The words you are searching are inside this book. To get more targeted content, please make full-text search by clicking here.

Free Flip-Book Chemistry Class 11th by Study Innovations. 534 Pages

Discover the best professional documents and content resources in AnyFlip Document Base.
Search
Published by rajusingh79, 2019-08-09 08:38:11

Free Flip-Book Chemistry Class 11th by Study Innovations. 534 Pages

Free Flip-Book Chemistry Class 11th by Study Innovations. 534 Pages

Keywords: IIT JEE study material, NEET Study Material, JEE mains Study Material, JEE advanced study material, AIIMS Study Material, IIT JEE Foundation study material, NEET Foundation study material, CBSE Study Material, Test Series, Question Bank, ICSE Study Material, School Exams study material, board exams study material, XII board exams Study Material, X board exams Study Material, Study Material, JEE mains, JEE advanced, Video Lectures, Study Innovations, online tuition, home tuition, online tutors, coaching & tutorials for English, Mathematics, Science, Physics, Chemistry, Biology.

(3) Macroscopic Properties of the System : Thermodynamics deals with matter in terms of bulk (large
number of chemical species) behaviour. The properties of the system which arise from the bulk behaviour of matter
are called macroscopic properties. The common examples of macroscopic properties are pressure, volume,
temperature, surface tension, viscosity, density, refractive index, etc.

The macroscopic properties can be sub – divided into two types

(i) Intensive properties : The properties which do not depend upon the quantity of matter present in the
system or size of the system are called intensive properties. Pressure, temperature, density, specific heat, surface
tension, refractive index, viscosity, melting point, boiling point, volume per mole, concentration etc. are the
examples of intensive properties of the system.

(ii) Extensive properties : The properties whose magnitude depends upon the quantity of matter present in
the system are called extensive properties. Total mass, volume, internal energy, enthalpy, entropy etc. are the
well known examples of extensive properties. These properties are additive in nature.

Note : Any extensive property if expressed as per mole or per gram becomes an intensive property. For

example, mass and volume are extensive properties, but density and specific volume, i.e. the mass per unit volume

and volume per unit mass respectively are intensive properties. Similarly, heat capacity is an extensive property but

specific heat is an intensive property.

(4) State of a system and State Variables : The state of a system means the condition of existence of
the system when its macroscopic properties have definite values. If any of the macroscopic properties of the system

changes, the state of the system is also said to change. Thus, the state of the system is fixed by its macroscopic properties.

Macroscopic properties which determine the state of a system are referred to as state variables or state

functions or thermodynamic parameters. The change in the state properties depends only upon the initial

and final states of the system, but is independent of the manner in which the change has been brought about.

In other words, the state properties do not depend upon a path followed.

Following are the state functions that are commonly used to describe the state of the system

(i) Pressure (P) (ii) Temperature (T) (iii) Volume (V) (iv) Internal energy (E)

(v) Enthalpy (H) (vi) Entropy (S) (vii) Free energy (G) (viii) Number of moles (n)

(5) Thermodynamic equilibrium : “A system is said to have attained a state of thermodynamic equilibrium
when it shows no further tendency to change its property with time”.

The criterion for thermodynamic equilibrium requires that the following three types of equilibrium exist

simultaneously in a system.

(i) Chemical Equilibrium : A system in which the composition of the system remains fixed and definite.

(ii) Mechanical Equilibrium : No chemical work is done between different parts of the system or between

the system and surrounding. It can be achieved by keeping pressure constant.

(iii) Thermal Equilibrium : Temperature remains constant i.e. no flow of heat between system and

surrounding.

(6) Thermodynamic process : When the thermodynamic system changes from one state to another, the
operation is called a process. The various types of the processes are

(i) Isothermal process : The process is termed isothermal if Isobaric

temperature remains fixed, i.e., operation is done at constant temperature. Pressure Isothermal
This can be achieved by placing the system in a constant temperature bath, Isochoric Adiabatic
i.e., thermostat. For an isothermal process dT = 0, i.e., heat is exchanged

with the surroundings and the system is not thermally isolated. Volume
Graphical representation of four
(ii) Adiabatic process : If a process is carried out under such
condition that no exchange of heat takes place between the system and basic thermodynamic process


surroundings, the process is termed adiabatic. The system is thermally isolated, i.e., dQ = 0. This can be done by
keeping the system in an insulated container, i.e., thermos flask. In adiabatic process, the temperature of the system
varies.

(iii) Isobaric process : The process is known as isobaric in which the pressure remains constant throughout
the change i.e., dP = 0.

(iv) Isochoric process : The process is termed as isochoric in which volume remains constant throughout
the change, i.e., dV = 0.

(v) Cyclic process : When a system undergoes a number of different processes and finally return to its initial
state, it is termed cyclic process. For a cyclic process dE = 0 and dH = 0.

(vi) Reversible process : A process which occurs infinitesimally slowly, i.e. opposing force is infinitesimally
smaller than driving force and when infinitesimal increase in the opposing force can reverse the process, it is said to
be reversible process.

(vii) Irreversible process : When the process occurs from initial to final state in single step in finite time and
cannot be reversed, it is termed an irreversible process. Amount of entropy increases in irreversible process.

Irreversible processes are spontaneous in nature. All natural processes are irreversible in nature

Difference between reversible and irreversible process

Reversible process than the Irreversible process

It is an ideal process and takes infinite time. It is a spontaneous process and takes finite time.
The driving force is infinitesimally greater The driving force is much greater than the opposing

opposing force. force.
It is in equilibrium at all stages. Equilibrium exists in the initial and final stages only.
Obtained work is maximum. Obtained work is not maximum
It is difficult to realise in practice. It can be performed in practice.

Internal energy, Heat and work.

(1) Internal energy (E) : “Every system having some quantity of matter is associated with a definite amount
of energy. This energy is known as internal energy.” The exact value of this energy is not known as it includes all
types of energies of molecules constituting the given mass of matter such as translational vibrational, rotational, the
kinetic and potential energy of the nuclei and electrons within the individual molecules and the manner in which the
molecules are linked together, etc.

E = E + E + E + E + E + ......translational
rotational vibrational bonding electronic

(i) Characteristics of internal energy

(a) Internal energy of a system is an extensive property.

(b) Internal energy is a state property.

(c) The change in the internal energy does not depend on the path by which the final state is reached.

(d) There is no change in internal energy in a cyclic process.

(e) The internal energy of an ideal gas is a function of temperature only.

(f) Internal energy of a system depends upon the quantity of substance, its chemical nature, temperature,
pressure and volume.

(g) The units of E are ergs (in CGS) or joules (in SI)

1 Joule = 107 ergs.

(ii) Change in internal energy ( ∆E ) : It is neither possible nor necessary to calculate the absolute value of
internal energy of a system. In thermodynamics, one is concerned only with energy change which occurs when the


system moves from one state to another. Let ∆E be the difference of energy of the initial state (Ein ) and the final
state (Ef ) , then, ∆E = E f − Ein ; ∆E is positive if E f > Ein and negative if Ef < Ein .

(2) Heat (q) and Work (w) : The energy of a system may increase or decrease in several ways but two
common ways are heat and work.

Heat is a form of energy. It flows from one system to another because of the difference in temperature. Heat
flows from higher temperature to lower temperature. Therefore, it is regarded as energy on the move.

Work is said to be performed if the point of application of force is displaced in the direction of the force. It is
equal to the force multiplied by the displacement (distance through which the force acts).

There are three main types of work which we generally come across. These are, Gravitational work, electrical
work and mechanical work. Gravitational work is said to be done when a body is raised through a certain height (h)
against the gravitational field (g).

Electrical work is important in systems where reaction takes place between ions whereas mechanical work is
performed when a system changes its volume in the presence of external pressure. Mechanical work is important
specially in systems that contain gases. This is also known as pressure - volume work.

Note : Heat is a random form of energy while work is an organised form of energy.

(i) Units of Heat and Work : The heat changes are measured in calories (cal), Kilo calories (kcal), joules (J)
or kilo joules (kJ). These are related as, 1 cal = 4.184 J; 1 kcal = 4.184 kJ

The S.I. unit of heat is joule (J) or kilojoule. The Joule (J) is equal to Newton – metre (1 J= 1 Nm).

Work is measured in terms of ergs or joules. The S.I. unit of work is Joule.

1 Joule = 107 ergs = 0.2390 cal.

1 cal > 1 joule > 1 erg

(ii) Sign conventions for heat and work : It is very important to understand the sign conventions for q

and w. The signs of w or q are related to the internal q(+ve) q(–ve) w (+ve) w (–ve)
energy change. When w or q is positive, it means that

energy has been supplied to the system as work or as

heat. The internal energy of the system in such a case

increases. System System System System

When w or q is negative, it means that energy Heat absorbed Heat given off Work done on Work done by
has left the system as work or as heat. The internal the system the system
energy of the system in such a case decreases.

Thus, to summarise,

Heat absorbed by the system = q positive ; Heat evolved by the system = q negative
Work done on the system = w positive ; Work done by the system = w negative.

Zeroth law of thermodynamics.

This law was formulated after the first and second laws of thermodynamics have been enunciated. This law
forms the basis of concept of temperature. This law can be stated as follows,

“If a system A is in thermal equilibrium with a system C and if B is also in thermal equilibrium with system C,
then A and B are in thermal equilibrium with each other whatever the composition of the system.”

A⇌C B⇌ C


A⇌B

First law of Thermodynamics.

First law of thermodynamics was proposed by Helmholtz and Robert Mayer. This law is also known as
law of conservation of energy. It states that,

“Energy can neither be created nor destroyed although it can be converted from one form into another.”

(1) Justification for the law : The first law of thermodynamics has no theoretical proof. This law is based
on human experience and has not yet been violated. The following observations justify the validity of this law

(i) The total energy of an isolated system remains constant although it can undergo a change from one form
to another.

(ii) It is not possible to construct a perpetual machine which can do work without the expenditure of energy, If
the law were not true, it would have been possible to construct such a machine.

(iii) James Joule (1850) conducted a large number of experiments regarding the conversion of work into
heat energy. He concluded that for every 4.183 joule of work done on the system, one calorie of heat is produced.
He also pointed out that the same amount of work done always produces same amount of heat irrespective of how
the work is done.

(iv) Energy is conserved in chemical reactions also. For example, the electrical energy equivalent to 286.4 kJ
mol-1 of energy is consumed when one mole of water decomposes into gaseous hydrogen and oxygen. On the other
hand, the same amount of energy in the form of heat is liberated when one mole of liquid water is produced from

gases hydrogen and oxygen.

H 2O(l) + 286.4 kJ → H 2 (g) + 1 O2 (g) ; H 2 (g) + 1 O2 (g) → H 2O(l) + 286.4 kJ
2 2

These examples justify that energy is always conserved though it may change its form.

(2) Mathematical expression for the law : The internal energy of a system can be changed in two ways
(i) By allowing heat to flow into the system or out of the system.

(ii) By doing work on the system or by the system.

Let us consider a system whose internal energy is E1 . Now, if the system absorbs q amount of heat, then the
internal energy of the system increases and becomes E1 + q .

If work (w) is done on the system, then its internal energy further increases and becomes E2 . Thus,

E2 = E1 + q + w or E2 − E1 = q + w or ∆E = q + w

i.e. (Change in internal energy) = (Heat added to the system) + (Work done on the system)

If a system does work (w) on the surroundings, its internal energy decreases. In this case, work is taken as
negative (–w). Now, q is the amount of heat added to the system and w is the work done by the system, then
change in internal energy becomes, ∆E = q + (−w) = q − w

i.e. (Change in internal energy) = (Heat added to the system) − (Work done by the system)

The relationship between internal energy, work and heat is a mathematical statement of first law of
thermodynamics.

(3) Some useful conclusions drawn from the law : ∆E = q + w


(i) When a system undergoes a change ∆E = 0 , i.e., there is no increase or decrease in the internal energy of
the system, the first law of thermodynamics reduces to

0 = q + w or q = −w

(heat absorbed from surroundings = work done by the system)
or w = −q

(heat given to surroundings = work done on the system)
(ii) If no work is done, w = 0 and the first law reduces to

∆E = q

i.e. increase in internal energy of the system is equal to the heat absorbed by the system or decrease in
internal energy of the system is equal to heat lost by the system.

(iii) If there is no exchange of heat between the system and surroundings, q = 0 , the first law reduces to

∆E = w
It shows if work is done on the system, its internal energy will increase or if work is done by the system its
internal energy will decrease. This occurs in an adiabatic process.
(iv) In case of gaseous system, if a gas expands against the constant external pressure P, let the volume
change be ∆V . The mechanical work done by the gas is equal to − P × ∆V .
Substituting this value in ∆E = q + w ; ∆E = q − P∆V

When, ∆V = 0 , ∆E = q or qv

The symbol qv indicates the heat change at constant volume.
(4) Limitations of the law : The first law of thermodynamics states that when one form of energy
disappears, an equivalent amount of another form of energy is produced. But it is silent about the extent to
which such conversion can take place. The first law of thermodynamics has some other limitations also.
(i) It puts no restriction on the direction of flow of heat. But, heat flows spontaneously from a higher to a lower
temperature.
(ii) It does not tell whether a specified change can occur spontaneously or not.
(iii) It does not tell that heat energy cannot be completely converted into an equivalent amount of work
without producing some changes elsewhere.

Enthalpy and Enthalpy change.

Heat content of a system at constant pressure is called enthalpy denoted by ‘H’.

From first law of thermodynamics, q = E + PV ……….(i)

Heat change at constant pressure can be given as

∆q = ∆E + P∆V ……….(ii)

At constant pressure heat can be replaced at enthalpy.

∆H = ∆E + P∆V ………..(iii)

Constant pressures are common in chemistry as most of the reactions are carried out in open vessels.

At constant volume, ∆V = 0 ; thus equation (ii) can be written as,

∆qv = ∆E
∴∆H = Heat change or heat of reaction (in chemical process) at constant pressure

∆E = Heat change or heat of reaction at constant volume.


