quantity or size of matter present in the

system. For example, mass, volume,

internal energy, enthalpy, heat capacity, etc.

are extensive properties.

Those properties which do not depend

on the quantity or size of matter present are

known as intensive properties. For

example temperature, density, pressure etc.

are intensive properties. A molar property,

, is the value of an extensive property χ of

the system for 1 mol of the substance. If n

is the amount of matter,

independent of the amount of matter. Other

examples are molar volume, V

and molar

heat capacity, C

. Let us understand the

distinction between extensive and intensive

properties by considering a gas enclosed in

a container of volume V and at temperature

T [Fig. 6.6(a)]. Let us make a partition such

that volume is halved, each part [Fig. 6.6

(b)] now has one half of the original volume,

, but the temperature will still remain the

same i.e., T. It is clear that volume is an

extensive property and temperature is an

intensive property.

In this sub-section, let us see how to measure

heat transferred to a system. This heat appears

as a rise in temperature of the system in case

of heat absorbed by the system.

The increase of temperature is proportional

to the heat transferred

q coeff T= × ∆

The magnitude of the coefficient depends

on the size, composition and nature of the

system. We can also write it as q = C ∆T

The coefficient, C is called the heat capacity.

Thus, we can measure the heat supplied

by monitoring the temperature rise, provided

we know the heat capacity.

When C is large, a given amount of heat

results in only a small temperature rise. Water

has a large heat capacity i.e., a lot of energy is

needed to raise its temperature.

C is directly proportional to amount of

substance. The molar heat capacity of a

substance, C













, is the heat capacity for

one mole of the substance and is the quantity

of heat needed to raise the temperature of

one mole by one degree celsius (or one

kelvin). Specific heat, also called specific heat

capacity is the quantity of heat required to

raise the temperature of one unit mass of a

Fig. 6.6(a) A gas at volume V and temperature T

Fig. 6.6 (b) Partition, each part having half the

volume of the gas

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THERMODYNAMICS 169

substance by one degree celsius (or one

kelvin). For finding out the heat, q, required

to raise the temperatures of a sample, we

multiply the specific heat of the substance,

c, by the mass m, and temperatures change,

∆T as

q c m T C T= × × =∆ ∆

(6.11)

(d) The Relationship between C

and C

for an Ideal Gas

At constant volume, the heat capacity, C is

denoted by C

and at constant pressure, this

is denoted by C

. Let us find the relationship

between the two.

We can write equation for heat, q

at constant volume as q

C T U

∆ ∆=

at constant pressure as q

C T H

∆ ∆=

The difference between C

and C

can be

derived for an ideal gas as:

For a mole of an ideal gas, ∆H = ∆U + ∆(pV

)

= ∆U + ∆(RT )

= ∆U + R∆T

∴ = +∆ ∆ ∆H U TR

(6.12)

On putting the values of ∆H and ∆U,

we have

C T C T T

p V

∆ ∆ ∆= + R

C C

p V

= + R

– C

= R (6.13)

6.3 MEASUREMENT OF

∆∆

∆U AND

∆∆

∆

∆H:

CALORIMETRY

We can measure energy changes associated

with chemical or physical processes by an

experimental technique called calorimetry. In

calorimetry, the process is carried out in a

vessel called calorimeter, which is immersed

in a known volume of a liquid. Knowing the

heat capacity of the liquid in which calorimeter

is immersed and the heat capacity of

calorimeter, it is possible to determine the heat

evolved in the process by measuring

temperature changes. Measurements are

made under two different conditions:

i) at constant volume, q

ii) at constant pressure, q

(a)

∆

∆∆

∆U Measurements

For chemical reactions, heat absorbed at

constant volume, is measured in a bomb

calorimeter (Fig. 6.7). Here, a steel vessel (the

bomb) is immersed in a water bath. The whole

device is called calorimeter. The steel vessel is

immersed in water bath to ensure that no heat

is lost to the surroundings. A combustible

substance is burnt in pure dioxygen supplied

Fig. 6.7 Bomb calorimeter

in the steel bomb. Heat evolved during the

reaction is transferred to the water around the

bomb and its temperature is monitored. Since

the bomb calorimeter is sealed, its volume does

not change i.e., the energy changes associated

with reactions are measured at constant

volume. Under these conditions, no work is

done as the reaction is carried out at constant

volume in the bomb calorimeter. Even for

reactions involving gases, there is no work

done as ∆V = 0. Temperature change of the

calorimeter produced by the completed

reaction is then converted to q

, by using the

known heat capacity of the calorimeter with

the help of equation 6.11.

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(b)

∆∆

∆

H Measurements

Measurement of heat change at constant pressure

(generally under atmospheric pressure) can be

done in a calorimeter shown in Fig. 6.8. We know

that

∆Η =

(at constant p) and, therefore, heat

absorbed or evolved, q

at constant pressure is

also called the heat of reaction or enthalpy of

reaction, ∆

In an exothermic reaction, heat is evolved,

and system loses heat to the surroundings.

Therefore, q

will be negative and ∆

H will also

be negative. Similarly in an endothermic

reaction, heat is absorbed, q

is positive and

∆

H will be positive.

Problem 6.6

1g of graphite is burnt in a bomb

calorimeter in excess of oxygen at 298 K

and 1 atmospheric pressure according to

the equation

C (graphite) + O

(g)

→

(g)

During the reaction, temperature rises

from 298 K to 299 K. If the heat capacity

Fig. 6.8 Calorimeter for measuring heat changes

at constant pressure (atmospheric

pressure).

of the bomb calorimeter is 20.7kJ/K,

what is the enthalpy change for the above

reaction at 298 K and 1 atm?

Solution

Suppose q is the quantity of heat from the

reaction mixture and C

is the heat capacity

of the calorimeter, then the quantity of heat

absorbed by the calorimeter.

q = C

× ∆T

Quantity of heat from the reaction will

have the same magnitude but opposite

sign because the heat lost by the system

(reaction mixture) is equal to the heat

gained by the calorimeter.

q C T

= kJ/K K

− × = − × −

= −

∆ 20 7 299 298

20 7

. ( )

(Here, negative sign indicates the

exothermic nature of the reaction)

Thus, ∆U for the combustion of the 1g of

graphite = – 20.7 kJK

–1

For combustion of 1 mol of graphite,

mol12 0 20 7

. .g kJ

−

× −

( )

= – 2.48 ×10

kJ mol

–1

, Since ∆ n

= 0,

∆ H = ∆ U = – 2.48 ×10

kJ mol

–1

6.4 ENTHALPY CHANGE,

∆

∆∆

∆

H OF A

REACTION – REACTION ENTHALPY

In a chemical reaction, reactants are converted

into products and is represented by,

Reactants → Products

The enthalpy change accompanying a

reaction is called the reaction enthalpy. The

enthalpy change of a chemical reaction, is given

by the symbol ∆

∆

H = (sum of enthalpies of products) – (sum

of enthalpies of reactants)

= −

∑∑

i products i reactants

H b H

(6.14)

Here symbol

∑

(sigma) is used for

summation and a

and b

are the stoichiometric

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THERMODYNAMICS 171

coefficients of the products and reactants

respectively in the balanced chemical

equation. For example, for the reaction

(g) + 2O

(g) → CO

(g) + 2H

O (l)

∆

r i

oducts i

reac ts

H a H b H= −

∑ ∑

Pr tan

= [H

(CO

,g) + 2H

O, l)]– [

(CH

, g)

+ 2H

, g)]

where H

is the molar enthalpy.

Enthalpy change is a very useful quantity.

Knowledge of this quantity is required when

one needs to plan the heating or cooling

required to maintain an industrial chemical

reaction at constant temperature. It is also

required to calculate temperature dependence

of equilibrium constant.