(1) In case of solids and liquids participating in a reaction,
∆H = ∆E(P∆V ≈ 0)

(2) Difference between ∆H and ∆E is significant when gases are involved in chemical reaction.
∆H = ∆E + P∆V
∆H = ∆E + ∆nRT
P∆V = ∆nRT

Here, ∆n = Number of gaseous moles of products – Number of gaseous moles of reactants.
Using the above relation we can interrelate heats of reaction at constant pressure and at constant volume.

Specific and Molar heat capacity.

(1) Specific heat (or specific heat capacity) of a substance is the quantity of heat (in calories, joules, kcal, or
kilo joules) required to raise the temperature of 1g of that substance through 1o C . It can be measured at constant
pressure (c p ) and at constant volume (cv ).

(2) Molar heat capacity of a substance is the quantity of heat required to raise the temperature of 1 mole of
the substance by 1o C .

∴ Molar heat capacity = Specific heat capacity × Molecular weight, i.e.,
Cv = cv × M and Cp = c p × M .

(3) Since gases on heating show considerable tendency towards expansion if heated under constant pressure
conditions, an additional energy has to be supplied for raising its temperature by 1o C relative to that required
under constant volume conditions, i.e.,

Cp > Cv or Cp = Cv + Work done on expansion, P∆V(= R)

where, Cp = molar heat capacity at constant pressure; Cv = molar heat capacity at constant volume.

Note :  Cp and Cv for solids and liquids are practically equal. However, they differ considerable in case

of gas because appreciable change in volume takes place with temperature.

(4) Some useful relations of Cp and Cv
(i) Cp − Cv = R = 2 calories = 8.314 J

(ii) Cv = 3 R (for monoatomic gas) and Cv = 3 + x (for di and polyatomic gas), where x varies from gas to gas.
2 2

(iii) Cp =γ (Ratio of molar capacities)
Cv

(iv) For monoatomic gas Cv = 3 calories whereas, Cp = Cv + R = 5calories

Cp 5 R
Cv 2
(v) For monoatomic gas, (γ ) = = 3 = 1.66
2
R

Cp 7 R
Cv 2
(vi) For diatomic gas (γ ) = = 5 = 1.40
2
R


(vii) For triatomic gas (γ ) = Cp = 8R = 1.33
Cv 6R

Values of Molar heat capacities of some gases,

Gas Cp Cv Cp– Cv Cp/Cv= γ Atomicity
He 5 3.01 1.99 1.661 1
N2 6.95 4.96 1.99 1.4 2
O2
CO2 6.82 4.83 1.99 1.4 2
H2S 8.75 6.71 2.04 1.30 3

8.62 6.53 2.09 1.32 3

Expansion of an Ideal gas.

(1) Isothermal Expansion : For an isothermal expansion, temperature remains fixed i.e. ∆E = 0 ( ∆E of
ideal gases depends only on temperature)

According to first law of thermodynamics,
∆E = q + w

∴q = −w

This shows that in isothermal expansion, the work is done by the system at the expense of heat absorbed.

Since for isothermal process, ∆E and ∆T are zero respectively, hence, ∆H = 0

(i) Work done in reversible isothermal expansion :

Consider an ideal gas enclosed in a cylinder fitted with a Pext Pext – dP
weightless and frictionless piston. The cylinder is not insulated. dV

The external pressure, Pext is equal to pressure of the gas, Pgas .

Let it be P.

Pext = Pgas = P Pgas Pgas
If the external pressure is decreased by an infinitesimal

amount dP, the gas will expand by an infinitesimal volume, dV.

As a result of expansion, the pressure of the gas within the cylinder falls to Pgas − dP , i.e., it becomes again equal to

the external pressure and, thus, the piston comes to rest. Such a process is repeated for a number of times, i.e., in

each step the gas expands by a volume dV.

Since the system is in thermal equilibrium with the surroundings, the infinitesimally small cooling produced
due to expansion is balanced by the absorption of heat from the surroundings and the temperature remains
constant throughout the expansion.

The work done by the gas in each step of expansion can be given as,

dw = −(Pext − dP)dV = −Pext .dV = −PdV
dP.dV, the product of two infinitesimal quantities, is neglected.

The total amount of work done by the isothermal reversible expansion of the ideal gas from volume V1 to

volume V2 is, given as,

w = −nRT log e V2 or w = −2.303nRT log 10 V2
V1 V1

At constant temperature, according to Boyle’s law,


P1V1 = P2V2 or V2 = P1 So, w = −2.303nRT log 10 P1
V1 P2 P2

Isothermal compression work of an ideal gas may be derived similarly and it has exactly the same value with
positive sign.

w compression = 2.303nRT log V1 = 2.303nRT log P2
V2 P1

(ii) Work done in irreversible isothermal expansion : Two types of irreversible isothermal expansions
are observed, i.e., (a) Free expansion and (b) Intermediate expansion. In free expansion, the external pressure is
zero, i.e., work done is zero when gas expands in vacuum. In intermediate expansion, the external pressure is less
than gas pressure. So, the work done when volume changes from V1 to V2 is given by

∫w = −PV2 × dV = −Pext (V2 − V1 )

V1 ext

Since Pext is less than the pressure of the gas, the work done during intermediate expansion is numerically
less than the work done during reversible isothermal expansion in which Pext is almost equal to Pgas .

Note : The work done by the system always depends upon the external pressure. The higher the value of

Pext , the more work is done by the gas. As Pext cannot be more than Pgas , otherwise compression

will occur, thus the largest value of Pext can be equal to Pgas . Under this condition when expansion

occurs, the maximum work is done by the gas on the surroundings.

(2) Adiabatic Expansion : In adiabatic expansion, no heat is allowed to enter or leave the system, hence,
q = 0.

According to first law of thermodynamics,
∆E = q + w

∴ ∆E = w

work is done by the gas during expansion at the expense of internal energy. In expansion, ∆E decreases
while in compression ∆E increases.

The molar specific heat capacity at constant volume of an ideal gas is given by

Cv =  dE  or dE = Cv .dT
 dT 
v

and for finite change ∆E = Cv∆T So, w = ∆E = Cv ∆T

The value of ∆T depends upon the process whether it is reversible or irreversible.

(i) Reversible adiabatic expansion : The following relationships are followed by an ideal gas under

reversible adiabatic expansion.

PV γ = constant
where, P = External pressure, V = Volume

γ = Cp
Cv

where, Cp = molar specific heat capacity at constant pressure, Cv = molar specific heat capacity at constant

volume.


 T1 γ =  P1 γ −1 =  P2 1−γ
T2 P2 P1

knowing γ , P1, P2 and initial temperature T1 , the final temperature T2 can be evaluated.

(ii) Irreversible adiabatic expansion : In free expansion, the external pressure is zero, i.e, work done is
zero. Accordingly, ∆E which is equal to w is also zero. If ∆E is zero, ∆T should be zero. Thus, in free expansion
(adiabatically), ∆T = 0 , ∆E = 0 , w = 0 and ∆H = 0 .

In intermediate expansion, the volume changes from V1 to V2 against external pressure, Pext .

w= −Pext (V2 − V1) = − Pext  RT2 − RT1  = − Pext  T2 P1 − T1 P2  × R
P2 P1 P1 P2

or w = Cv (T2 − T1) = − RPext  T2 P1 − T1 P2 
P1 P2

Example 1. The work done in ergs for the reversible expansion of one mole of an ideal gas from a volume of 10 litres to

Solution:(b) 20 litres at 25o C is [CMC Vellore 1991]
Example 2.
Solution:(c) (a) 2.303 × 298 × 0.082log 2 (b) − 298 × 107 × 8.31 × 2.303 log 2
Example 3.
(c) 2.303 × 298 × 0.082 log 0.5 (d) 2.303 × 298 × 2 log 2
Solution: (a)
w = −2.303 nRT log V2 = −2.303 × 1 × 8.31 × 107 × 298 log 20 = − 298 × 107 × 8.31× 2.303 log 2
Example 4. V1 10
Solution: (c)
Example 5. An ideal gas expands from 10 −3 m3 to 10 −2 m3 at 300 K against a constant pressure of 105 Nm−2 . The work
Solution:(a)
done is

(a) − 10 3 kJ (b) 10 2 kJ (c) 0.9 kJ (d) −900 kJ

w = −P(V2 − V1) = −105 (10 −2 − 10 −3 ) = −105 × 10 −2 (1 − 0.1) = −0.9 × 10 3 J = −0.9 kJ

At 27°C one mole of an ideal gas is compressed isothermally and reversibly from a pressure of 2 atm to 10

atm. The values of ∆E and q are (R = 2) [BHU 2001]

(a) 0, –965.84 cal (b) – 965.84 cal, + 965.84 cal

(c) + 865.58 cal, –865.58 cal (d) –865.58 cal, –865.58 cal

w = −2.303nRT log P1 = 2.303 × 1 × 2 × 300 log 2 = 965.84
P2 10

At constant temperature ∆E = 0

∆E = q + w ; q = −w = −965.84cal

The work done by a system is 8 joule, when 40 joule heat is supplied to it, what is the increase in internal

energy of system [BHU 2001]

(a) 25 J (b) 30 J (c) 32 J (d) 28 J

q = 40J

w = −8 J (work done by the system)
∆E = q + w = 40 − 8 = 32J

One mole of an ideal gas is allowed to expand reversibly and adiabatically from a temperature of 27o C. If the

work done during the process is 3 kJ, the final temperature will be equal to (Cv = 20 JK −1 ) [KCET 2000]

(a) 150 K (b) 100 K (c) 26.85 K (d) 295 K

Work done during adiabatic expansion = Cv (T2 − T1 )

−3000 = 20(T2 − 300) or T2 = 150 K .


Joule – Thomson effect.

The phenomenon of producing lowering of temperature when a gas is made to expand adiabatically from a

region of high pressure into a region of low pressure, is known as Porous plug
Joule-Thomson effect.
DT ∝ DP V2
During this process, enthalpy remains constant. It is, V1

therefore, called isoenthalpic process.

The actual change in temperature on expansion of the gas is P2 T1 T2 P2
expressed in terms of Joule–Thomson coefficient (µ), defined as

µ =  ∂T 
 ∂P H
X Y

i.e. it is the temperature change in degrees produced by a

drop of one atmospheric pressure when the gas expands under conditions of constant enthalpy.

(1) For cooling, µ = +ve (because dT and dP both will be – ve)

(2) For heating, µ = −ve (because dT = +ve, dP = – ve)

(3) µ = 0 means dT = 0 for any value of dP, i.e., there is neither cooling nor heating.

The temperature at which a real gas shows no cooling or heating effect on adiabatic expansion (i.e. µ = 0), is
called Inversion temperature. Below this temperature it shows cooling effect while above this temperature, it
shows heating effect.

Most of the gases have inversion temperature above room temperature. H2 and He have low inversion
temperatures (being – 80°C and –240°C respectively). That is why they show heating effect at room temperature.

Spontaneous and Non-spontaneous processes.

(1) Definition : A process which can take place by itself under the given set of conditions once it has been
initiated if necessary, is said to be a spontaneous process. In other words, a spontaneous process is a process
that can occur without work being done on it. The spontaneous processes are also called feasible or probable
processes.

On the other hand, the processes which are forbidden and are made to take place only by supplying energy
continuously from outside the system are called non-spontaneous processes. In other words, non spontaneous
processes can be brought about by doing work.

Note : A spontaneous process is unidirectional and irreversible.

(2) Examples of Spontaneous and Non-spontaneous processes Pinhole Insulation

(i) A river flows from a mountain towards the sea is
spontaneous process.

(ii) Heat flows from a conductor at high temperature to Vacuum
another at low temperature is spontaneous process.

(iii) A ball rolls down the hill is spontaneous process. Gas under pressure

(iv) The diffusion of the solute from a concentrated solution Spontaneous expansion of an ideal gas into vacuum
to a dilute solution occurs when these are brought into contact is
spontaneous process.

(v) Mixing of different gases is spontaneous process.


(vi) Heat flows from a hot reservoir to a cold reservoir is spontaneous process.

(vii) Electricity flows from high potential to low potential is spontaneous process.

(viii) Expansion of an ideal gas into vacuum through a pinhole is spontaneous process.

All the above spontaneous processes becomes non-spontaneous when we reverse them by doing work.

(3) Spontaneous process and Enthalpy change : A spontaneous process is accompanied by decrease in
internal energy or enthalpy, i.e., work can be obtained by the spontaneous process. It indicates that only exothermic
reactions are spontaneous. But the melting of ice and evaporation of water are endothermic processes which also
proceeds spontaneously. It means, there is some other factor in addition to enthalpy change (∆H) which explains
the spontaneous nature of the system. This factor is entropy.

Second law of thermodynamics.

All the limitations of the first law of thermodynamics can be remove by the second law of thermodynamics.
This law is generalisation of certain experiences about heat engines and refrigerators. It has been stated in a number
of ways, but all the statements are logically equivalent to one another.

(1) Statements of the law
(i) Kelvin statement : “It is impossible to derive a continuous supply of work by cooling a body to a
temperature lower than that of the coldest of its surroundings.”

(ii) Clausius statement : “It is impossible for a self acting machine, unaided by any external agency, to
convert heat from one body to another at a higher temperature or Heat cannot itself pass from a colder body to a
hotter body, but tends invariably towards a lower thermal level.”

(iii) Ostwald statement : “It is impossible to construct a machine functioning in cycle which can convert heat
completely into equivalent amount of work without producing changes elsewhere, i.e., perpetual motions are not
allowed.”

(iv) Carnot statement: “It is impossible to take heat from a hot reservoir and convert it completely into work
by a cyclic process without transferring a part of it to a cold reservoir.”

(2) Proof of the law : No rigorous proof is available for the second law. The formulation of the second law is
based upon the observations and has yet to be disproved. No deviations of this law have so far been reported.
However, the law is applicable to cyclic processes only.

Conversion of heat into work: The Carnot cycle.

Carnot, a French engineer, in 1824 employed merely theoretical and an imaginary reversible cycle known as

carnot cycle to demonstrate the maximum convertibility of heat into work.
The system consists of one mole of an ideal gas enclosed in a cylinder fitted with a piston, which is subjected
to a series of four successive operations. The four operations are

(1) Isothermal reversible expansion, q2 = −w1 = RT2 log e V2 .....(i)
V1

(2) Adiabatic reversible expansion, ∆E = −w2 = −Cv(T2 − T1) .....(ii)

(3) Isothermal reversible compression, − q1 = w3 = RT1 log e V4 .....(iii)
V3

(4) Adiabatic reversible compression, w4 = Cv(T2 − T1) .....(iv)


The net heat absorbed, q, by the ideal gas in the cycle is given by

q = q2 + (−q1 ) = RT2 log e V2 + RT1 log e V4 = RT2 log e V2 − RT1 log e V3 ......(v)
V1 V3 V1 V4

According to the expression governing adiabatic changes,

T2  V3 γ −1
T1 V2
= (For adiabatic expansion) (P1,V1) (P2,V2 ,T2)

T1 =  V1 γ −1 (For adiabatic compression) A 1 B
T2 V4 q2 2
4
or V3 = V4 or V2 = V3 Pressure D q1 3 C
V2 V1 V1 V4
(P4,V4) (P3,V3, T1)
V3
Substituting the value of V4 in eq. (v),

q = q2 − q1 = RT2 log e V2 − RT1 log e V2 Volume
V1 V1

q2 = R(T2 − T1)log e V2 ......(vi)
V1

Similarly, net work done by the gas is given by

w = −w1 − w2 + w3 + w4

So, w = R(T2 − T1)log e V2 ......(vii)
V1

Thus, q = w . For cyclic process, the essential condition is that net work done is equal to heat absorbed. This

condition is satisfied in a carnot cycle. Dividing Equation (vii) by (vi) we get,

w = T2 − T1 = Thermodynamic efficiency
q2 T2

Thus, the larger the temperature difference between high and low temperature reservoirs, the more the heat
converted into work by the heat engine.

Since T2 − T1 < 1 , it follows that w < q2 . This means that only a part of heat absorbed by the system at the
T2

higher temperature is transformed into work. The rest of the heat is given out to surroundings. The efficiency of the

heat engine is always less then 1. This has led to the following enunciation of the second law of thermodynamics.
It is impossible to convert heat into work without compensation.

Example 6. An engine operating between 150°C and 25°C takes 500 J heat from a higher temperature reservoir if there
Solution: (a)
are no frictional losses then work done by engine is [MH CET 1999]

(a) 147.7J (b) 157.75 J (c) 165.85J (d) 169.95J

T2 = 150 + 273 = 423K

T1 = 25 + 273 = 298K

q2 = 500J

w = T2 − T1
q2 T2


w = 500 423 − 298  = 147.7 J
423 

Example 7. Find out the working capacity of an engine which is working between 110° to 25°C, if temperature of boyler is
increased up to 140°C and sink temperature equal to its.

(a) I − 20.2%, II − 18% (b) I − 22.2%, II - 27.8% (c) I − 28%, II − 30% (d) None of these

Solution:(b) First condition

T2 = 110o C = 110 + 273K = 383K η=?

T1 = 25o C = 25 + 273K = 298K

η = T2 − T1 = 383 − 298 = 0.222 or 22.2%
T2 383

Second condition

T2 = 140o C = 140 + 273K = 413K η=?

T1 = 25o C = 25 + 273K = 298K

η = T2 − T1 = 413 − 298 = 0.278 or 27.8%
T2 413

Entropy and Entropy change.

(1) Definition : Entropy is a thermodynamic state quantity which is a measure of randomness or disorder of
the molecules of the system.

Entropy is represented by the symbol “S”. It is difficult to define the actual entropy of a system. It is more
convenient to define the change of entropy during a change of state.

The entropy change of a system may be defined as the integral of all the terms involving heat exchanged (q)
divided by the absolute temperature (T) during each infinitesimally small change of the process carried out
reversibly at constant temperature.

∆S = S final − Sinitial = qrev
T

If heat is absorbed, then ∆S = +ve and if heat is evolved, then ∆S = −ve.

(2) Units of entropy : Since entropy change is expressed by a heat term divided by temperature, it is
expressed in terms of calorie per degree, i.e.,cal deg-1 . In SI units, the entropy is expressed in terms of joule per

degree Kelvin, i.e., JK −1 .

(3) Characteristics of entropy : The important characteristics of entropy are summed up below

(i) Entropy is an extensive property. Its value depends upon the amount of the substance present in the
system.

(ii) Entropy of a system is a state function. It depends upon the state variables (T, p, V,n).

(iii) The change in entropy in going from one state to another is independent of the path.
(iv) The change in entropy for a cyclic process is always zero.


(v) The total entropy change of an isolated system is equal to the entropy change of system and entropy
change of the surroundings. The sum is called entropy change of universe.

∆Suniverse = −∆Ssys + ∆SSurr

(a) In a reversible process, ∆Suniverse = 0 and, therefore

∆Ssys = −∆SSurr

(b) In an irreversible process, ∆Suniverse > 0 . This means that there is increase in entropy of universe is
spontaneous changes.

(vi) Entropy is a measure of unavailable energy for useful work.

Unavailable energy = Entropy × Temperature

(vii) Entropy, S is related to thermodynamic probability (W) by the relation,

S = k log e W and S = 2.303k log10W ; where, k is Boltzmann's constant

(4) Entropy changes in system & surroundings and total entropy change for Exothermic and

Endothermic reactions : Heat increases the thermal motion of Entropy of the Entropy of the
the atoms or molecules and increases their disorder and hence their surroundings surroundings

entropy. In case of an exothermic process, the heat escapes into the increases decreases

surroundings and therefore, entropy of the surroundings increases

on the other hand in case of endothermic process, the heat enters Heat Heat

the system from the surroundings and therefore. The entropy of the

surroundings decreases. (a) Exothermic reaction (b) Endothermic reaction

In general, there will be an overall increase of the total entropy Entropy changes in (a) exothermic and (b) endothermic processes
(or disorder) whenever the disorder of the surroundings is greater than the decrease in disorder of the system. The

process will be spontaneous only when the total entropy Entropy of
increases. vaporisation

(5) Entropy change during phase transition : The change of Entropy, S Liquid Gas
matter from one state (solid, liquid or gas) to another is called phase Entropy of
transition. Such changes occur at definite temperature such as melting Boiling
point (solid to liquid). boiling point (liquid to vapours) etc, and are fusion point
accompanied by absorption or evolution of heat.
Solid
When a solid changes into a liquid at its fusion temperature, there is

absorption of heat (latent heat). Let ∆H f be the molar heat of fusion. Melting
The entropy change will be point

∆S f = ∆H f Temperature, T
Tf
Increase in disorder or entropy when
a solid change to liquid to gas

Similarly, if the latent heat of vaporisation and sublimation are

denoted by ∆Hvap and ∆H sub , respectively, the entropy of vaporisation and sublimation are given by


∆Svap = ∆H vap and ∆Ssub = ∆H sub
Tb Ts

Since ∆H f , ∆Hvap and ∆HSub are all positive, these processes are accompanied by increase of entropy.

The reverse processes are accompanied by decrease in entropy.

Note :  Entropy increases not only in phase transition but also when the number of moles of

products is greater than the number of moles of reactants. ( nproduct > nreactant i.e.∆n = +ve )

(6) Entropy change for an ideal gas : In going from initial to final state, the entropy change, ∆S for an
ideal gas is given by the following relations,

(i) When T and V are two variables, ∆S = nCv ln T2 + nR ln V2 . Assuming Cv is constant
T1 V1

(ii) When T and p are two variables, ∆S = nCP ln T2 − nR ln p2 . Assuming Cp , is constant
T1 p1

(a) Thus, for an isothermal process (T constant), ∆S = nR ln V2 or = −nR ln p2
V1 p1

(b) For isobaric process (p constant), ∆S = nCp ln T2
T1

(c) For isochoric process (V constant), ∆S = n Cv ln T2
T1

(d) Entropy change during adiabatic expansion : In such process q=0 at all stages. Hence ∆S = 0 . Thus,

reversible adiabatic processes are called isoentropic process.

Example 8. The enthalpy of vapourisation of a liquid is 30 kJ mol–1 and entropy of vapourisation is 75mol–1 K. The boiling
point of the liquid at 1 atm is [IIT JEE 2004]

(a) 250 K (b) 400 K (c) 450 K (d) 600 K

Solution: (b) ∆Sv = ∆H v 30kJ = 30000 J
Tb

∆Hv = 30000 J ; ∆Sv = 75 mole −1 K ; Tb = 30000 = 400K
75

Example 9. If 900 J/g of heat is exchanged at boiling point of water, then what is increase in entropy [BHU 1998]

(a) 43.4 J/mole (b) 87.2 J/mole (c) 900 J/mole (d) Zero

Solution: (a) Boiling point (Tb ) = 100°C = 373K ; ∆Hv = 900J / g

∆Svap = ∆H v ; Molecular weight of water = 18
T

∆Svap = 900 × 18 = 43.4 JK −1mol−1
373

Example 10. The standard entropies of CO2 (g), C(s) and O2(g) are 213.5, 5.690 and 205 JK–1 respectively. The standard

entropy of formation of CO2 (g) is [CPMT 2001]

(a) 1.86 JK–1 (b) 1.96 JK–1 (c) 2.81 JK–1 (d) 2.86 JK–1

Solution: (c) Formation of CO2 is, C(s) + O2(g) → CO2(g)

∆S 0 = ∆S 0(product) − ∆S 0(reactants) = 213.5 − [5.690 + 205] = 2.81 JK −1


Free energy and Free energy change.

Gibb's free energy (G) is a state function and is a measure of maximum work done or useful work done from
a reversible reaction at constant temperature and pressure.

(1) Characteristics of free energy
(i) The free energy of a system is the enthalpy of the system minus the product of absolute temperature and
entropy i.e., G = H − TS
(ii) Like other state functions E, H and S, it is also expressed as ∆G . Also ∆G = ∆H − T∆Ssystem where ∆S is

entropy change for system only. This is Gibb's Helmholtz equation.

(iii) At equilibrium ∆G = 0

(iv) For a spontaneous process decrease in free energy is noticed i.e., ∆G = −ve .

(v) At absolute zero, T∆S is zero. Therefore if ∆G is – ve, ∆H should be – ve or only exothermic reactions
proceed spontaneously at absolute zero.

(vi) ∆Gsystem = T∆Suniverse

(vii) The standard free energy change, ∆Go = −2.303RT log10 K, where K is equilibrium constant.

(a) Thus if K > 1, then ∆Go = −ve thus reactions with equilibrium constant K>1 are thermodynamically
spontaneous.

(b) If K<1, then ∆Go = +ve and thus reactions with equilibrium constant K<1 are thermodynamically
spontaneous in reverse direction.

(2) Criteria for spontaneity of reaction : For a spontaneous change ∆G = −ve and therefore use of
∆G = ∆H − T∆S, provides the following conditions for a change to be spontaneous.

∆H ∆S ∆G Reaction characteristics Example
Reaction is spontaneous at all temperatures 2O3(g) → 3O2(g)
– + Always negative

+ – Always positive Reaction is non spontaneous at all 3O2(g) → 2O3(g)
temperatures

– – Negative at low Reaction is spontaneous at low temperature CaO(s) + CO2(g) → CaCO3(s)
temperature but positive at but becomes non spontaneous at high

high temperature temperature

+ + Positive at low temperature Reaction is non spontaneous at low CaCO3(s) → CaO(s) + CO2(g)
but negative at high temperature but becomes spontaneous at

temperature high temperature

Example 11. For a reaction at 25°C enthalpy change and entropy change are – 11.7×103 J mol–1 and − 105J mol−1K−1
Solution: (b)
Example 12. respectively, what is the Gibb's free energy. [BHU 2001]
Solution: (c)
(a) 15.05 kJ (b) 19.59 kJ (c) 2.55 kJ (d) 22.55 kJ

∆G = ∆H − T∆S, T = 25 + 273 = 298K

∆G = −11.7 ×103 − 298 × (−105) = 19590J = 19.59 kJ

For reaction Ag2O(s) → 2 Ag(s) + 1 O2(g) the value of ∆H = 30.56kJ mol−1 and ∆S = 0.066kJ mol−1 K−1.
2

Temperature at which free energy change for reaction will be zero is [MH CET 1999]

(a) 373 K (b) 413 K (c) 463 K (d) 493 K

∆G = ∆H − T∆S


∆H = 30.56kJ mol−1 ; ∆S = 0.066 kJmol−1 K −1 ; ∆G = 0 at equilibrium; T = ?

∴∆H = T∆S or 30.56 = T × 0.066

T = 463 K

Example 13. The free energy change for the following reactions are given below

C2H2(g) + 5 O2(g) → 2CO2(g) + H2O(l); ∆Go = −1234kJ
2

C(s) + O2(g) → CO2(g); ∆Go = −394kJ

H2(g) + 1 O2(g) → H2O (l); ∆Go = −237kJ
2

What is the standard free energy change for the reaction H2(g) + 2C(s) → C2H2(g) [Kerala PMT 2002]

(a) – 209 kJ (b) – 2259 kJ (c) +2259 kJ (d) 209 kJ

Solution : (a) ∆G = ∆G0(Product) − ∆G0(Reactants) = (−1234) − [−237 + 2(−394)] = −1234 + 1025 = −209kJ
Example 14. Calculate the free energy per mole when liquid water boils against 1 atm pressure (∆Hvap = 2.0723 kJ / g)

[DPMT 1995]

(a) 0 (b) 0.2 (c) 0.3 (d) 0.4

Solution: (a) ∆H = 2.0723 ×18kJ / mole = 37.3kJ / mole ; T = 373K ; ∆Gvap =?

∆S = ∆Hvap = 37.3kJ = 0.1kJK −1mol−1
T 373

∆Gvap = ∆Hvap − T∆Svap = 37.3 − 373 × 0.1 = 0

Example 15. ∆G0 for the reaction x + y = z is – 4.606 kcal. The value of equilibrium constant of the reaction at 227o C is

(R = 2.0 kcal mol−1K−1) [Roorkee 1999]

(a) 100 (b) 10 (c) 2 (d) 0.01

Solution: (a) R = 2.0 kcal mol−1K−1 ; T = 500 K ; ∆G0 = −4.606 kcal ;

∆G0 = −2.303RT log K
−4.606 = −2.303 × 0.002 × 500 log K
log K = 2, K = 100

Third law of thermodynamics.