(a) Standard Enthalpy of Reactions

Enthalpy of a reaction depends on the

conditions under which a reaction is carried

out. It is, therefore, necessary that we must

specify some standard conditions. The

standard enthalpy of reaction is the

enthalpy change for a reaction when all

the participating substances are in their

standard states.

The standard state of a substance at a

specified temperature is its pure form at

1 bar. For example, the standard state of liquid

ethanol at 298 K is pure liquid ethanol at

1 bar; standard state of solid iron at 500 K is

pure iron at 1 bar. Usually data are taken

at 298 K.

Standard conditions are denoted by adding

the superscript

0 to the symbol

∆H, e.g., ∆H

(b) Enthalpy Changes during Phase

Transformations

Phase transformations also involve energy

changes. Ice, for example, requires heat for

melting. Normally this melting takes place at

constant pressure (atmospheric pressure) and

during phase change, temperature remains

constant (at 273 K).

O(s) → H

O(l); ∆

fus

= 6.00 kJ moI

–1

Here ∆

fus

is enthalpy of fusion in standard

state. If water freezes, then process is reversed

and equal amount of heat is given off to the

surroundings.

The enthalpy change that accompanies

melting of one mole of a solid substance

in standard state is called standard

enthalpy of fusion or molar enthalpy of

fusion,

∆

∆∆

∆

fus

Melting of a solid is endothermic, so all

enthalpies of fusion are positive. Water requires

Table 6.1 Standard Enthalpy Changes of Fusion and Vaporisation

and T

are melting and boiling points, respectively)

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heat for evaporation. At constant temperature

of its boiling point T

and at constant pressure:

O(l) → H

O(g); ∆

vap

= + 40.79 kJ moI

–1

∆

vap

is the standard enthalpy of vaporisation.

Amount of heat required to vaporize

one mole of a liquid at constant

temperature and under standard pressure

(1bar) is called its standard enthalpy of

vaporization or molar enthalpy of

vaporization,

∆∆

∆

vap

Sublimation is direct conversion of a solid

into its vapour. Solid CO

or ‘dry ice’ sublimes

at 195K with ∆

sub

=25.2 kJ mol

–1

;

naphthalene sublimes slowly and for this

∆

sub

= 73.0 kJ mol

–1

Standard enthalpy of sublimation,

∆

sub

is the change in enthalpy when one

mole of a solid substance sublimes at a

constant temperature and under standard

pressure (1bar).

The magnitude of the enthalpy change

depends on the strength of the intermolecular

interactions in the substance undergoing the

phase transfomations. For example, the strong

hydrogen bonds between water molecules hold

them tightly in liquid phase. For an organic

liquid, such as acetone, the intermolecular

dipole-dipole interactions are significantly

weaker. Thus, it requires less heat to vaporise

1 mol of acetone than it does to vaporize 1 mol

of water. Table 6.1 gives values of standard

enthalpy changes of fusion and vaporisation

for some substances.

Problem 6.7

A swimmer coming out from a pool is

covered with a film of water weighing

about 18g. How much heat must be

supplied to evaporate this water at

298 K ? Calculate the internal energy of

vaporisation at 298K.

∆

vap

for water

at 298K= 44.01kJ mol

–1

Solution

We can represent the process of

evaporation as

H O( ) H O(g)

1mol

vaporisation

1  →

No. of moles in 18 g H

O(l) is

18g

1mol

18 g mol

−

= =

Heat supplied to evaporate18g water at

298 K = n × ∆

vap

= (1 mol) × (44.01 kJ mol

–1

)

= 44.01 kJ

(assuming steam behaving as an ideal

gas).

∆ = ∆ − ∆ = ∆ − ∆

vap vap vap g

U H p V H n RT

V V

∆ ∆

∆

vap

H T

− =

−

− − − −

n R

JK mol K

44 01

1 8 314 298 10

1 1 3 1

( )( . )( )( )

kJ J

vvap

= −

44 01 2 48

41 5

. .

kJ kJ

Problem 6.8

Assuming the water vapour to be a perfect

gas, calculate the internal energy change

when 1 mol of water at 100°C and 1 bar

pressure is converted to ice at 0°C. Given

the enthalpy of fusion of ice is 6.00 kJ mol

heat capacity of water is 4.2 J/g°C

The change take place as follows:

Step - 1 1 mol H

O (l, 100°C) à 1

mol (l, 0°C) Enthalpy

change ∆H

Step - 2 1 mol H

O (l, 0°C) à 1 mol

O( S, 0°C) Enthalpy

change ∆H

Total enthalpy change will be -

∆H = ∆H

+ ∆H

∆H

= - (18 x 4.2 x 100) J mol

-1

= - 7560 J mol

-1

= - 7.56 k J mol

-1

∆

= - 6.00 kJ mol

-1

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THERMODYNAMICS 173

aggregation (also known as reference

states) is called Standard Molar Enthalpy

of Formation. Its symbol is

∆∆

∆

, where

the subscript ‘ f ’ indicates that one mole of

the compound in question has been formed

in its standard state from its elements in their

most stable states of aggregation. The reference

state of an element is its most stable state of

aggregation at 25°C and 1 bar pressure.

For example, the reference state of dihydrogen

is H

gas and those of dioxygen, carbon and

sulphur are O

gas, C

graphite

and S

rhombic

respectively. Some reactions with standard

molar enthalpies of formation are as follows.

Table 6.2 Standard Molar Enthalpies of Formation (

∆

∆∆

∆

) at 298K of a

Few Selected Substances

Therefore,

∆H = - 7.56 kJ mol

-1

(-6.00 kJ mol

-1

)

= -13.56 kJ mol

-1

There is negligible change in the volume

during the change form liquid to solid

state.

Therefore, p∆v = ∆ng RT = 0

∆H = ∆U = - 13.56kJ mol

-1

The standard enthalpy change for the

formation of one mole of a compound from

its elements in their most stable states of

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CHEMISTRY174

(g) + ½O

(g) → H

O(1);

∆

= –285.8 kJ mol

–1

C (graphite, s) + 2H

(g) → Ch

(g);

∆

= –74.81 kJ mol

–1

2C (graphite, s)+3H

(g)+ ½O

(g) → C

OH(1);

∆

= – 277.7kJ mol

–1

It is important to understand that a

standard molar enthalpy of formation, ∆

is just a special case of ∆

, where one mole

of a compound is formed from its constituent

elements, as in the above three equations,

where 1 mol of each, water, methane and

ethanol is formed. In contrast, the enthalpy

change for an exothermic reaction:

CaO(s) + CO

(g) → CaCo

(s);

∆

= – 178.3kJ mol

–1

is not an enthalpy of formation of calcium

carbonate, since calcium carbonate has been

formed from other compounds, and not from

its constituent elements. Also, for the reaction

given below, enthalpy change is not standard

enthalpy of formation, ∆

for HBr(g).

(g) + Br

(l) → 2HBr(g);

∆

= – 178.3kJ mol

–1

Here two moles, instead of one mole of the

product is formed from the elements, i.e., .

∆

= 2∆

Therefore, by dividing all coefficients in the

balanced equation by 2, expression for

enthalpy of formation of HBr (g) is written as

½H

(g) + ½Br

(1) → HBr(g);

∆

= – 36.4 kJ mol

–1

Standard enthalpies of formation of some

common substances are given in Table 6.2.

By convention, standard enthalpy for

formation, ∆

, of an element in reference

state, i.e., its most stable state of aggregation

is taken as zero.

Suppose, you are a chemical engineer and

want to know how much heat is required to

decompose calcium carbonate to lime and

carbon dioxide, with all the substances in their

standard state.

CaCO

(s) → CaO(s) + CO

(g); ∆

= ?

Here, we can make use of standard enthalpy

of formation and calculate the enthalpy

change for the reaction. The following general

equation can be used for the enthalpy change

calculation.