This law was first formulated by German chemist Walther Nernst in 1906. According to this law,
“The entropy of all perfectly crystalline solids is zero at the absolute zero temperature. Since entropy is a
measure of disorder, it can be interpreted that at absolute zero, a perfectly crystalline solid has a perfect order of its
constituent particles.”

The most important application of the third law of thermodynamics is that it helps in the calculation of
absolute entropies of the substance at any temperature T.

S = 2.303Cp log T

Where Cp is the heat capacity of the substance at constant pressure and is supposed to remain constant in the
range of 0 to T.


Limitations of the law

(1) Glassy solids even at 0K has entropy greater than zero.

(2) Solids having mixtures of isotopes do not have zero entropy at 0K. For example, entropy of solid
chlorine is not zero at 0K.

(3) Crystals of Co, N2O, NO, H2O, etc. do not have perfect order even at 0K thus their entropy is not equal
to zero.

Thermochemistry

“Thermochemistry is a branch of physical chemistry which is concerned with energy changes
accompanying chemical transformation. It is also termed as chemical energetics. It is based on the first law of
thermodynamics.”

Thermochemical equations.

A balanced chemical equation which tells about the amount of heat evolved or absorbed during the reaction
is called a thermochemical equation. A complete thermochemical equation supplies the following information’s.

(1) It tells about the physical state of the reactants and products. This is done by inserting symbol (s), (l) and
(g) for solid, liquid and gaseous state respectively along side the chemical formulae.

(2) It tells about the allotropic form (if any) of the reactant by inserting the respective allotropic name, for
example, C (graphite) (s).

(3) The aqueous solution of the substance is indicated by the word aq.

(4) It tells whether a reaction proceeds with the evolution of heat or with the absorption of heat, i.e. heat
change involved in the system.

Remember that like algebraic equations, thermochemical equations can be reversed, added, subtracted and
multiplied.

Exothermic and Endothermic reactions.

(1) Exothermic reactions : The chemical reactions which proceed with the evolution of heat energy are
called exothermic reactions. The heat energy produced during the reactions is indicated by writing +q or more
precisely by giving the actual numerical value on the products side. In general exothermic reactions may be

represented as, A + B → C + D + q (heat energy)

In the exothermic reactions the enthalpy of the products will be less than the enthalpy of the reactants, so that
the enthalpy change is negative as shown below

∆H = H p − Hr ; H p < Hr ; ∆H = − ve

Examples : (i) C(s) + O2 (g) → CO2 (g) + 393.5kJ (at constant temperature and pressure)

or C(s) + O2 (g) → CO2 (g); ∆H = −393.5kJ

(ii) H 2 (g) + 1 O2 (g) → H 2 O(l); ∆H = −285.8kJ
2

(iii) N 2 (g) + 3H 2 (g) → 2NH 3 (g); ∆H = −92.3kJ
(iv) 2SO2 (g) + O2 (g) → 2SO3 (g); ∆H = −694.6kJ


(v) CH 4 (g) + 2O2 (g) → CO2 (g) + 2H 2O; ∆H = −890.3kJ
(vi) Fermentation is also an example of exothermic reaction.

(2) Endothermic reactions : The chemical reactions which proceed with the absorption of heat energy are
called endothermic reactions. Since the heat is added to the reactants in these reactions, the heat absorbed is
indicated by either putting (–) or by writing the actual numerical value of heat on the reactant side

A + B → C + D − q (heat energy)

The heat absorbed at constant temperature and constant pressure measures enthalpy change. Because of
the absorption of heat, the enthalpy of products will be more than the enthalpy of the reactants. Consequently, ∆H
will be positive (+ve) for the endothermic reactions.

∆H = H p − Hr ; H p > Hr ; ∆H = +ve

Example : (i) N 2 (g) + O2 (g) + 180.5 kJ → 2NO(g)
or N 2 (g) + O2 (g) → 2NO(g); ∆H = +180.5 kJ

(ii) C(s) + 2S(s) → CS2 (l) ∆H = +92.0kJ
(iii) H 2 (g) + I 2 (g) → 2HI(g); ∆H = +53.9kJ
(iv) 2HgO(s) → 2Hg(l) + O2 (g); ∆H = +180.4kJ
(v) SnO2 (s) + 2C(s) → Sn(s) + 2CO2 (g); ∆H = +360.0kJ
(vi) Fe + S → FeS
(vii) Preparation of ozone by passing silent electric discharged through oxygen is the example of

endothermic reaction.

(viii) Evaporation of water is also the example of endothermic reaction.

The enthalpy changes for exothermic and Hr Reactant Hp Products
endothermic reactions are shown in figure. H H
∆H=Hp–Hr = – ve ∆H=Hp–Hr = + ve
Similarly, if we consider heat change at Hp (Hp<Hr) (Hp>Hr)
constant volume and temperature, ∆E is – ve, now
it may be concluded that Products Hr Reactant

For exothermic reaction : ∆H or ∆E = −ve Reaction path Reaction path

For endothermic reactions : ∆H or ∆E = +ve Enthalpy change during an Enthalpy change during an
exothermic reaction endothermic reaction

Heat of reaction or Enthalpy of reaction.

Heat of reaction is defined as the amount of heat evolved or absorbed when quantities of the substances
indicated by the chemical equation have completely reacted. The heat of reaction (or enthalpy of reaction) is
actually the difference between the enthalpies of the products and the reactants when the quantities of the reactants
indicated by the chemical equation have completely reacted. Mathematically,

Enthalpy of reaction (heat of reaction) = ∆H = ΣH P − ΣH R

(1) Factors which influence the heat of reaction : There are a number of factors which affect the
magnitude of heat of reaction.


(i) Physical state of reactants and products : Heat energy is involved for changing the physical state of a
chemical substance. For example in the conversion of water into steam, heat is absorbed and heat is evolved when
steam is condensed. Considering the following two reactions

H 2 (g) + 1 O2 (g) = H 2O(g); ∆H = − 57.8 kcal
2

H 2 (g) + 1 O2 (g) = H 2O(l); ∆H = − 68.32 kcal
2

It is observed that there is difference in the value of ∆H if water is obtained in gaseous or liquid state. ∆H
value in second case is higher because heat is evolved when steam condenses. Hence, physical sate always affects
the heat of reaction.

(ii) Allotropic forms of the element : Heat energy is also involved when one allotropic form of an element
is converted into another. Thus, the value of ∆H depends on the allotropic form used in the reaction. For example,
the value of ∆H is different when carbon in the form of diamond or in amorphous form is used.

C (diamond) + O2 (g) → CO2 (g); ∆H = − 94.3 kcal

C (amorphous) + O2 (g) → CO2 (g); ∆H = − 97.6 kcal
The difference between the two values is equal to the heat absorbed when 12g of diamond is converted into
12g of amorphous carbon. This is termed as heat of transition.

(iii) Temperature : Heat of reaction has been found to depend upon the temperature at which reaction is
occurring. The variation of the heat of reaction with temperature can be ascertained by using Kirchhoff's equation.

∆H T2 − ∆H T1 = ∆CP
T2 − T1

Kirchhoff's equation at constant volume may be given as,

∆ET2 − ∆ET1 = ∆Cν
T2 − T1

(iv) Reaction carried out at constant pressure or constant volume : When a chemical reaction occurs

at constant volume, the heat change is called the internal energy of reaction at constant volume. However, most of

the reactions are carried out at constant pressure; the enthalpy change is then termed as the enthalpy of reaction at

constant pressure. The difference in the values is negligible when solids and liquids are involved in a chemical

change. But, in reactions which involve gases, the difference in two values is considerable.

∆E + ∆nRT = ∆H or qv + ∆nRT = q p

∆E = qv = heat change at constant volume; ∆H = q p = heat change at constant pressure,

∆n = total number of moles of gaseous product - total number of moles of gaseous reactants.

(2) Types of heat of reaction

(i) Heat of formation : It is the quantity of heat evolved or absorbed (i.e. the change in enthalpy) when one

mole of the substance is formed from its constituent elements under given conditions of temperature and pressure. It

is represented by ∆H f . When the temperature and pressure are as 25o C and 1 atmospheric pressure. The heat of

formation under these conditions is called standard heat of formation. It is usually represented by ∆H 0 .
f

The standard heat of formation of 1 mole of NH 3 (g) and 1 mole of HCl (g).

1 N 2 (g) + 3 H 2 (g) → NH 3 (g); ∆H(g) = −11kcal
2 2


1 H 2 (g) + 1 Cl 2 (g) → HCl ; ∆H f = −22 kcal
2 2

It may be calculated by

∆H 0 = Sum of standard heats of  − Sum of standard heats of
   
 Formation of products   Formation of reactants 

∆H 0 = [Σ∆H 0 (products) – Σ∆H ]0
(Reactants)

(ii) Heat of combustion : It is the amount of heat evolved or absorbed (i.e. change in enthalpy) when one
mole of the substance is completely burnt in air or oxygen. For example

CH 4 (g) + 2O2 (g) → CO2 (g) + 2H 2O(l); ∆H = − 192 kcal

C2 H 6 (g) + 3.5 O2 (g) → 2CO2 (g) + 3H 2O(l); ∆H = − 372.8 kcal

It may be calculated by

∆H 0 = Sum of the standard heats of  − Sum of the standard heats 
   
 Combustion of products   Combustion of reactants 

∆H 0 = [Σ∆H 0 (Products) - ΣH 0 (Reactants) ]
f f

Note : Heat of combustion increases with increase in number of carbon and hydrogen.

Heat of combustion of carbon is equal to the intrinsic energy of CO2 .

The enthalpy or heat of combustion have a number of applications. Some of these are described below,

(a) Calorific value of foods and fuels : Energy is needed for the working of all machines. Even human body is
no exception. Coal, petroleum, natural gas etc. serve as the principal sources of energy for man-made machines,
the food which we eat serves as a source of energy to our body.

The energy released by the combustion of foods or fuels is usually compared in terms of their combustion
energies per gram. It is known as calorific value. The amount of heat produced in calories or Joules when one
gram of a substance (food or fuel) is completely burnt or oxidised.

When methane burns, 890.3 kJ mol–1 of energy is released.

CH 4 (g) + 2O2 (g) → CO2 (g) + 2H 2O(l); ∆H CH4 = −890.3 kJ
1 mole (16g)

So, the calorific value of methane = − 890.3 = −55.6kJ / g
16

Calorific values of some important food stuffs and fuels

Fuel Calorific value (kJ/g) Food Calorific value (kJ/g)
Wood 17 Milk 3.1
Charcoal 33 Egg 6.7
Kerosene 48 Rice 16.7
Methane 55 Sugar 17.3
L.P.G. 55 Butter 30.4
Hydrogen 150 Ghee 37.6

Out of the fuels listed, hydrogen has the highest calorific value. The calorific value of proteins is quite low.


(b) Enthalpies of formation : Enthalpies of formation of various compounds, which are not directly obtained,
can be calculated from the data of enthalpies of combustions easily by the application of Hess's law.

Heat of reaction = Σ Heat of combustion of reactants – Σ Heat of combustion of products.

(iii) Heat of neutralisation : It is the amount of heat evolved (i.e., change in enthalpy) when one equivalent
of an acid is neutralised by one equivalent of a base in fairly dilute solution, e.g., Neutralisation reactions are
always exothermic reaction and the value of ∆H is (−ve) .

HCl(aq.) + NaOH (aq.) → NaCl(aq.) + H2O; ∆H = − 13.7 kcal

The heat of neutralisation of a strong acid against a strong base is always constant (13.7 kcal or 57 kJ mole−1) .
It is because in dilute solutions all strong acids and bases ionise completely and thus the heat of neutralisation in
such cases is actually the heat of formation of water from H + and OH− ions, i.e.,

H + + OH − → H2O; ∆H = − 13.7 kcal

In case of neutralisation of a weak acid or a weak base against a strong base or acid respectively, since a part of the

evolved heat is used up in ionising the weak acid or base, it is always less than 13.7 kcal mole−1 (57 kJ mole−1). For

example, heat of neutralisation of HCN (a weak acid) and NaOH (a strong alkali) is − 2.9 kcal because 10.8 kcal of

heat is absorbed for the ionisation of HCN (i.e., the heat of dissociation or ionisation of HCN is 10.8 kcal ) Similarly Heat

of neutralization of NH4OH and HCl is less then 13.7 kcal.

HCN (aq.) + NaOH (aq.) → NaCN(aq.) + H2O; ∆H = − 2.9 kcal

HCN (aq.) ⇌ H+ + CN–; ∆H = 10.8 Kcal

(iv) Heat of solution : It is the amount of heat evolved or absorbed (i.e., change in enthalpy) when one mole
of the solute is dissolved completely in excess of the solvent (usually water). For example,

NH4Cl(s) + H2O(l) → NH4Cl(aq.); ∆H = + 3.90 kcal

BaCl2(s) + H2O(l) → BaCl2(aq.); ∆H = − 2.70 kcal

(v) Heat of hydration : It is the amount of heat evolved or absorbed (i.e change in enthalpy) when 1 mole of
an anhydrous or a partially hydrated salt combines with the required number of moles of water to form a specific

hydrate. For example, CuSO4(s) + 5H2O(l) → CuSO4. 5H2O(s); ∆H = − 18.69

(vi) Heat of vapourisation : When a liquid is allowed to evaporate, it absorbs heat from the surroundings
and evaporation is accompanied by increase in enthalpy. For example: 10.5 k cals is the increase in enthalpy when

one mole of water is allowed to evaporate at 25o C . When the vapours are allowed to condense to liquid state, the
heat is evolved and condensation of vapour is accompanied by decrease in enthalpy. The value for the

condensation of one mole of water vapour at 25o C is also 10.5 k cals.

The evaporation and condensation can be represented as,

H2O(l) → H2O(g); ∆H = + 10.5 kcals (+43.93 kJ)

H2O(g) → H2O(l); ∆H = − 10.5 kcals (−43.93 kJ)
Thus the change in enthalpy when a liquid changes into vapour state or when vapour changes into liquid state
is called heat of vapourisation.

(vii) Heat of fusion : When a solid is allowed to melt, it changes into liquid state with the absorption of heat
(increase in enthalpy) and when a liquid is allowed to freeze, it changes into solid with the evolution of heat
(decrease in enthalpy). The change in enthalpy of such type of transformations is called enthalpy of fusion. For
example,


H2O(ice) → H2O(liquid); ∆H = + 1.44 kcals (+ 6.02 kJ)
H 2O (liquid) → H 2O (ice); ∆H = − 1.44 kcals (− 6.02 kJ)

Note :  The enthalpy of fusion of ice per mole is 6 kJ.

(viii) Heat of precipitation : It is defined as the amount of heat liberated in the precipitation of one mole of
a sparingly soluble substance when solutions of suitable electrolytes are mixed, for example

Ba 2+ + SO 2− (aq) → BaSO4 (s): ∆H = − 4.66 kcal
4

(ix) Heat of sublimation : Sublimation is a process in which a solid on heating changes directly into gaseous

state below its melting point.

Heat of sublimation of a substance is the amount of heat absorbed in the conversion of 1 mole of a solid
directly into vapour phase at a given temperature below its melting point.

I 2 (s) → I 2 (g) ; ∆H = + 62.39 kJ
Most solids that sublime are molecular in nature e.g. iodine and naphthalene etc.

∆H sub. = ∆H fusion + ∆H vaporisation

(3) Experimental determination of the heat of reaction : The heat evolved or absorbed in a chemical
reaction is measured by carrying out the reaction in an apparatus called calorimeter. The principle of

measurement is that heat given out is equal to heat taken, i.e., Q = (W + m) × s × (T2 − T1 ),

Where Q is the heat of the reaction (given out), W is the water equivalent of the calorimeter and m is the mass

of liquid in the calorimeter and s its specific heat, T2 is the final temperature and T1 the initial temperature of the

system. Different types of calorimeters are used but two of the common Firing leads Thermometer
types are,

(i) Water calorimeter and (ii) Bomb calorimeter Oxygen
Bomb calorimeter : This is commonly used to find the heat of Stirrer inlet

combustion of organic substances. It consists of a sealed combustion

chamber called a bomb. A weigh quantity of the substance in a dish
along with oxygen under about 20 atmospheric pressure is placed in the Sample

bomb which is lowered in water contained in an insulated copper

vessel. The vessel is fitted with a stirrer and a sensitive thermometer.

The arrangement is shown in fig. Bomb Oxygen under
The temperature of the water is noted and the substance is ignited pressure

by an electric current. After combustion the rise in temperature of the Water
system is noted. The heat of combustion can be calculated from the

heat gained by water and calorimeter. Bomb calorimeter

Since the reaction in a bomb calorimeter proceeds at constant
volume, the heat of combustion measured is ∆E

∆E = (W + m)(t 2 − t1)× s × M kcal
w1

Where M is the molecular mass of the substance, w1 is the weight of substance taken, W is the water
equivalent of calorimeter, m is the mass of liquid in the calorimeter and s is the specific heat of liquid.

∆H can be calculated from the relation, ∆H = ∆E + ∆nRT


Example 16. The heat of formations of CO(g) and CO2(g) are –26.4 kcal and −94.0 kcal respectively. The heat of
Solution:(b)
Example 17. combustion of carbon monoxide will be [MP PET/PMT 1988; EAMCET 1993]
Solution:(c)
Example 18. (a) + 26.4 kcal (b) – 67.6 kcal (c) –120.6 kcal (d) + 52.8 kcal
Solution:(b)
CO + 1 O2 → CO2 ; ∆H =?
Example 19. 2

Solution:(d) ∆HCO2 = −94.0 kcal , ∆HCO = − 26.4 kcal
Example 20.
Solution:(b) ∆H = ∆H(CO2) − ∆H(CO) = −94.0 − (−26.4) = − 67.6 kcal

Reaction H2(g) + I2(g) → 2HI(g); ∆H = −12.40 kcal . According to this, the heat of formation of HI will be

(a) 12.4 kcal (b) –12.4 kcal (c) –6.20 kcal [MP PET 1990]

(d) 6.20 kcal

H2(g) + I2(g) → 2HI(g)

For 2M, ∆H = −12.40kcal

1M, −12.40 = −6.20 kcal
2

C(diamond) + O2(g) → CO2(g); ∆H = − 395 kJ

C(graphite) + O2(g) → CO2(g); ∆H = − 393.5 kJ

From the data, the ∆H when diamond is formed from graphite is [CBSE 1989; BHU 1987]

(a) –1.5kJ (b) + 1.5kJ (c) +3.0kJ (d) –3.0kJ

C(graphite) → C(diamond)

∆H(diamond) = −395 kJ ; ∆H(graphite) = −393.5 kJ,

∆H = ∆Hgraphite – ∆Hdiamond =– 393.5 − (−395.0)

∆H = +1.5 kJ

The enthalpy of combustion of benzene from the following data will be

(i) 6C(s) + 3H 2(g) → C6 H6 (l); ∆H = + 45.9 kJ

(ii) H2(g) + 1 O2 (g) → H 2O(l); ∆H = −285.9 kJ
2

(iii) C(s) + O2(g) → CO2(g); ∆H = − 393.5 kJ

(a) + 3172.8 kJ (b) –1549.2 kJ (c) –3172.8 kJ (d) –3264.6 kJ

C6 H6 + 15 O2 → 6CO2 + 3H2O; ∆H = ?
2

∆H = ∆H(Product) − ∆H(Reactant) = [(6 × −393.5) + (3 × −285.9)]−(45.9) = −3264.6 kJ

The enthalpy of formation of H2O(l) is – 285.77 kJ mol–1 and enthalpy of neutralisation of strong acid and
strong base is –56.07 kJ mol–1, what is the enthalpy of formation of OH– ion

(a) + 229.70kJ (b) –229.70kJ (c) +226.70kJ (d) –22.670kJ

H +(aq) + OH−(aq) → H2O(l); ∆H = −56.07kJ

∆H = ∆H f (H2O) − [∆H f (H +) + ∆H f (OH)−] [∆H f (H +) = 0]

−56.07 = −285.77 − (0 + x)

∴ x = −285.77 + 56.07 = −229.70 kJ


Example 21. The heat of combustion of glucose is given by C6H12O6 + 6O2 → 6CO2 + 6H2O ; ∆H = −2840 kJ . Which of
Solution:(b) the following energy is required for the production of .18 gm of glucose by the reverse reaction.

Example 22. (a) 28.40 kJ (b) 2.84 kJ (c) 5.68 kJ (d) 56.8 kJ

∆Hcomb = −2840 kJ for 180g of C6H12O6

∴ ∆Hcomb for 0.18gm glucose = −2840 × 0.18 = −2.84kJ
180

For the reverse reaction ∆Hcomb of glucose = 2.84 kJ

One gram sample of NH 4 NO3 is decomposed in a bomb calorimeter. The temperature of the calorimeter

increases by 6.12K. The heat capacity of the system is 1.23kJ/g/deg. What is the molar heat of decomposition

for NH4 NO3. [AIIMS 2003]

(a) – 7.53 kJ/mole (b) – 398.1 kJ / mole (c) −16.1 kJ / mole (d) −602 kJ / mole

Solution:(a) Heat evolved = 1.23× 6.12
Example 23.
∴ Molar heat capacity = 1.23 × 6.12 = 7.5276

Molar heat of decomposition = −7.53kJ / mole

If ∆H 0 for H2O2 and H2O are – 188kJ/mole and −286kJ / mole. What will be the enthalpy change of the
f

reaction 2H2O2(l) → 2H2O(l) + O2(g) [MP PMT 1992]

(a) −196 kJ / mole (b) 146 kJ / mole (c) −494 kJ / mole (d) −98 kJ / mole

Solution:(a) H2 + O2 → H2O2 ; ∆H 0 = −188 kJ / mole
f

H2 + 1 O2 → H2O ∆H 0 = −286kJ / mole
2 f

∆H = ∆H 0(product) − ∆H 0(Reactants) = (2 × −286) − (2 × −188) = −572 + 376 = −196

Example 24. The heat of combustion of carbon is –94 kcal at 1 atm pressure. The intrinsic energy of CO2 is
Solution:(b)
(a) + 94 kcal (b) –94 kcal (c) + 47 kcal (d) –47 kcal

C(s) + O2(g) → CO2(g); ∆H = −94 kcal

∆H = ∆E + ∆ng RT ; ∆E = ?

∆ ng = 1 − 1 = 0 ; ∆H = ∆E ; ∆E = −94kcal

Example 25. When 1 mole of ice melts at 0 °C and a constant pressure of 1 atm, 1440 cal of heat are absorbed by the
system. The molar volume of ice and water are 0.0196 and 0.0180 L, respectively calculate ∆E

(a) 1430.03 cal (b) 1500.0 cal (c) 1450.0 cal (d) 1440.03 cal

Solution:(d) q = 1440cal

∴∆H = 1440 cal (absorbed)

given H2O(s) ⇌ H2O(l)

P∆V = 1× (0.0180–0.0196) =– 0.0016 L atm
 (1L atm = 24.206 cal)
∴− 0.0016L atm = −0.039cal

∆H = ∆E + P∆V ⇒ ∆E = ∆H − P∆V = 1440 − (−0.039) = 1440.039 cal


Example 26. The difference between heats of reaction at constant pressure and constant volume for the reaction

2C6 H 6 (l) + 15O2 (g) → 12CO2 (g) + 6H 2O(l) at 25 o C in kJ is [IIT 1991; AFMC 1994]

(a) –7.43 (b) +3.72 (c) −3.72 (d) +7.43

Solution : (a) ∆ng = 12 − 15 = −3

qP − qv = ∆ng RT = −3 × 8.314 × 298 = −7.43kJ.
1000

Laws of thermochemistry .

(1) Levoisier and Laplace law : According to this law enthalpy of decomposition of a compound is
numerically equal to the enthalpy of formation of that compound with opposite sign, For example,

C(s) + O2 → CO2 (g); ∆H = −94.3 kcal ; CO2 (g) → C(s) + O2 (g); ∆H = +94.3kcal

(2) Hess's law (the law of constant heat summation) : This law was presented by Hess in 1840. According
to this law “If a chemical reaction can be made to take place in a number of ways in one or in several steps, the total
enthalpy change (total heat change) is always the same, i.e. the total enthalpy change is independent of intermediate
steps involved in the change.” The enthalpy change of a chemical reaction depends on the initial and final stages
only. Let a substance A be changed in three steps to D with enthalpy change from A to B, ∆H1 calorie, from B to
C, ∆H 2 calorie and from C to D, ∆H3 calorie. Total enthalpy change from A to D will be equal to the sum of
enthalpies involved in various steps,

Total enthalpy change ∆H steps = ∆H1 + ∆H 2 + ∆H 3

Now if D is directly converted into A, let the enthalpy change be ∆H direct . According to Hess's law
∆H steps + ∆H direct = 0, i.e. ∆H steps must be equal to ∆H direct numerically but with opposite sign. In case it is not so,
say ∆H steps (which is negative) is more that ∆H direct (which is positive), then in one cycle, some energy will be
created which is not possible on the basis of first law of thermodynamics. Thus, ∆H steps must be equal to ∆H direct
numerically.

(i) Experimental verification of Hess's law

(a) Formation of carbon dioxide from carbon

First method : carbon is directly converted into CO2 (g).

C(s) + O2 (g) = CO2 (g); ∆H = −94.0 kcal

Second method : Carbon is first converted into CO(g) and then CO(g) into CO2 (g), i.e. conversion has
been carried in two steps,

C(s) + 1 O2 = CO(g) ; ∆H = −26.0 kcal
2

CO(g) + 1 O2 = CO2 (g); ∆H = − 68.0 kcal
2

Total enthalpy change C(s) to CO2 (g); ∆H = −94.0 kcal
(b) Formation of ammonium chloride from ammonia and hydrochloric acid:


First method : NH 3 (g) + HCl = NH 4 Cl(g); ∆H = − 42.2 kcal
NH4Cl(g)+aq= NH4Cl(aq); ∆H = + 4.0 kcal

NH3(g) + HCl(g) + aq = NH4Cl(aq); ∆H = −38.2kcal
Second method : NH3(g) + aq = NH3(aq); ∆H = −8.4 kcal

HCl(g) + aq = HCl(aq); ∆H = −17.3kcal
NH3(aq) + HCl(aq)= NH4Cl(aq); ∆H = −12.3kcal

NH3(g) + HCl(g) + aq = NH4Cl(aq); ∆H = −38.0 kcal
Conclusions
• The heat of formation of compounds is independent of the manner of its formation.
• The heat of reaction is independent of the time consumed in the process.
• The heat of reaction depends on the sum of enthalpies of products minus sum of the enthalpies of

reactants.
• Thermochemical equations can be added, subtracted or multiplied like algebraic equations.
(ii) Applications of Hess's law
(a) For the determination of enthalpies of formation of those compounds which cannot be prepared directly
from the elements easily using enthalpies of combustion of compounds.
(b) For the determination of enthalpies of extremely slow reactions.
(c) For the determination of enthalpies of transformation of one allotropic form into another.
(d) For the determination of bond energies.

∆H reaction = Σ Bond energies of reactants – Σ Bond energies of products.
(e) For the determination of resonance energy.
(f) For the determination of lattice energy.

Bond energy or Bond enthalpies.

When a bond is formed between atoms, energy is released. Obviously same amount of energy will be
required to break the bond. The energy required to break the bond is termed bond dissociation energy. The
more precise definition is,

“The amount of energy required to break one mole of bond of a particular type between the atoms in the
gaseous state, i.e., to separate the atoms in the gaseous state under 1 atmospheric pressure and the specified
temperature is called bond dissociation energy.”

For example, H − H(g) → 2H(g); ∆H = + 433 kJ mol −1

Cl − Cl(g) → 2Cl (g); ∆H = + 242.5 kJ mol −1

H − Cl(g) → H(g) + Cl(g); ∆H = + 431kJ mol −1

I − I(g) → 2I(g); ∆H = + 15.1kJ mol −1

H − I(g) → H(g) + I(g); ∆H = + 299 kJ mol −1


The bond dissociation energy of a diatomic molecule is also called bond energy. However, the bond
dissociation energy depends upon the nature of bond and also the molecule in which the bond is present. When a
molecule of a compound contains more than one bond of the same kind, the average value of the dissociation
energies of a given bond is taken. This average bond dissociation energy required to break each bond in a
compound is called bond energy.