∆

∑

∆

(products)

–

∑

∆

(reactants)

(6.15)

where a and b represent the coefficients of the

products and reactants in the balanced

equation. Let us apply the above equation for

decomposition of calcium carbonate. Here,

coefficients ‘a’ and ‘b’ are 1 each. Therefore,

∆

= ∆

= [CaO(s)]+ ∆

[CO

(g)]

– ∆

= [CaCO

(s)]

=1 (–635.1 kJ mol

–1

) + 1(–393.5 kJ mol

–1

)

–1(–1206.9 kJ mol

–1

)

= 178.3 kJ mol

–1

Thus, the decomposition of CaCO

(s) is an

endothermic process and you have to heat it

for getting the desired products.

(d) Thermochemical Equations

A balanced chemical equation together with

the value of its ∆

H is called a thermochemical

equation. We specify the physical state

(alongwith allotropic state) of the substance in

an equation. For example:

OH(l) + 3O

(g) → 2CO

(g) + 3H

O(l);

∆

= – 1367 kJ mol

–1

The above equation describes the

combustion of liquid ethanol at constant

temperature and pressure. The negative sign

of enthalpy change indicates that this is an

exothermic reaction.

It would be necessary to remember the

following conventions regarding thermo-

chemical equations.

1. The coefficients in a balanced thermo-

chemical equation refer to the number of

moles (never molecules) of reactants and

products involved in the reaction.

2. The numerical value of ∆

refers to the

number of moles of substances specified

by an equation. Standard enthalpy change

∆

will have units as kJ mol

–1

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THERMODYNAMICS 175

To illustrate the concept, let us consider

the calculation of heat of reaction for the

following reaction :

Fe O s H g Fe s H O l

2 3 2 2

3 2 3

(

)

( )

→

(

)

(

)

From the Table (6.2) of standard enthalpy of

formation (∆

), we find :

∆

l) = –285.83 kJ mol

–1

;

∆

(Fe

,s) = – 824.2 kJ mol

–1

;

Also ∆

(Fe, s) = 0 and

∆

, g) = 0 as per convention

Then,

∆

= 3(–285.83 kJ mol

–1

)

– 1(– 824.2 kJ mol

–1

)

= (–857.5 + 824.2) kJ mol

–1

= –33.3 kJ mol

–1

Note that the coefficients used in these

calculations are pure numbers, which are

equal to the respective stoichiometric

coefficients. The unit for ∆

kJ mol

–1

, which means per mole of reaction.

Once we balance the chemical equation in a

particular way, as above, this defines the mole

of reaction. If we had balanced the equation

differently, for example,

2 3 2 2

Fe O s H g Fe s H O l

( )

→

( )

then this amount of reaction would be one

mole of reaction and ∆

would be

∆

(–285.83 kJ mol

–1

)

–

(–824.2 kJ mol

–1

)

= (– 428.7 + 412.1) kJ mol

–1

= –16.6 kJ mol

–1

= ½ ∆

It shows that enthalpy is an extensive quantity.

3. When a chemical equation is reversed, the

value of ∆

is reversed in sign. For

example

(g) + 3H

(g) → 2NH

(g);

∆

= – 91.8 kJ. mol

–1

2NH

(g) → N

(g) + 3H

(g);

∆

= + 91.8 kJ mol

–1

(e) Hess’s Law of Constant Heat

Summation

We know that enthalpy is a state function,

therefore the change in enthalpy is

independent of the path between initial state

(reactants) and final state (products). In other

words, enthalpy change for a reaction is the

same whether it occurs in one step or in a

series of steps. This may be stated as follows

in the form of Hess’s Law.

If a reaction takes place in several steps

then its standard reaction enthalpy is the

sum of the standard enthalpies of the

intermediate reactions into which the

overall reaction may be divided at the same

temperature.

Let us understand the importance of this

law with the help of an example.

Consider the enthalpy change for the

reaction

C (graphite,s) +

(g) → CO (g); ∆

= ?

Although CO(g) is the major product, some

gas is always produced in this reaction.

Therefore, we cannot measure enthalpy change

for the above reaction directly. However, if we

can find some other reactions involving related

species, it is possible to calculate the enthalpy

change for the above reaction.

Let us consider the following reactions:

C (graphite,s) + O

(g) → CO

(g);

∆

= – 393.5 kJ mol

–1

(i)

CO (g) +

(g) → CO

(g)

∆

= – 283.0 kJ mol

–1

(ii)

We can combine the above two reactions

in such a way so as to obtain the desired

reaction. To get one mole of CO(g) on the right,

we reverse equation (ii). In this, heat is

absorbed instead of being released, so we

change sign of ∆

value

(g) → CO (g) +

(g);

∆

= + 283.0 kJ mol

–1

(iii)

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Adding equation (i) and (iii), we get the

desired equation,

C graphite s O g CO g, ;

(

)

(

)

→

(

)

for which ∆

= (– 393.5 + 283.0)

= – 110.5 kJ mol

–1

In general, if enthalpy of an overall reaction

A→B along one route is ∆

H and ∆

, ∆

∆

..... representing enthalpies of reactions

leading to same product, B along another

route,then we have

∆

H = ∆

+ ∆

... (6.16)

It can be represented as:

6.5 ENTHALPIES FOR DIFFERENT TYPES

OF REACTIONS

It is convenient to give name to enthalpies

specifying the types of reactions.

(a) Standard Enthalpy of Combustion

(symbol :

∆∆

∆

)

Combustion reactions are exothermic in

nature. These are important in industry,

rocketry, and other walks of life. Standard

enthalpy of combustion is defined as the

enthalpy change per mole (or per unit amount)

of a substance, when it undergoes combustion

and all the reactants and products being in

their standard states at the specified

temperature.

Cooking gas in cylinders contains mostly

butane (C

). During complete combustion

of one mole of butane, 2658 kJ of heat is

released. We can write the thermochemical

reactions for this as:

C H g O g CO g H O

4 10 2 2 2

4 5 1( ) ( ) ( ) ( );+ → +

∆

= – 2658.0 kJ mol

–1

Similarly, combustion of glucose gives out

2802.0 kJ/mol of heat, for which the overall

equation is :

C H O g O g CO g H O

6 12 6 2 2 2

6 6 6 1( ) ( ) ( ) ( );+ → +

∆

= – 2802.0 kJ mol

–1

Our body also generates energy from food

by the same overall process as combustion,

although the final products are produced after

a series of complex bio-chemical reactions

involving enzymes.

∆H

∆

Problem 6.9

The combustion of one mole of benzene

takes place at 298 K and 1 atm. After

combustion, CO

(g) and H

O (1) are

produced and 3267.0 kJ of heat is

liberated. Calculate the standard

enthalpy of formation, ∆

of benzene.

Standard enthalpies of formation of

(g) and

H O(l)

are –393.5 kJ mol

–1

and – 285.83 kJ mol

–1

respectively.

Solution

The formation reaction of benezene is

given by :

6 3

2 6 6

C graphite H g C H l

( )

→

(

)

;

∆

= ? ... (i)

The enthalpy of combustion of 1 mol of

benzene is :

C H l O CO g H O l

6 6 2 2 2

6 3

( )

+ →

( )

;

∆

= – 3267 kJ mol

–1

... (ii)

The enthalpy of formation of 1 mol of

(g) :

C graphite O g CO g

( )

→

( )

2 2

;

∆

= – 393.5 kJ mol

–1

... (iii)

The enthalpy of formation of 1 mol of

O(l) is :

H g O g H O l

2 2 2

(

)

(

)

→

( )

;

∆

= – 285.83 kJ mol

–1

... (iv)

multiplying eqn. (iii) by 6 and eqn. (iv)

by 3 we get:

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THERMODYNAMICS 177

6 6 6

2 2

C graphite O g CO g

( )

→

( )

;

∆

= – 2361 kJ mol

–1

3 1

2 2 2

H g O g H O

( )

→

( )

;

∆

= – 857.49 kJ mol

–1

Summing up the above two equations :

6 3

2 2 2

C graphite H g O g CO g

H O l

( )

→

( )

;

∆

= – 3218.49 kJ mol

–1

... (v)

Reversing equation (ii);

6 3

2 2 6 6 2

CO g H O l C H l O

( )

→

( )

+ ;

∆

= – 3267.0 kJ mol

–1

... (vi)

Adding equations (v) and (vi), we get

6 3

2 6 6

C graphite H g C H l

( )

→

(

)

;

∆

= – 48.51 kJ mol

–1

... (iv)

In this case, the enthalpy of atomization is

same as the enthalpy of sublimation.

∆∆

∆

bond

)

Chemical reactions involve the breaking and

making of chemical bonds. Energy is required

to break a bond and energy is released when

a bond is formed. It is possible to relate heat

of reaction to changes in energy associated

with breaking and making of chemical bonds.

With reference to the enthalpy changes

associated with chemical bonds, two different

terms are used in thermodynamics.

(i) Bond dissociation enthalpy

(ii) Mean bond enthalpy

Let us discuss these terms with reference

to diatomic and polyatomic molecules.

Diatomic Molecules: Consider the following

process in which the bonds in one mole of

dihydrogen gas (H

) are broken:

(g) → 2H(g) ; ∆

H–H

= 435.0 kJ mol

–1

The enthalpy change involved in this process

is the bond dissociation enthalpy of H–H bond.

The bond dissociation enthalpy is the change

in enthalpy when one mole of covalent bonds

of a gaseous covalent compound is broken to

form products in the gas phase.

Note that it is the same as the enthalpy of

atomization of dihydrogen. This is true for all

diatomic molecules. For example:

(g) → 2Cl(g) ; ∆

Cl–Cl

= 242 kJ mol

–1

(g) → 2O(g) ; ∆

O=O

= 428 kJ mol

–1

In the case of polyatomic molecules, bond

dissociation enthalpy is different for different

bonds within the same molecule.

Polyatomic Molecules: Let us now consider

a polyatomic molecule like methane, CH

. The

overall thermochemical equation for its

atomization reaction is given below:

CH g C g H g

4( ) ( ) ( );→ +

∆

= 1665 kJ mol

–1

In methane, all the four C – H bonds are

identical in bond length and energy. However,

the energies required to break the individual

C – H bonds in each successive step differ :

(b) Enthalpy of Atomization

(symbol:

∆∆

∆

)

Consider the following example of atomization

of dihydrogen

(g) → 2H(g); ∆

= 435.0 kJ mol

–1

You can see that H atoms are formed by

breaking H–H bonds in dihydrogen. The

enthalpy change in this process is known as

enthalpy of atomization, ∆

. It is the

enthalpy change on breaking one mole of

bonds completely to obtain atoms in the gas

phase.

In case of diatomic molecules, like

dihydrogen (given above), the enthalpy of

atomization is also the bond dissociation

enthalpy. The other examples of enthalpy of

atomization can be

(g) → C(g) + 4H(g); ∆

= 1665 kJ mol

–1

Note that the products are only atoms of C

and H in gaseous phase. Now see the following

reaction:

Na(s) → Na(g) ; ∆

= 108.4 kJ mol

–1

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(g) → CH

(g)+H(g);∆

bond

= +427 kJ mol

–1

(g) → CH

(g)+H(g);∆

bond

= +439 kJ mol

–1

(g)

→ CH(g)+H(g);∆

bond

= +452 kJ mol

–1

CH(g) → C(g)+H(g);∆

bond

= +347 kJ mol

–1

Therefore,

(g) → C(g)+4H(g);∆

= 1665 kJ mol

–1

In such cases we use mean bond enthalpy

of C – H bond.

For example in CH

, ∆

C–H

is calculated as:

∆

C–H

= ¼ (∆

) = ¼ (1665 kJ mol

–1

)

= 416 kJ mol

–1

We find that mean C–H bond enthalpy in

methane is 416 kJ/mol. It has been found that

mean C–H bond enthalpies differ slightly from

compound to compound, as in

Cl,CH

, etc, but it does not differ

in a great deal*. Using Hess’s law, bond

enthalpies can be calculated. Bond enthalpy

values of some single and multiple bonds are

given in Table 6.3. The reaction enthalpies are

very important quantities as these arise from

the changes that accompany the breaking of

old bonds and formation of the new bonds.

We can predict enthalpy of a reaction in gas

phase, if we know different bond enthalpies.

The standard enthalpy of reaction, ∆

related to bond enthalpies of the reactants and

products in gas phase reactions as:

∆

−

∑

bond enthalpies

reactants

products

(6.17)**

This relationship is particularly more

useful when the required values of ∆

are

not available. The net enthalpy change of a

reaction is the amount of energy required to

break all the bonds in the reactant molecules

minus the amount of energy required to break

all the bonds in the product molecules.

Remember that this relationship is

H C N O F Si P S Cl Br I

435.8 414

389 464 569 293 318 339 431 368 297 H

347 293 351 439 289 264 259 330 276 238 C

159 201 272 - 209 - 201 243 - N

138 184 368 351 - 205 - 201 O

155 540 490 327 255 197 - F

176 213 226 360 289 213 Si

213 230 331 272 213 P

213 251 213 - S

243 218 209 Cl

192 180 Br

151 I

Table 6.3(a) Some Mean Single Bond Enthalpies in kJ mol

–1

at 298 K

N = N 418 C = C 611 O = O 498

N ≡ N 946 C ≡ C 837

C = N 615 C = O 741

C ≡ N 891 C ≡ O 1070

Table 6.3(b) Some Mean Multiple Bond Enthalpies in kJ mol

–1

at 298 K

If we use enthalpy of bond formation, (∆

bond

), which is the enthalpy change when one mole of a particular type of

bond is formed from gaseous atom, then ∆

∑

∆

bonds of products

–

∑

∆

bonds of reactants

Note that symbol used for bond dissociation enthalpy and mean bond enthalpy is the same.

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THERMODYNAMICS 179

approximate and is valid when all substances

(reactants and products) in the reaction are in

gaseous state.

(d) Lattice Enthalpy

The lattice enthalpy of an ionic compound is

the enthalpy change which occurs when one

mole of an ionic compound dissociates into its

ions in gaseous state.

Na Cl s Na g Cl g

+ − + −

( )

→ +

( )

( ) ;

∆

lattice

= +788 kJ mol

–1

Since it is impossible to determine lattice

enthalpies directly by experiment, we use an

indirect method where we construct an

enthalpy diagram called a Born-Haber Cycle

(Fig. 6.9).

Let us now calculate the lattice enthalpy

of Na

–

(s) by following steps given below :

Na s Na g( ) ( )→

, sublimation of sodium

metal, ∆

sub

= 108.4 kJ mol

–1

Na g Na g e g( ) ( ) ( )→ +

+ −1

, the ionization of

sodium atoms, ionization enthalpy

∆

= 496 kJ mol

–1

Cl g Cl g( ) ( )→

, the dissociation of

chlorine, the reaction enthalpy is half the

bond dissociation enthalpy.

∆

bond

= 121 kJ mol

–1

Cl g e g Cl g( ) ( ) ( )+ →

−1

electron gained by

chlorine atoms. The electron gain enthalpy,

∆

= – 348.6 kJ mol

–1

You have learnt about ionization enthalpy

and electron gain enthalpy in Unit 3. In

fact, these terms have been taken from

thermodynamics. Earlier terms, ionization

energy and electron affinity were in

practice in place of the above terms (see

the box for justification).