Bond energy is also called, the heat of formation of the bond from gaseous atoms constituting the bond with
reverse sign.

H(g) + Cl(g) → H − Cl (g); ∆H = − 431kJ mol −1

Bond energy of H − Cl = − (enthalpy of formation) = −(−431) = + 431kJ mol −1

Consider the dissociation of water molecule which consists of two O − H bonds. The dissociation occurs in
two stages.

H 2O(g) → H(g) + OH(g); ∆H = 497.89 kJ mol −1
OH(g) → H(g) + O(g); ∆H = 428.5 kJ mol −1

The average of these two bond dissociation energies gives the value of bond energy of O − H.

Bond energy of O−H bond = 497.8 + 428.5 = 463.15 kJ mol −1
2

Similarly, the bond energy of N − H bond in NH3 is equal to one – third of the energy of dissociation of
NH3 and those of C − H bond in CH4 is equal to one – fourth of the energy of dissociation of CH 4 .

Bond energy of C − H = 1664 = 416 kJ mol −1
4

[CH 4 (g) → C(g) + 4H(g); ∆H = 1664 kJ mol −1 ]
Applications of bond energy
(1) Heat of a reaction = Σ Bond energy of reactants – Σ Bond energy of products.

Note :  In case of atomic species, bond energy is replaced by heat of atomization.

Order of bond energy in halogen Cl > Br > F2 > I 2

(2) Determination of resonance energy : When a compound shows resonance, there is considerable
difference between the heat of formation as calculated from bond energies and that determined experimentally.

Resonance energy = Experimental or actual heat of formation ∼ Calculated heat of formation.

Example 27. If enthalpies of methane and ethane are respectively 320 and 360 calories then the bond energy of C – C
Solution:(d)
bond is [UPSEAT 2003]

(a) 80 calories (b) 40 calories (c) 60 calories (d) – 120 calories

CH4 → 4C − H; ∆H = 320

1EC − H = 320 = 80 calories then 6 EC − H = 480 cal.
4

C2H6 → EC−C + 6EC−H ; ∆H = 360 cal.
360 = EC−C + 480


EC−C = 360 − 480 = −120cal

Example 28. Given that C(g) + 4H(g) → CH4(g); ∆H = −166 kJ . The bond energy C − H will be [AMU 2002]
Solution:(b)
Example 29. (a) 208 kJ/mol (b) – 41.5 kJ/mol (c) 832 kJ/mol (d) None of these
Solution:(a)
C(g) + 4H(g) → CH4(g); ∆H = −166kJ
Example 30.
Solution: (d) Bond energy for C – H = − 166 = −41.5 kJ / mole
4
Example 31.
Solution:(c) Calculate resonance energy of N2O from the following data. observed ∆H o (N 2O) = 82 kJ mol −1
f

B.E. of N ≡ N ⇒ 946 kJ mol−1; B.E. of N = N ⇒ 418 kJ mol−1

B.E. of O = O ⇒ 498 kJ mol −1 ; B.E. of N = 0 ⇒ 607 kJ mol−1 [Roorkee 1991]

(a) – 88 kJ mol–1 (b) – 44 kJ mol–1 (c) – 22 kJ mol–1 (d) None of these

N2(g) + 1 O2(g) → N2O
2

N ≡ N + 1 O = O → N = N = O
2

Calculated ∆H o ( N 2O) = [B.E.( N ≡ N ) + 1 B.E.(O = O)] − [B.E.(N=N) + B.E.N =O]
f 2

= 946 + 498  − [418 + 607] = +170 kJ / mole
2 

Resonance energy = observed ∆H 0 − calculated ∆H 0 = 82 − 170 = −88kJ mol−1
f f

If at 298K the bond energies of C − H, C − C, C = C and H − H bonds are respectively 414, 347, 615 and

435 kJ mol–1, the value of enthalpy change for the reaction H2C = CH2(g) + H2(g) → H3C − CH3(g) at 298 K

will be [AIEEE 2003]

(a) + 250 kJ (b) –250 kJ (c) + 125 kJ (d) – 125 kJ

CH2 = CH2(g) + H2(g) → H3C − CH3(g)

4 EC−H ⇒ 414 × 4 = 1656 6EC−H ⇒ 414 × 6 = 2484

1 EC=C ⇒ 615 × 1 = 615 1EC−C ⇒ 347 × 1 = 347

1EH−H ⇒ 435 × 1 = 435

4∆HC−H + ∆HC=C + ∆HH −H = 2706 → 6∆HC−H + 1∆HC−C = 2831

∆H = 2706 − 2831 = −125kJ

The bond dissociation energies of gaseous H 2 , Cl2 and HCl are 104, 58 and 103 kcal respectively. The

enthalpy of formation of HCl gas would be [MP PET 1997; MP PMT 1999, 2001]

(a) – 44 kcal (b) 44 kcal (c) –22 kcal (d) 22 kcal

1 H2 + 1 Cl2 → HCl
2 2

∆H = ΣB.E.(Reactants) − ΣB.E.(Products)

∆H = 1 B.E.(H2) + 1 BE(Cl2) − B.E.(HCl)
 2 2

= 1 (104) + 1 (58) − 103
2 2

= 81 − 103 = −22 kcal


∆H = −22 kcal.

Example 32. Given the bond energies N ≡ N, H − H and N–H bonds are 945, 436 and 391 kJ mole–1 respectively the
Solution:(a)
enthalpy of the following reaction N2(g) + 3H2(g) → 2NH3(g) is [EAMCET 1992; JIPMER 1997]

(a) – 93 kJ (b) 102 kJ (c) 90 kJ (d) 105 kJ

N ≡ N + 3H − H H
945+3 × 436 |
→
energy absorbed 2N—H

|

H

2×(3×391) =2346

energy released

Net energy (enthalpy) = BEreactants – BEproducts = 2253 – 2346 =– 93
∆H = −93kJ

***


Chemical equilibrium

Whenever we hear the word Equilibrium immediately a picture arises in our mind an object under the
influence of two opposing forces. For chemical reactions also this is true. A reaction also can exist in a state
of equilibrium balancing forward and backward reactions.

Equilibrium and its dynamic nature.

(1) Definition : “Equilibrium is the state at which the concentration of Products
reactants and products do not change with time. i.e. concentrations of reactants
Concentration
and products become constant.”

(2) Characteristics : Following are the important characteristics of Reactants
equilibrium state,

(i) Equilibrium state can be recognised by the constancy of certain Time Equilibrium
measurable properties such as pressure, density, colour, concentration etc. state

by changing these conditions of the system, we can control the extent to which a reaction proceeds.

(ii) Equilibrium state can only be achieved in close vessel, but if the process is carried out in an open vessel
equilibrium state cannot be attained because in an open vessel, the reverse process will not take place.

(iii) Equilibrium state is reversible in nature.

(iv) Equilibrium state is also dynamic in nature. Dynamic means moving and at a Inlet
microscopic level, the system is in motion. The dynamic state of equilibrium can be Constant
compared to water tank having an inlet and outlet. Water in tank can remain at the
same level if the rate of flow of water from inlet (compared to rate of forward reaction) level
is made equal to the rate of flow of water from outlet (compared to rate of backward
reaction). Thus, the water level in the tank remains constant, though both the inlet and Outlet
outlet of water are working all the time.

(v) At equilibrium state,

Rate of forward reaction = Rate of backward reaction

(vi) At equilibrium state, ∆G = 0, so that ∆H = T∆S. Rate of reaction ∆G= 0
(3) Types : Equilibrium in a system implies the existence of the following Equilibrium state
types of equilibria simultaneously,

(i) Thermal equilibrium : There is no flow of heat from one part to Time
another i.e. T = constant.

(ii) Mechanical equilibrium : There is no flow of matter from one part to another i.e. P = constant.

(iii) Physical equilibrium : There is the substance exist in three states: solid, liquid and gaseous.

(iv) Chemical equilibrium : There is no change in composition of any part of the system with time.

Physical equilibrium.

The physical equilibrium is a state of equilibrium between the same chemical species in different phases (solid,
liquid and gaseous). The various equilibria which can exist in any physical system are,


Solid ⇌ Liquid
⇌ Vapour
Liquid ⇌ Gas(vapour)
Solid ⇌ Saturated solution of solid in a liquid
Solid ⇌ Saturated solution of gas in a liquid
Gas(vapour
)

(1) Solid-liquid equilibrium Rate of transfer of molecules from water to ice
Rate of transfer of molecules from ice to water =

Rate of melting of ice = Rate of freezing of water

Free energy change and solid-liquid equilibrium in water : For ice-water system free energy change

(∆G), at 273 K, and one atmosphere pressure is zero i.e., ∆G = 0; Ice ⇌ Water; H 2O(s) ⇌ H 2O(l)

(i) At temperature higher than 273 K, and 1 atm pressure, ∆G < 0. Thus, the process in the forward direction
would become favourable and ice will melt to give more water.

(ii) At temperature less than 273 K, and 1 atm pressure, ∆G > 0. Thus, the reverse reaction will become
favourable, and more ice will be formed from liquid water.

(2) Liquid-vapour equilibrium : A liquid placed in an open container disappears completely. After some
time vapours of the liquid held in an open container can escape out to the atmosphere. Thus, when vapour of
liquid exists in equilibrium with the liquid, then

Rate of vaporisation = Rate of condensation,

H 2O(l) ⇌ H 2O(v) Y Vaporisation
Conditions necessary for a liquid-vapour equilibrium

(i) The system must be a closed system i.e., the amount of matter in Condensation
the system must remain constant.
Rate

(ii) The system must be at a constant temperature. Time X
(iii) The visible properties of the system should not change with time.

(3) Solid-vapour equilibrium : Certain solid substances on heating get converted directly into vapour
without passing through the liquid phase. This process is called sublimation. The vapour when cooled, gives back

the solid, it is called disposition.

Solid ⇌ Vapour

The substances which undergo sublimation are camphor, iodine,

ammonium chloride etc.

For exampie, Ammonium chloride whenhcehoaoetl ated sublimes. Saturated
solution of
NH 4 Cl (s) NH 4 Cl (v) Sugar molecules
sugar
(4) Equilibrium between a solid and its solution : When a
saturated solution is in contact with the solid solute, there exists a dynamic Solid
equilibrium between the solid and the solution phase. sugar


Solid substance ⇌ Solution of the substance
Example : Sugar and sugar solution. In a saturated solution, a dynamic equilibrium is established between

dissolved sugar and solid sugar.

Sugar (s) ⇌ Sugar (aq)
At the equilibrium state, the number of sugar molecules going into the solution from the solid sugar is equal to
the number of molecules precipitating out from the solution, i.e., at equilibrium,
Rate of dissolution of solid sugar = Rate of precipitation of sugar from the solution.
(5) Equilibrium between a gas and its solution in a liquid : Gases dissolve in liquids. The solubility of a
gas in any liquid depends upon the,
(i) Nature of the gas and liquid.

(ii) Temperature of the liquid.

(iii) Pressure of the gas over the surface of the solution.

Henry’s law : “At a certain temperature, the mass of a gas which dissolves in a definite volume of a liquid is
proportional to the pressure of the gas in equilibrium with the solution.”

m ∝ P or m = KP ; (where K is the proportionality constant)

Thus, at a constant temperature, the ratio of the molar concentration of the gas in the solution and into the
atmosphere is constant.

Limitations of Henry’s law

• Henry’s law is applicable to ideal gases only. Henry’s law should be applied only at low pressures because
real gases behave like ideal gases at low pressures.

• Henry’s law is not applicable to gases which react chemically with the solvent.
• Henry’s law will not apply to the solution of gases like ammonia (NH3 ) and hydrogen chloride (HCl) in
water because these gases react chemically with water.

Note :  A chilled soda water bottle fizzes out when opened because, soda water is a solution of carbon

dioxide gas, CO2 (g) in water at high pressure. As soon as the bottle is opened under normal

atmospheric conditions, the dissolved gas escapes out to reach a new equilibrium state, so that the

pressure of the gas inside the bottle becomes equal to the atmospheric pressure. At low pressure, the

solubility of the gas in water decreases.

Chemical equilibrium.

The equilibrium between different chemical species present in the same or different phases is called chemical
equilibrium. There are two types of chemical equilibrium.

(1) Homogeneous equilibrium : The equilibrium reactions in which all the reactants and the products are in
the same phase are called homogeneous equilibrium reactions.

Example : (i) C2 H5OH (l) + CH 3COOH (l) ⇌ CH 3COOC2 H5 (l) + H 2O(l)
liquid phase

(ii) N 2 (g) + 3H 2 (g) ⇌ 2NH 3 (g)
gas phase

(iii) 2SO2 (g) + O2 (g) ⇌ 2SO3 (g)
gas phase

(2) Heterogeneous equilibrium : The equilibrium reactions in which the reactants and the products are
present in different phases are called heterogeneous equilibrium reactions.


Example : (i) CaCO3 (s) ⇌ CaO(s) + CO2 (g)

(ii) 2NaHCO3 (s) ⇌ Na2CO3 (s) + CO2 (g) + H 2O(g)

(iii) H 2O(l) ⇌ H 2O(g)

(iv) Ca(OH)2 (s) + H 2O(l) ⇌ Ca 2+ (aq) + 2OH − (aq)

Note :  The equilibrium expression for heterogeneous reactions does not include the concentrations

of pure solids because their concentrations remain constant.

Reversible and irreversible reactions.

A chemical reaction is said to have taken place when the concentration of reactants decreases, and the
concentration of the products increases with time. The chemical reactions are classified on the basis of the extent to
which they proceed, into the following two classes;

(1) Reversible reactions : Reactions in which only a part of the total amount of reactants is converted into
products are termed as reversible reactions.