Fig. 6.9 Enthalpy diagram for lattice enthalpy

of NaCl

Na g Cl g Na Cl s

+ − + −

+ →( ) ( ) ( )

The sequence of steps is shown in Fig. 6.9,

and is known as a Born-Haber cycle. The

Ionization Energy and Electron Affinity

Ionization energy and electron affinity are

defined at absolute zero. At any other

temperature, heat capacities for the

reactants and the products have to be

taken into account. Enthalpies of reactions

for

M(g) → M

(g) + e

–

(for ionization)

M(g) + e

–

→ M

–

(g) (for electron gain)

at temperature, T is

∆

(T ) = ∆

(0) +

∫

∆

The value of C

for each species in the

above reaction is 5/2 R (C

= 3/2R)

So, ∆

= + 5/2 R (for ionization)

∆

= – 5/2 R (for electron gain)

Therefore,

∆

(ionization enthalpy)

= E

(ionization energy) + 5/2 RT

∆

(electron gain enthalpy)

= – A( electron affinity) – 5/2 RT

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CHEMISTRY180

The enthalpy of solution of AB(s), ∆

sol

, in

water is, therefore, determined by the selective

values of the lattice enthalpy,∆

lattice

and

enthalpy of hydration of ions, ∆

hyd

∆

sol

= ∆

lattice

+ ∆

hyd

For most of the ionic compounds, ∆

sol

positive and the dissociation process is

endothermic. Therefore the solubility of most

salts in water increases with rise of

temperature. If the lattice enthalpy is very

high, the dissolution of the compound may not

take place at all. Why do many fluorides tend

to be less soluble than the corresponding

chlorides? Estimates of the magnitudes of

enthalpy changes may be made by using tables

of bond energies (enthalpies) and lattice

energies (enthalpies).

(f) Enthalpy of Dilution

It is known that enthalpy of solution is the

enthalpy change associated with the addition

of a specified amount of solute to the specified

amount of solvent at a constant temperature

and pressure. This argument can be applied

to any solvent with slight modification.

Enthalpy change for dissolving one mole of

gaseous hydrogen chloride in 10 mol of water

can be represented by the following equation.

For convenience we will use the symbol aq. for

water

HCl(g) + 10 aq. → HCl.10 aq.

∆H = –69.01 kJ / mol

Let us consider the following set of enthalpy

changes:

(S-1) HCl(g) + 25 aq. → HCl.25 aq.

∆H = –72.03 kJ / mol

(S-2) HCl(g) + 40 aq. → HCl.40 aq.

∆H = –72.79 kJ / mol

(S-3) HCl(g) + ∞ aq.

→ HCl. ∞ aq.

∆H = –74.85 kJ / mol

The values of ∆H show general dependence

of the enthalpy of solution on amount of solvent.

As more and more solvent is used, the enthalpy

of solution approaches a limiting value, i.e, the

value in infinitely dilute solution. For

hydrochloric acid this value of ∆H is given

above in equation (S-3).

importance of the cycle is that, the sum of

the enthalpy changes round a cycle is

zero. Applying Hess’s law, we get,

∆

lattice

= 411.2 + 108.4 + 121 + 496 – 348.6

∆

lattice

= + 788kJ

for NaCl(s) d Na

(g) + Cl

–

(g)

Internal energy is smaller by 2RT ( because

∆n

= 2) and is equal to + 783 kJ mol

–1

Now we use the value of lattice enthalpy to

calculate enthalpy of solution from the

expression:

∆

sol

= ∆

lattice

+ ∆

hyd

For one mole of NaCl(s),

lattice enthalpy = + 788 kJ mol

–1

and ∆

hyd

= – 784 kJ mol

–1

( from the

literature)

∆

sol

= + 788 kJ mol

–1

– 784 kJ mol

–1

= + 4 kJ mol

–1

The dissolution of NaCl(s) is accompanied

by very little heat change.

(e) Enthalpy of Solution (symbol :

∆∆

∆

sol

)

Enthalpy of solution of a substance is the

enthalpy change when one mole of it dissolves

in a specified amount of solvent. The enthalpy

of solution at infinite dilution is the enthalpy

change observed on dissolving the substance

in an infinite amount of solvent when the

interactions between the ions (or solute

molecules) are negligible.

When an ionic compound dissolves in a

solvent, the ions leave their ordered positions on

the crystal lattice. These are now more free in

solution. But solvation of these ions (hydration

in case solvent is water) also occurs at the same

time. This is shown diagrammatically, for an

ionic compound, AB (s)

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THERMODYNAMICS 181

If we subtract the first equation (equation

S-1) from the second equation (equation S-2)

in the above set of equations, we obtain–

HCl.25 aq. + 15 aq. → HCl.40 aq.

∆H = [ –72.79 – (–72.03)] kJ / mol

= – 0.76 kJ / mol

This value (–0.76kJ/mol) of ∆H is enthalpy

of dilution. It is the heat withdrawn from the

surroundings when additional solvent is

added to the solution. The enthalpy of dilution

of a solution is dependent on the original

concentration of the solution and the amount

of solvent added.

6.6 SPONTANEITY

The first law of thermodynamics tells us about

the relationship between the heat absorbed

and the work performed on or by a system. It

puts no restrictions on the direction of heat

flow. However, the flow of heat is unidirectional

from higher temperature to lower

temperature. In fact, all naturally occurring

processes whether chemical or physical will

tend to proceed spontaneously in one

direction only. For example, a gas expanding

to fill the available volume, burning carbon

in dioxygen giving carbon dioxide.

But heat will not flow from colder body to

warmer body on its own, the gas in a container

will not spontaneously contract into one

corner or carbon dioxide will not form carbon

and dioxygen spontaneously. These and many

other spontaneously occurring changes show

unidirectional change. We may ask ‘what is the

driving force of spontaneously occurring

changes ? What determines the direction of a

spontaneous change ? In this section, we shall

establish some criterion for these processes

whether these will take place or not.

Let us first understand what do we mean

by spontaneous reaction or change ? You may

think by your common observation that

spontaneous reaction is one which occurs

immediately when contact is made between

the reactants. Take the case of combination of

hydrogen and oxygen. These gases may be

mixed at room temperature and left for many

years without observing any perceptible

change. Although the reaction is taking place

between them, it is at an extremely slow rate.

It is still called spontaneous reaction. So

spontaneity means ‘having the potential to

proceed without the assistance of external

agency’. However, it does not tell about the

rate of the reaction or process. Another aspect

of spontaneous reaction or process, as we see

is that these cannot reverse their direction on

their own. We may summarise it as follows:

A spontaneous process is an

irreversible process and may only be

reversed by some external agency.

(a) Is Decrease in Enthalpy a Criterion

for Spontaneity ?

If we examine the phenomenon like flow of

water down hill or fall of a stone on to the

ground, we find that there is a net decrease in

potential energy in the direction of change. By

analogy, we may be tempted to state that a

chemical reaction is spontaneous in a given

direction, because decrease in energy has

taken place, as in the case of exothermic

reactions. For example:

(g) +

(g) = NH

(g) ;

∆

= – 46.1 kJ mol

–1

(g) +

(g) = HCl (g) ;

∆

= – 92.32 kJ mol

–1

(g) +

(g) → H

O(l) ;

∆

= –285.8 kJ mol

–1

The decrease in enthalpy in passing from

reactants to products may be shown for any

exothermic reaction on an enthalpy diagram

as shown in Fig. 6.10(a).

Thus, the postulate that driving force for a

chemical reaction may be due to decrease in

energy sounds ‘reasonable’ as the basis of

evidence so far !

Now let us examine the following reactions:

(g) + O

(g) → NO

(g);

∆

= +33.2 kJ mol

–1

C(graphite, s) + 2 S(l) → CS

(l);

∆

= +128.5 kJ mol

–1

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CHEMISTRY182

These reactions though endothermic, are

spontaneous. The increase in enthalpy may

be represented on an enthalpy diagram as

shown in Fig. 6.10(b).

Fig. 6.10 (a) Enthalpy diagram for exothermic

reactions

Fig. 6.10 (b) Enthalpy diagram for endothermic

reactions

Fig. 6.11 Diffusion of two gases

respectively and separated by a movable

partition [Fig. 6.11 (a)]. When the partition is

withdrawn [Fig.6.11( b)], the gases begin to

diffuse into each other and after a period of

time, diffusion will be complete.

Let us examine the process. Before

partition, if we were to pick up the gas

molecules from left container, we would be

sure that these will be molecules of gas A and

similarly if we were to pick up the gas

molecules from right container, we would be

sure that these will be molecules of gas B. But,

if we were to pick up molecules from container

when partition is removed, we are not sure

whether the molecules picked are of gas A or

gas B. We say that the system has become less

predictable or more chaotic.

We may now formulate another postulate:

in an isolated system, there is always a

tendency for the systems’ energy to become

more disordered or chaotic and this could be

a criterion for spontaneous change !

At this point, we introduce another

thermodynamic function, entropy denoted as

S. The above mentioned disorder is the

manifestation of entropy. To form a mental

Therefore, it becomes obvious that while

decrease in enthalpy may be a contributory

factor for spontaneity, but it is not true for all

cases.

(b) Entropy and Spontaneity

Then, what drives the spontaneous process in

a given direction ? Let us examine such a case

in which ∆H = 0 i.e., there is no change in

enthalpy, but still the process is spontaneous.

Let us consider diffusion of two gases into

each other in a closed container which is

isolated from the surroundings as shown in

Fig. 6.11.

The two gases, say, gas A and gas B are

represented by black dots and white dots

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THERMODYNAMICS 183

picture, one can think of entropy as a measure

of the degree of randomness or disorder in the

system. The greater the disorder in an isolated

system, the higher is the entropy. As far as a

chemical reaction is concerned, this entropy

change can be attributed to rearrangement of

atoms or ions from one pattern in the reactants

to another (in the products). If the structure

of the products is very much disordered than

that of the reactants, there will be a resultant

increase in entropy. The change in entropy

accompanying a chemical reaction may be

estimated qualitatively by a consideration of

the structures of the species taking part in the

reaction. Decrease of regularity in structure

would mean increase in entropy. For a given

substance, the crystalline solid state is the

state of lowest entropy (most ordered), The

gaseous state is state of highest entropy.

Now let us try to quantify entropy. One way

to calculate the degree of disorder or chaotic

distribution of energy among molecules would

be through statistical method which is beyond

the scope of this treatment. Other way would

be to relate this process to the heat involved in

a process which would make entropy a

thermodynamic concept. Entropy, like any

other thermodynamic property such as

internal energy U and enthalpy H is a state

function and ∆S is independent of path.

Whenever heat is added to the system, it

increases molecular motions causing

increased randomness in the system. Thus

heat (q) has randomising influence on the

system. Can we then equate ∆S with q ? Wait !

Experience suggests us that the distribution

of heat also depends on the temperature at

which heat is added to the system. A system

at higher temperature has greater randomness

in it than one at lower temperature. Thus,

temperature is the measure of average

chaotic motion of particles in the system.

Heat added to a system at lower temperature

causes greater randomness than when the

same quantity of heat is added to it at higher

temperature. This suggests that the entropy

change is inversely proportional to the

temperature. ∆

S is related with q and T for a

reversible reaction as :

∆S =

rev

(6.18)

The total entropy change ( ∆S

total

) for the

system and surroundings of a spontaneous

process is given by

∆ ∆ ∆S S S

total system surr

= + > 0

(6.19)

When a system is in equilibrium, the

entropy is maximum, and the change in

entropy, ∆

S = 0.

We can say that entropy for a spontaneous

process increases till it reaches maximum and

at equilibrium the change in entropy is zero.

Since entropy is a state property, we can

calculate the change in entropy of a reversible

process by

∆

sys

sys rev,

We find that both for reversible and

irreversible expansion for an ideal gas, under

isothermal conditions, ∆U = 0, but ∆S

total

i.e.,

∆ ∆S S

sys surr

is not zero for irreversible

process. Thus, ∆U does not discriminate

between reversible and irreversible process,

whereas ∆S does.

Problem 6.10

Predict in which of the following, entropy

increases/decreases :

(i) A liquid crystallizes into a solid.

(ii) Temperature of a crystalline solid is

raised from 0 K to 115 K.

iii NaHCO s Na CO s

CO g H O g

(

)

(

)

→

(

)

(

)

(

)

3 2 3

2 2

(iv)

H g H g

( )

→

( )

Solution

(i) After freezing, the molecules attain an

ordered state and therefore, entropy

decreases.

(ii) At 0 K, the contituent particles are

static and entropy is minimum. If

temperature is raised to 115 K, these

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CHEMISTRY184

begin to move and oscillate about their

equilibrium positions in the lattice and

system becomes more disordered,

therefore entropy increases.

(iii) Reactant, NaHCO

is a solid and it

has low entropy. Among products

there are one solid and two gases.

Therefore, the products represent a

condition of higher entropy.

(iv) Here one molecule gives two atoms

i.e., number of particles increases

leading to more disordered state.

Two moles of H atoms have higher

entropy than one mole of dihydrogen

molecule.

Problem 6.11

For oxidation of iron,

4 3 2

2 2 3

Fe s O g Fe O s

(

)

( )

→

(

)

entropy change is – 549.4 JK

–1

mol

–1

298 K. Inspite of negative entropy change

of this reaction, why is the reaction

spontaneous?

(∆

for this reaction is

–1648 × 10

J mol

–1

)

Solution

One decides the spontaneity of a reaction

by considering

∆ ∆ ∆S S S

total sys surr

( )

. For calculating

∆S

surr

, we have to consider the heat

absorbed by the surroundings which is

equal to – ∆

. .

. At temperature T, entropy

change of the surroundings is

= −

− ×

( )

−

1648 10

298

3 1

J mol

= 5530 JK

–1

mol

–1

Thus, total entropy change for this

reaction

∆

r total

S = +

−

( )

5530

549 4

1 1

JK mol

– –

= 4980.6 JK

–1

mol

–1

This shows that the above reaction is

spontaneous.

We have seen that for a system, it is the total

entropy change, ∆S

total

which decides the

spontaneity of the process. But most of the

chemical reactions fall into the category of

either closed systems or open systems.

Therefore, for most of the chemical reactions

there are changes in both enthalpy and

entropy. It is clear from the discussion in

previous sections that neither decrease in

enthalpy nor increase in entropy alone can

determine the direction of spontaneous change

for these systems.

For this purpose, we define a new

thermodynamic function the Gibbs energy or

Gibbs function, G, as

G = H – TS (6.20)

Gibbs function, G is an extensive property

and a state function.

The change in Gibbs energy for the system,

∆G

sys

can be written as

∆ ∆ ∆ ∆G H T S S T

sys sys sys sys

= − −

At constant temperature,

∆ =

∴ −∆ ∆ ∆G H T S

sys sys sys

Usually the subscript ‘system’ is dropped

and we simply write this equation as

∆ ∆ ∆G H T S= −

(6.21)

Thus, Gibbs energy change = enthalpy

change – temperature × entropy change, and

is referred to as the Gibbs equation, one of the

most important equations in chemistry. Here,

we have considered both terms together for

spontaneity: energy (in terms of ∆H) and

entropy (∆S, a measure of disorder) as

indicated earlier. Dimensionally if we analyse,

we find that ∆G has units of energy because,

both ∆H and the T∆S are energy terms, since

T∆

S = (K) (J/K) = J.

Now let us consider how ∆G is related to

reaction spontaneity.