(i) Characteristics of reversible reactions
(a) These reactions can be started from either side,
(b) These reactions are never complete,
(c) These reactions have a tendency to attain a state of equilibrium,

(d) This sign (⇌) represents the reversibility of the reaction,
(e) Free energy change in a reversible reaction is zero (∆G = 0),
(ii) Examples of reversible reactions
(a) Neutralisation between an acid and a base either of which or both are weak e.g.,

CH 3COOH + Na OH ⇌ CH 3COONa + H 2O

(b) Salt hydrolysis, e.g., Fe Cl3 + 3H 2O ⇌ Fe (OH )3 + 6HCl

(c) Thermal decomposition, e.g., PCl5(g) ⇌ PCl 3(g) + Cl 2(g) − Q

CaCO3(s) ⇌ CaO(s) + CO2(g) ; 2HI(g) ⇌ H 2(g) + I 2 (g)

(d) Esterification, e.g., CH3COOH + C2H5OH ⇌ CH3COOC2H5 + H2O
(e) Evaporation of water in a closed vessel, e.g., H 2O(l) ⇌ H 2O(g) − Q

(f) Other reactions, e.g., N 2(g) + 3H 2(g) ⇌ 2NH 3(g) + Q ; 2SO2(g) + O2(g) ⇌ 2SO3(g) + Q

(2) Irreversible reactions : Reactions in which the entire amounts of the reactants are converted into
products are termed as irreversible reactions.

(i) Characteristics of irreversible reactions
(a) These reactions proceed only in one direction (forward direction),
(b) These reactions can proceed to completion,
(c) The arrow (→) is placed between reactants and products,
(d) In an irreversible reaction, ∆G < 0,

(ii) Examples of irreversible reactions


(a) Neutralisation between strong acid and strong base e.g. NaOH + HCl → NaCl + H 2O + 13.7 kcal

(b) Double decomposition reactions or precipitation reactions e.g.

BaCl 2(aq) + H 2 SO4 (aq) → BaSO4 (s) ↓ +2HCl(aq) ; AgNO3(aq) + NaCl(aq) → AgCl(g) ↓ + NaNO3(aq)

(c) Thermal decomposition, e.g. 2KClO3(s) MnO2,∆ → 2KCl(s) + 3O2 ↑

2Pb(NO3 )2 heat → 2PbO + 4 NO2 + O2 ↑ ; NH 4 NO2 heat → N 2 ↑ +2H 2O ↑

(d) Redox reactions, e.g., SnCl 2(aq) + 2FeCl 3(aq) → SnCl4 (aq) + 2FeCl 2(aq)

(e) Other reactions, e.g., NaCl + H 2SO4 → NaHSO4 + HCl ↑

Law of mass action.

On the basis of observations of many equilibrium reactions, two Norwegian chemists Guldberg and Waage
suggested (1864) a quantitative relationship between the rates of reactions and the concentration of the reacting
substances. This relationship is known as law of mass action. It states that

“The rate of a chemical reaction is directly proportional to the product of the molar concentrations of the
reactants at a constant temperature at any given time.”

The molar concentration i.e. number of moles per litre is also called active mass. It is expressed by enclosing the
symbols of formulae of the substance in square brackets. For example, molar concentration of A is expressed as [A].

Let us consider a simple reaction between the species A and B : A + B → Products

According to law of mass action, rate of reaction, r ∝ [A] [B] = k [A] [B]

Where [A] and [B] are the molar concentrations of the reactants A and B respectively, k is a constant of
proportionality for the forward reaction and is known as rate constant. The rate constant is also called velocity
constant. Now, if the concentration of each of the reactants involved in the reaction is unity, i.e., [A] = [B] = 1,
then, rate of reaction, r = k × 1 × 1 or r = k

Thus, the rate constant of a reaction at a given temperature may be defined as “the rate of the reaction when
the concentration of each of the reactants is unity.”

For a general reaction, aA + bB + cC → Products

The law of mass action may be written as : Rate of reaction, r = k[A]a [B]b [C]c

Thus, the law of mass action may be restated as, “The rate of a chemical reaction at any particular
temperature is proportional to the product of the molar concentrations of reactants with each concentration term
raised to the power equal to the number of molecules of the respective reactants taking part in the reaction.”

The number of molecules of a reactant taking part in a reaction is also called its stoichiometric coefficient.
For example, a, b and c…. in the above equation are called stoichiometric coefficients of A, B and C….
respectively.


Equilibrium constant.

(1) Equilibrium constant in terms of law of mass action : The law of mass action may be applied to a
reversible reaction to derive a mathematical expression for equilibrium constant known as law of chemical
equilibrium.

Let us consider a simple reversible reaction, A + B ⇌ X + Y in which an equilibrium exists between the
reactants (A and B) and the products (X and Y). The forward reaction is,

A+B→X+Y

According to law of mass action,

Rate of forward reaction ∝ [A][B] = k f [A][B]

Where k f is the rate constant for the forward reaction and [A] and [B] are molar concentrations of reactants A
and B respectively.

Similarly, the backward reaction is ; X + Y → A + B

Rate of backward reaction ∝ [X][Y] = kb [X][Y]

Where kb is the rate constant for the backward reaction and [X] and [Y] are molar concentrations of products
X and Y respectively.

At equilibrium, the rates of two opposing reactions become equal. Therefore, at equilibrium,

Rate of forward reaction = Rate of backward reaction

k f [A][B] = kb [X][Y]

kf = [X][Y] or K = [X][Y]
kb [ A][B] [ A][B]

The combined constant K, which is equal to k f / kb is called equilibrium constant and has a constant value
for a reaction at a given temperature. The above equation is known as law of chemical equilibrium.

For a general reaction of the type : aA + bB ⇌ cC + dD

The equilibrium constant may be represented as : K = [C]c [D]d
[A]a [B]b

where the exponents a, b, c and d have the same values as those in the balanced chemical equation. Thus,
the equilibrium constant may be defined as,

“The ratio between the products of molar concentrations of the products to that of the molar concentrations of
the reactants with each concentration term raised to a power equal to its stoichiometric coefficient in the balanced
chemical equation at a constant temperature.”

(2) Characteristics of equilibrium constant
(i) The value of equilibrium constant is independent of the original concentration of reactants.

For example, the equilibrium constant for the reaction,

Fe 3+ (aq) + SCN − (aq) = FeSCN 2+ (aq) ; K = [FeSCN 2+ ] = 138.0 Lmol −1 (at 298 K)
[Fe 3+ ][SCN − ]

Whatever may be the initial concentrations of the reactants, Fe 3+ and SCN − ions, the value of K comes out
to be 138.0 L mol–1 at 298 K.


(ii) The equilibrium constant has a definite value for every reaction at a particular temperature. However, it
varies with change in temperature.

For example, the equilibrium constant for the reaction between hydrogen and iodine to form hydrogen iodide
is 48 at 717 K.

H 2(g) + I 2(g) = 2HI(g); K = [HI]2 = 48
[H 2 ][I 2 ]

For this reaction, the value of K is fixed as long as the temperature remains constant.

(iii) For a reversible reaction, the equilibrium constant for the forward reaction is inverse of the equilibrium

constant for the backward reaction.

For example, if equilibrium constant, K, for the reaction of combination between hydrogen and iodine at 717 K is 48

H 2(g) + I 2(g) ⇌ 2HI (g); K′ = [HI]2 = 48
[H 2 ][I 2 ]

Then, the equilibrium constant for the decomposition of hydrogen iodide is the inverse of the above
equilibrium constant.

2HI(g) ⇌ H 2 (g) + I 2 (g) ; K = [H 2 ][I 2 ] = 1 = 1 = 0.02
[HI]2 K′ 48

In general, K = K ′ 1forward reaction

backward reaction

(iv) The value of an equilibrium constant tells the extent to which a reaction proceeds in the forward or reverse

direction. If value of K is large, the reaction proceeds to a greater extent in the forward direction and if it is small,

the reverse reaction proceeds to a large extent and the progress in the forward direction is small.

(v) The equilibrium constant is independent of the presence of catalyst. This is so because the catalyst affects

the rates of forward and backward reactions equally.

(vi) The value of equilibrium constant changes with the change of temperature. Thermodynamically, it can be

shown that if K1 and K2 be the equilibrium constants of a reaction at absolute temperatures T1 and T2 . If ∆H is
the heat of reaction at constant volume, then

log K2 − log K1 = − ∆H  1 − 1 (Van’t Hoff equation)
 
2.303 R  T2 T1 

The effect of temperature can be studied in the following three cases

(a) When ∆H = 0 i.e., neither heat is evolved nor absorbed

log K2 − log K1 = 0 or log K2 = log K1 or K2 = K1

Thus, equilibrium constant remains the same at all temperatures.

(b) When ∆H = +ve i.e., heat is absorbed, the reaction is endothermic. The temperature T2 is higher than T1 .

log K2 − log K1 = +ve or log K2 > log K1 or K2 > K1

The value of equilibrium constant is higher at higher temperature in case of endothermic reactions.

(c) When ∆H = –ve, i.e., heat is evolved, the reaction is exothermic. The temperature T2 is higher than T1 .

log K2 − log K1 = −ve or log K1 > log K2 or K1 > K2

The value of equilibrium constant is lower at higher temperature in the case of exothermic reactions.

(vii) The value of the equilibrium constant depends upon the stoichiometry of the chemical equation.


Examples :

(a) If the equation (having equilibrium constant K) is divided by 2, then the equilibrium constant for the new

equation is the square root of K i.e. K . For example, the thermal dissociation of SO3 can be represented in two
ways as follows,

2SO3 (g) ⇌ 2SO2 (g) + O2 (g) and SO3 (g) ⇌ SO2 (g) + 1 / 2O2 (g)

K = [SO2 ]2 [O2 ] and K′ = [SO2][O2]1 / 2 ; K′ = K or (K)1/ 2
[SO3 ]2 [SO3 ]

(b) Similarly, if a particular equation is multiplied by 2, the equilibrium constant for the new reaction (K′) will

be the square of the equilibrium constant (K) for the original reaction i.e., K′ = K 2

(c) If the chemical equation for a particular reaction is written in two steps having equilibrium constants K1
and K2 , then the equilibrium constants are related as K = K1 × K2

For example, the reaction N 2(g) + 2O2(g) ⇌ 2NO2(g) with equilibrium constant (K) can be written in two steps :

N 2 (g) + O2 (g) ⇌ 2NO (g) ; (Equilibrium constant = K1 )

2NO(g) + O2(g) ⇌ 2NO2(g) ; (Equilibrium constant = K2 )

Now, K1 [NO]2 and K2 = [NO2 ]2
= [N 2 ][O2 ] [NO]2[O2 ]

Therefore, K1 × K2 = [NO]2 × [NO2 ]2 = [NO2 ]2 =K
[N 2 ][O2 ] [NO]2[O2 ] [N 2 ][O2 ]2

(3) Types of equilibrium constant : Generally two types of equilibrium constants are used,

(i) Kc → It is used when the various species are generally expressed in terms of moles/litre or in terms of molar
concentrations.

(ii) Kp → It is used when in gaseous reactions, the concentration of gases expressed in terms of their partial
pressures.

Kp is not always equal to Kc. Kp and Kc are related by the following expression, K p = Kc (RT)∆n

where, R = Gas constant = 0.0831 bar dm3 mol −1 k −1 ; T = Temperature in Kelvin

∆n = number of moles of gaseous products – number of moles of gaseous reactants in chemical equation

(4) Unit of equilibrium constant : Equilibrium constant K has no units i.e., dimensionless if the total
number of moles of the products is exactly equal to the total number of moles of reactants. On the other hand if the
number of mioles of products and reactants are not equal, K has specific units.

Units of Kp and Kc and the value of ∆n

Value of ∆n Relation between Kp and Kc Units of Kp Units of Kc

0 Kp = Kc No unit No unit

>0 Kp > Kc (atm)∆n (mole l–1)∆n

<0 Kp < Kc (atm)∆n (mole l–1)∆n


(5) Applications of equilibrium constant : Knowing the value of the equilibrium constant for a chemical
reaction is important in many ways. For example, it judge the extent of the reaction and predict the direction of the
reaction.

(i) Judging the extent of reaction

We can make the following generalisations concerning the composition of equilibrium mixture.

(a) If Kc > 103 , products predominate over reactants. If Kc is very large, the reaction proceeds almost all
the way to completion.

(b) If Kc < 10−3 , reactants predominate over products. If Kc is very small, the reaction proceeds hardly at all.

(c) If Kc is in the range 10−3 to 103 , appreaciable concentration of both reactants and products are present.
This is illustrated as follows,

Kc

Reaction proceeds 10–3 103 Reaction proceeds
hardly at all to completion

Both reactants and
products are present

at equilibrium

(ii) Reaction quotient and predicting the direction of reaction : The concentration ratio, i.e., ratio of
the product of concentrations of products to that of reactants is also known as concentration quotient and is
denoted by Q.

Concentration quotient, Q = [X][Y] .
[ A][B]

It may be noted that Q becomes equal to equilibrium constant (K) when the reaction is at the equilibrium
state. At equilibrium, Q = K = Kc = K p . Thus,

(a) If Q > K, the reaction will proceed in the direction of reactants (reverse reaction).

(b) If Q < K, the reaction will proceed in the direction of the products (forward reaction).

(c) If Q = K, the reaction mixture is already at equilibrium.

Thus, a reaction has a tendency to form products if Q < K and to form reactants if Q > K.

This has also been shown in figure,

Q Kc Q Kc Q Kc

Reactants → Products Equilibrium Reactants ← Products


(6) Calculation of equilibrium constant : We have studied in the characteristics of the equilibrium constant
that its value does not depend upon the original concentrations of the reactants and products involved in the
reaction. However, its value depends upon their concentrations at the equilibrium point. Thus, if the equilibrium
concentrations of the species taking part in the reaction be known, then the value of the equilibrium constant and
vice versa can be calculated.