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THERMODYNAMICS 185

We know,

∆S

total

= ∆S

sys

+ ∆

surr

If the system is in thermal equilibrium with

the surrounding, then the temperature of the

surrounding is same as that of the system.

Also, increase in enthalpy of the surrounding

is equal to decrease in the enthalpy of the

system.

Therefore, entropy change of

surroundings,

∆

surr

sys

= −

∆ ∆

∆

S S

total sys

= + −













sys

Rearranging the above equation:

T∆S

total

= T∆S

sys

– ∆

sys

For spontaneous process,

∆S

total

> 0

, so

T∆S

sys

– ∆H

sys

> Ο

⇒ − −

( )

>∆ ∆H T S

sys sys

Using equation 6.21, the above equation can

be written as

–∆G > O

∆ ∆ ∆G H T S= − < 0

(6.22)

∆H

sys

is the enthalpy change of a reaction,

T∆S

sys

is the energy which is not available to

do useful work. So ∆G is the net energy

available to do useful work and is thus a

measure of the ‘free energy’. For this reason, it

is also known as the free energy of the reaction.

∆G

gives a criteria for spontaneity at

constant pressure and temperature.

(i) If ∆G

is negative (< 0), the process is

spontaneous.

(ii) If ∆G

is positive (> 0), the process is non

spontaneous.

Note : If a reaction has a positive enthalpy

change and positive entropy change, it can be

spontaneous when T∆

S is large enough to

outweigh ∆H. This can happen in two ways;

(a) The positive entropy change of the system

can be ‘small’ in which case T must be large.

(b) The positive entropy change of the system

can be ’large’, in which case T may be small.

The former is one of the reasons why reactions

are often carried out at high temperature.

Table 6.4 summarises the effect of temperature

on spontaneity of reactions.

(d) Entropy and Second Law of

Thermodynamics

We know that for an isolated system the

change in energy remains constant. Therefore,

increase in entropy in such systems is the

natural direction of a spontaneous change.

This, in fact is the second law of

thermodynamics. Like first law of

thermodynamics, second law can also be

stated in several ways. The second law of

thermodynamics explains why spontaneous

exothermic reactions are so common. In

exothermic reactions heat released by the

reaction increases the disorder of the

surroundings and overall entropy change is

positive which makes the reaction

spontaneous.

(e) Absolute Entropy and Third Law of

Thermodynamics

Molecules of a substance may move in a

straight line in any direction, they may spin

like a top and the bonds in the molecules may

stretch and compress. These motions of the

molecule are called translational, rotational

and vibrational motion respectively. When

temperature of the system rises, these motions

become more vigorous and entropy increases.

On the other hand when temperature is

lowered, the entropy decreases. The entropy

of any pure crystalline substance

approaches zero as the temperature

approaches absolute zero. This is called

third law of thermodynamics. This is so

because there is perfect order in a crystal at

absolute zero. The statement is confined to

pure crystalline solids because theoretical

arguments and practical evidences have

shown that entropy of solutions and super

cooled liquids is not zero at 0 K. The

importance of the third law lies in the fact that

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CHEMISTRY186

Table 6.4 Effect of Temperature on Spontaneity of Reactions

∆

Description*

– + – Reaction spontaneous at all temperatures

– – – (at low T ) Reaction spontaneous at low temperature

– – + (at high T ) Reaction nonspontaneous at high temperature

+ + + (at low T ) Reaction nonspontaneous at low temperature

+ + – (at high T ) Reaction spontaneous at high temperature

+ – + (at all T ) Reaction nonspontaneous at all temperatures

The term low temperature and high temperature are relative. For a particular reaction, high temperature could even

mean room temperature.

it permits the calculation of absolute values

of entropy of pure substance from thermal

data alone. For a pure substance, this can be

done by summing

rev

increments from 0 K

to 298 K. Standard entropies can be used to

calculate standard entropy changes by a

Hess’s law type of calculation.

6.7 GIBBS ENERGY CHANGE AND

EQUILIBRIUM

We have seen how a knowledge of the sign and

magnitude of the free energy change of a

chemical reaction allows:

(i) Prediction of the spontaneity of the

chemical reaction.

(ii) Prediction of the useful work that could

be extracted from it.

So far we have considered free energy

changes in irreversible reactions. Let us now

examine the free energy changes in reversible

reactions.

‘Reversible’ under strict thermodynamic

sense is a special way of carrying out a

process such that system is at all times in

perfect equilibrium with its surroundings.

When applied to a chemical reaction, the

term ‘reversible’ indicates that a given

reaction can proceed in either direction

simultaneously, so that a dynamic

equilibrium is set up. This means that the

reactions in both the directions should

proceed with a decrease in free energy, which

seems impossible. It is possible only if at

equilibrium the free energy of the system is

minimum. If it is not, the system would

spontaneously change to configuration of

lower free energy.

So, the criterion for equilibrium

A + B l C + D ; is

∆

G = 0

Gibbs energy for a reaction in which all

reactants and products are in standard state,

∆

is related to the equilibrium constant of

the reaction as follows:

0 = ∆

+ RT ln K

or ∆

= – RT ln K

or ∆

= – 2.303 R

T log K (6.23)

We also know that

(6.24)

For strongly endothermic reactions, the

value of ∆

may be large and positive. In

such a case, value of K will be much smaller

than 1 and the reaction is unlikely to form

much product. In case of exothermic

reactions, ∆

is large and negative, and ∆

is likely to be large and negative too. In such

cases, K will be much larger than 1. We may

expect strongly exothermic reactions to have

a large K, and hence can go to near

completion. ∆

also depends upon ∆

, if

the changes in the entropy of reaction is also

taken into account, the value of K or extent

of chemical reaction will also be affected,

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THERMODYNAMICS 187

depending upon whether ∆

is positive or

negative.

Using equation (6.24),

(i) It is possible to obtain an estimate of ∆G

from the measurement of ∆H

and ∆S

and then calculate K at any temperature

for economic yields of the products.

(ii) If K is measured directly in the

laboratory, value of ∆G

at any other

temperature can be calculated.

Using equation (6.24),

– .

. .

13 6 10

2 303 8 314 298

( )

(

)

J mol

JK mol K

–1

–1 –1

= 2.38

Hence K = antilog 2.38 = 2.4 × 10

Problem 6.14

At 60°C, dinitrogen tetroxide is 50

per cent dissociated. Calculate the

standard free energy change at this

temperature and at one atmosphere.

Solution

(g) 2NO

(g)

If N

is 50% dissociated, the mole

fraction of both the substances is given

x x

N O

2 4

1 0 5

2 0 5

1 0 5

−

p p

N O NO

2 4 2

0 5

1 5

= × =

atm,

atm.

The equilibrium constant K

is given by

N O

2 4

( )

1 5

1 5 0 5

( . ) ( . )

= 1.33 atm.

Since

∆

= –RT ln K

∆

= (– 8.314 JK

–1

mol

–1

) × (333 K)

× (2.303) × (0.1239)

= – 763.8 kJ mol

–1

Problem 6.12

Calculate ∆

for conversion of oxygen

to ozone, 3/2 O

(g) → O

(g) at 298 K. if K

for this conversion is 2.47 × 10

–29

Solution

We know ∆

= – 2.303 RT log K

and

R = 8.314 JK

–1

mol

–1

Therefore, ∆

– 2.303 (8.314 J K

–1

mol

–1

)

× (298 K) (log 2.47 × 10

–29

)

= 163000 J mol

–1

= 163 kJ mol

–1

Problem 6.13

Find out the value of equilibrium constant

for the following reaction at 298 K.

Standard Gibbs energy change, ∆

the given temperature is –13.6 kJ mol

–1

Solution

We know, log K =

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CHEMISTRY188

SUMMARY

Thermodynamics deals with energy changes in chemical or physical processes and enables us

to study these changes quantitatively and to make useful predictions. For these purposes, we

divide the universe into the system and the surroundings. Chemical or physical processes

lead to evolution or absorption of heat (q), part of which may be converted into work (w). These

quantities are related through the first law of thermodynamics via ∆U = q + w. ∆U, change in

internal energy, depends on initial and final states only and is a state function, whereas q and

w depend on the path and are not the state functions. We follow sign conventions of q and w by

giving the positive sign to these quantities when these are added to the system. We can measure

the transfer of heat from one system to another which causes the change in temperature. The

magnitude of rise in temperature depends on the heat capacity (C) of a substance. Therefore,

heat absorbed or evolved is q = C∆T. Work can be measured by w = –p

∆V, in case of expansion

of gases. Under reversible process, we can put p

= p

for infinitesimal changes in the volume

making w

rev

= – p dV. In this condition, we can use gas equation, pV = nRT.

At constant volume, w = 0, then ∆U = q

, heat transfer at constant volume. But in study of

chemical reactions, we usually have constant pressure. We define another state function

enthalpy. Enthalpy change, ∆H = ∆U + ∆n

RT, can be found directly from the heat changes at

constant pressure, ∆H = q

There are varieties of enthalpy changes. Changes of phase such as melting, vaporization

and sublimation usually occur at constant temperature and can be characterized by enthalpy

changes which are always positive. Enthalpy of formation, combustion and other enthalpy

changes can be calculated using Hess’s law. Enthalpy change for chemical reactions can be

determined by

∆ ∆ ∆

r i f i f

H a H b H=

( )

−

( )

∑∑

products reactions

and in gaseous state by

∆

= Σ bond enthalpies of the reactants – Σ bond enthalpies of the products

First law of thermodynamics does not guide us about the direction of chemical reactions

i.e., what is the driving force of a chemical reaction. For isolated systems,

∆U = 0. We define another state function, S, entropy for this purpose. Entropy is a measure of

disorder or randomness. For a spontaneous change, total entropy change is positive. Therefore,

for an isolated system, ∆U = 0, ∆S > 0, so entropy change distinguishes a spontaneous change,

while energy change does not. Entropy changes can be measured by the equation

∆S =

rev

for a reversible process.

rev

is independent of path.

Chemical reactions are generally carried at constant pressure, so we define another state

function Gibbs energy, G, which is related to entropy and enthalpy changes of the system by

the equation:

∆

G = ∆

H – T ∆

For a spontaneous change, ∆G

sys

< 0 and at equilibrium, ∆G

sys

= 0.

Standard Gibbs energy change is related to equilibrium constant by

∆

= – RT ln K.

K can be calculated from this equation, if we know ∆

which can be found from

. Temperature is an important factor in the equation. Many reactions which

are non-spontaneous at low temperature, are made spontaneous at high temperature for systems

having positive entropy of reaction.

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THERMODYNAMICS 189

EXERCISES

6.1 Choose the correct answer. A thermodynamic state function is a quantity

(i) used to determine heat changes

(ii) whose value is independent of path

(iii) used to determine pressure volume work

(iv) whose value depends on temperature only.

6.2 For the process to occur under adiabatic conditions, the correct condition is:

(i) ∆T = 0

(ii) ∆p = 0

(iii) q = 0

(iv) w = 0

6.3 The enthalpies of all elements in their standard states are:

(i) unity

(ii) zero

(iii) < 0

(iv) different for each element

6.4 ∆U

of combustion of methane is – X kJ mol

–1

. The value of ∆H

(i) = ∆U

(ii) > ∆U

(iii) < ∆U

(iv) = 0

6.5 The enthalpy of combustion of methane, graphite and dihydrogen at 298 K

are, –890.3 kJ mol

–1

–393.5 kJ mol

–1

, and –285.8 kJ mol

–1

respectively. Enthalpy

of formation of CH

(g) will be

(i) –74.8 kJ mol

–1

(ii) –52.27 kJ mol

–1

(iii) +74.8 kJ mol

–1

(iv) +52.26 kJ mol

–1

6.6 A reaction, A + B → C + D + q is found to have a positive entropy change. The

reaction will be

(i) possible at high temperature

(ii) possible only at low temperature

(iii) not possible at any temperature

(v) possible at any temperature

6.7 In a process, 701 J of heat is absorbed by a system and 394 J of work is done

by the system. What is the change in internal energy for the process?

6.8 The reaction of cyanamide, NH

CN (s), with dioxygen was carried out in a bomb

calorimeter, and ∆U was found to be –742.7 kJ mol

–1

at 298 K. Calculate enthalpy

change for the reaction at 298 K.

CN(g) +

(g) → N

(g) + CO

(g) + H

O(l)

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CHEMISTRY190

6.9 Calculate the number of kJ of heat necessary to raise the temperature of

60.0 g of aluminium from 35°C to 55°C. Molar heat capacity of Al is 24 J mol

–1

6.10 Calculate the enthalpy change on freezing of 1.0 mol of water at10.0°C to ice at

–10.0°C. ∆

fus

H = 6.03 kJ mol

–1

at 0°C.

O(l)] = 75.3 J mol

–1

O(s)] = 36.8 J mol

–1

6.11 Enthalpy of combustion of carbon to CO

is –393.5 kJ mol

–1

. Calculate the heat

released upon formation of 35.2 g of CO

from carbon and dioxygen gas.

6.12 Enthalpies of formation of CO(g), CO

(g), N

O(g) and N

(g) are –110, – 393, 81

and 9.7 kJ mol

–1

respectively. Find the value of ∆

H for the reaction:

(g) + 3CO(g) → N

O(g) + 3CO

(g)

6.13 Given

(g) + 3H

(g) → 2NH

(g) ; ∆

= –92.4 kJ mol

–1

What is the standard enthalpy of formation of NH

gas?

6.14 Calculate the standard enthalpy of formation of CH

OH(l) from the following

data:

OH (l) +

(g) → CO

(g) + 2H

O(l) ; ∆

= –726 kJ mol

–1

C(graphite) + O

(g) → CO

(g) ; ∆

= –393 kJ mol

–1

(g) +

(g) → H

O(l) ; ∆

= –286 kJ mol

–1

6.15 Calculate the enthalpy change for the process

CCl

(g) → C(g) + 4 Cl(g)

and calculate bond enthalpy of C – Cl in CCl

(g).

∆

vap

(CCl

) = 30.5 kJ mol

–1

∆

(CCl

) = –135.5 kJ mol

–1

∆

–1

, where ∆

is enthalpy of atomisation

∆

(Cl

) = 242 kJ mol

–1

6.16 For an isolated system, ∆U = 0, what will be ∆S ?

6.17 For the reaction at 298 K,

2A + B → C

∆H = 400 kJ mol

–1

and ∆S = 0.2 kJ K

–1

mol

–1

At what temperature will the reaction become spontaneous considering ∆H

and ∆S to be constant over the temperature range.

6.18 For the reaction,

2 Cl(g) → Cl

(g), what are the signs of ∆H and ∆S ?

6.19 For the reaction

2 A(g) + B(g) → 2D(g)

∆U

= –10.5 kJ and ∆S

= –44.1 JK

–1

Calculate ∆G

for the reaction, and predict whether the reaction may occur

spontaneously.

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THERMODYNAMICS 191

6.20 The equilibrium constant for a reaction is 10. What will be the value of ∆G

R = 8.314 JK

–1

mol

–1

, T = 300 K.

6.21 Comment on the thermodynamic stability of NO(g), given

(g) +

(g) → NO(g) ; ∆

= 90 kJ mol

–1

NO(g) +

(g) → NO

(g) : ∆

= –74 kJ mol

–1

6.22 Calculate the entropy change in surroundings when 1.00 mol of H

O(l) is formed

under standard conditions. ∆

= –286 kJ mol

–1

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