Homogeneous equilibria and equations for equilibrium constant (Equilibrium pressure is P atm in a V L flask)

∆n = 0 ; Kp = Kc ∆n < 0 ; Kp < Kc ∆n > 0; Kp > Kc

H2 + I2 ⇌ 2HI N2 + 3H2 ⇌ 2NH3 2SO2 + O2 ⇌ 2 SO3 PCl ⇌ PCl3 + Cl2
(g) (g) (g) (g) (g) (g) 5
(g) (g) (g) (g) (g) (g)

Initial mole 1 1 0 1 3 0 21 0 1 00
(1– x) (1–x) (3–3x) 2x (2–2x) (1–x) 2x
Mole at (1–x) (1–x) x x
2x 2
Equilibrium

Total mole at (4 – 2x) (3 – x) (1 + x)
equilibrium

Active masses  1 − x   1 − x  2x  1 − x  3  1 − x   2x   2 − 2x   1 − x   2x   1 − x   x   x 
 V   V  V  V   V   V   V   V   V   V   V   V 

Mole fraction  1 − x   1 − x  2x 1− x 3  1 − x  x  2 − 2x   1 − x   2x   1 − x   x   x 
 2   2  2 2  2 − x  (2 − x)  3−x  3−x 3−x 1+ x  +   + 
2(2 − x) 1 x 1 x

Partial pressure p  1 − x  p  1 − x  p  2x  P 1 − x  P  3(1 − x)  Px P  2 − 2x  P 1 − x  P  2x  P 1 − x  P x  P x 
 2   2   2  2(2 − x) _ 2(2 − x) (2 − x)  3−x   3 − x   3−x  1+ x 1+ x 1+ x
Kc
Kp 4x2 4x2V 2 x 2V x2

(1 − x) 2 27 (1 − x) 4 (1 − x) 3 (1 − x)V

4x2 16x 2 (2 − x) 2 x2 (3 − x) Px 2
27(1 − x) 4 P 2 P (1 − x) 3
(1 − x) 2 (1 − x2)

Heterogeneous equilibria and equation for equilibrium constant (Equilibrium pressure is P atm)

NH4 HS(s) ⇌ NH3(g) + H2S(g) C(s) + CO2(g) ⇌ 2CO(g) NH2CO2NH4(s) ⇌ 2NH3(g) + CO2(g)

Initial mole 1 0 11 0 1 00
0
Mole at equilibrium
(1–x) x (1–x) (1–x) (1–x) 2x x
Total moles at equilibrium x 2x 2x 3x
(solid not included)
Mole fraction (1+x)

x = 1 1  1 − x  21
2x 2 2  1 + x  33

 2x  2P
 1+ x  3

Partial pressure PP P 1 − x  4P3
22  1 + x  27

P 2x  P
 1+ x  3

Kp P2 4P2x2
(1 − x 2)
4


(7) Equilibrium constant and standard free energy change : Standard free energy change of a reaction
and its equilibrium constant are related to each other at temperature T by the following relation,

∆Go = −2.303 RT log K

when, ∆Go = −ve , the value of equilibrium constant will be large positive quantity and
when, ∆Go = +ve , the value of equilibrium constant is less than 1 i.e., low concentration of products at
equilibrium state.

Example : 1 In the reversible reaction A + B ⇌ C + D, the concentration of each C and D at equilibrium was 0.8
Solution: (d) mole/litre, then the equilibrium constant Kc will be
[MP PET 1986]

(a) 6.4 (b) 0.64 (c) 1.6 (d) 16.0

Suppose 1 mole of A and B each taken then 0.8 mole/litre of C and D each formed remaining concentration
of A and B will be (1 – 0.8) = 0.2 mole/litre each.

Kc = [C][D] = 0.8 × 0.8 = 16.0
[ A][B] 0.2 × 0.2

Example : 2 For the system A(g) + 2B(g) ⇌ C(g), the equilibrium concentrations are (A) 0.06 mole/litre (B) 0.12 mole/litre
Solution: (a) (C) 0.216 mole/litre. The Keq for the reaction is
Example : 3 [CPMT 1983]
Solution: (b)
Example : 4 (a) 250 (b) 416 (c) 4 × 10 −3 (d) 125
Solution: (a)
For reaction A + 2B ⇌ C; K eq [C] = 0.216 = 250
Example : 5 = [A][B]2 0.06 × 0.12 × 0.12

Molar concentration of O2 is 96 gm, it contained in 2 litre vessel, active mass will be

(a) 16 mole/litre (b) 1.5 mole/litre (c) 4 mole/litre (d) 24 mole/litre

weight weight 96 3
M.wt. × Volume 32 × 2 2
Active mass = M.wt. = = = = 1.5 mol / litre
Volume

2 moles of PCl5 were heated in a closed vessel of 2 litre capacity. At equilibrium, 40% of PCl5 is dissociated

into PCl3 and Cl2. The value of equilibrium constant is [MP PMT 1989]

(a) 0.266 (b) 0.53 (c) 2.66 (d) 5.3

At start, PCl5 ⇌ PCl 3 + Cl 2
2 00

At equilibrium. 2 × 60 2 × 40 2 × 40
100 100 100

Volume of cantainer = 2 litre

2× 40 × 2× 40
100 ×2 100 ×2
∴ Kc = 2 × 60 = 0.266

100 × 2

A mixture of 0.3 mole of H 2 and 0.3 mole of I 2 is allowed to react in a 10 litre evacuated flask at 500o C .
The reaction is H2 + I2 ⇌ 2HI the Kc is found to be 64. The amount of unreacted I 2 at equilibrium is

[KCET 1990]

(a) 0.15 mole (b) 0.06 mole (c) 0.03 mole (d) 0.2 mole


Solution: (b) Kc = [HI]2 ] ; 64 = x2
[H 2 ][I 2 0.03 × 0.03

x 2 = 64 × 9 × 10 −4 ; x = 8 × 3 × 10−2 = 0.24
x is the amount of HI at equilibrium. Amount of I 2 at equilibrium will be
0.30 − 0.24 = 0.06 mole

Example : 6 The rate constant for forward and backward reactions of hydrolysis of ester are 1.1 × 10 −2 and 1.5 × 10 −3 per
minute respectively. Equilibrium constant for the reaction,

CH 3COOC2 H 5 + H 2O ⇌ CH 3COOH + C2 H 5 OH is [AIIMS 1999]

(a) 4.33 (b) 5.33 (c) 6.33 (d) 7.33

Solution: (d) Kf = 1.1 × 10 −2 , Kb = 1.5 × 10 −3 ; Kc = Kf = 1.1 × 10 −2 = 7.33
Kb 1.5 × 10 −3

Example : 7 For the reaction PCl3 (g) + Cl2 (g) ⇌ PCl5 at 250o C , the value of Kc is 26, then the value of K p on the
Solution: (a)
same temperature will be [MNR 1990; MP PET 2001]

(a) 0.61 (b) 0.57 (c) 0.83 (d) 0.46

∆ng = 1 – 2 = –1

∴ K p = K c (RT)∆n ;  K p = Kc (RT)−1

Example : 8 Since R = 0.0821 litre atm k–1 mol–1, T = 250oC = 250 + 273 = 523 K p = 26(0.0821× 523)−1 = 0.61
If Kp for reaction A(g) + 2B(g) ⇌ 3C(g) + D(g) is 0.05 atm at 1000K its Kc in term of R will be [CBSE PMT 1989]

(a) 5 × 10 −4 (b) 5 (c) 5 × 10 −5 (d) None of these
R R R

Solution: (c) Kp = Kc (RT)∆n ⇒ 5 × 10 −2 = Kc (R × 1000)1 ⇒ Kc = 5 × 10 −5
R

Example : 9 If the equilibrium constant of the reaction 2HI ⇌ H2 + I2 is 0.25, then the equilibrium constant of the reaction

H2 + I2 ⇌ 2HI would be [MP PMT 1989, 95]

(a) 1.0 (b) 2.0 (c) 3.0 (d) 4.0

Solution: (d) Kc for the IInd reaction is reverse of Ist for reaction 2HI ⇌ H 2 + I 2 is 0.25Kc′ for reaction, H 2 + I 2 ⇌

2HI will be K c′ = 1 = 1 = 4 .
Kc 0.25

Example : 10 If equilibrium constant for reaction 2AB ⇌ A2 + B2 , is 49, then the equilibrium constant for reaction

AB ⇌ 1 A2 + 1 B2 , will be [MP PMT 2003; EAMCET 1998]
2 2

(a) 7 (b) 20 (c) 49 (d) 21

Solution: (a) 2AB ⇌ A2 + B2

Kc = [A2 ][B2 ] = 49
[ AB] 2


For reaction AB ⇌ 1 A2 + 1 B2
2 2

K c′ [A2 ]1/ 2[B2 ]1/ 2
[ AB]
=

Kc′ = Kc = 49 = 7

Example : 11 For the reaction 2NO2(g) ⇌ 2NO(g) + O2(g) Kc = 1.8 × 10 −6 at 185o C . At 185o C , the value of Kc′ for the

reaction NO(g) + 1 O2(g) ⇌ NO2(g) is [UPSEAT 2000]
2

(a) 0.9 × 106 (b) 7.5 × 10 2 (c) 1.95 × 10−3 (d) 1.95 × 103

Solution: (b) Reaction is reversed and halved.

∴ Kc′ = 1; Kc′ = 1 = 7.5 × 10 2
Kc 1.8 × 10−6

Example : 12 In an equilibrium reaction for which ∆G0 = 0 the equilibrium constant K p should be [BHU 1987]

(a) 0 (b) 1 (c) 2 (d) 10

Solution: (b) If ∆G0 = 0 and ∆G0 = −2.303RT log K p
log K p = 0 , K p = 1

Example : 13 ∆G0 for HI(g) ≅ +1.7kJ / mole what is the equilibrium constant at 25o C for 2HI(g) ⇌ H 2(g) + I 2(g)

[KCET 1992]

(a) 24.0 (b) 3.9 (c) 2.0 (d) 0.5

Solution: (d) ∆G0 = −2.303RT log K p = −2.303 × 8.314 × 10−3 × 298 log K p

1.7 = −2.303 × 8.314 × 10−3 × 298 log Kp

K p = 0.5

Example : 14 It is found that the equilibrium constant increases by a factor of four when the temperature is increased from
25o C to 40o C . The value of ∆H o is

(a) 25.46 kJ mol −1 (b) 171.67 kJ mol −1 (c) 89.43 kJ mol −1 (d) 71.67 kJ mol −1

Solution: (d) Using the equation,

log (K p )40o C = ∆H  1 − 1  ,
(K p ) 25 o C 2.303R T1 T2

we get log 4 = ∆H  1 25 − 1 40 
2.303 × 8.314  273 + 273 + 

∴ ∆H = 71.67 kJ mol −1

Example : 15 Kc for the reaction A (g) + B (g) ⇌ 2C(g) is 3.0 at 400K. In an experiment “a” mol of A is mixed with 3 mol
of B in a 1-L vessel. At equilibrium 3 mol of C is formed. The value of ‘a’ will be,

(a) 4.5 mol (b) 9.5 mol (c) 2.5 mol (d) 3.5 mol


Solution: (d) A(g) + B(g) ⇌ 2C(g)
a3 0
a−x 3−x 2x

From the equation, 2x = 3 ⇒ x = 1.5

Kc = (a − 4x2 − x) ; 3= (a 4 × (1.5)2 ; 3= 4 × 2.25 ⇒ a = 3.5
x)(b − 1.5)(3 − 1.5) (a − 1.5)(1.5)

Example : 16 For the reaction AB(g) ⇌ A(g) + B(g), AB is 33% dissociated at a total pressure of P. Then

(a) P = K p (b) P = 4K p (c) P = 3K p (d) p = 8K p

Solution: (d) AB(g) ⇌ A(g) + B(g)
1 00
−1 / 3 +1 / 3 +1 / 3
2/3 1/3 1/3

(∑ )n eq / m = 2 + 1 + 1 = 4
3 3 3 3

PA PB 1/ 3 P 1/ 3 P 1
PAB 4/3 4/3 8
Kp = = = P
2 / 3
4 / 3 P

P = 8Kp

Example : 17 The total pressure observed at equilibrium in the dissociation of solid ammonium carbamate at a certain
temperature is 2.0 atm. The equilibrium constant K p is

(a) 4.185 (b) 1.185 (c) 2.276 (d) 1.072

Solution: (b) NH 4 COONH 2 (s) ⇌ 2NH 3 (g) + CO2 (g)
2 1
3 P 3 P

Kp = PN2H3 PCO2 =  2 P  2  1 P  = 4 P 3 = 4 × (2)3 = 1.185
 3   3  27 27

Example : 18 At the equilibrium of the reaction N 2O4 (g) ⇌ 2NO2 (g) , the observed molar mass of N 2O4 is 77.70 g. The
percentage dissociation of N 2O4 is

(a) 28.4 (b) 46.7 (c) 22.4 (d) 18.4

Solution: (d) α = M Th − M obs ; Molar mass of N 2O4 = 92 g mol −1
M obs (n − 1)

Here, n = 2; α = 92.00 − 77.70 = 0.184 = 18.4%
77.70(2 − 1)

Example : 19 In the equilibrium H 2O(g) ⇌ H 2(g) + 1 O2 (g) , the extent of dissociation of water when p = 1 atm and
2

K = 2.08 × 10 −3 is approximately

(a) 2% (b) 0.2% (c) 20% (d) 1%


Solution: (a) For the equilibrium H 2O (g) → H 2(g) + 1 O2 (g)
2

Kp = α 3/2P1/2 P = 1atm
2

α = ( 2.K p )2 / 3 = 0.0205 ≈ 2%

Factors which Change the State of Equilibrium : Le-Chatelier's Principle.

Le-Chatelier and Braun (1884), French chemists, made certain generalizations to explain the effect of
changes in concentration, temperature or pressure on the state of system in equilibrium. When a system is subjected
to a change in one of these factors, the equilibrium gets disturbed and the system readjusts itself until it returns to
equilibrium. The generalization is known as Le-Chatelier's principle. It may stated as :

“Change in any of the factors that determine the equilibrium conditions of a system will shift the equilibrium in
such a manner to reduce or to counteract the effect of the change.”

The principle is very helpful in predicting qualitatively the effect of change in concentration, pressure or
temperature on a system in equilibrium. This is applicable to all physical and chemical equilibria.

(1) Effect of change of concentration : According to Le-Chatelier's principle, “If concentration of one or
all the reactant species is increased, the equilibrium shifts in the forward direction and more of the products are
formed. Alternatively, if the concentration of one or all the product species is increased, the equilibrium shifts in the
backward direction forming more reactants.”

Thus,

Increase in concentration of any of the reactants Shifts the → Forward direction
equilibrium to

Increase in concentration of any of the products Shifts the → Backward direction
equilibrium to

(2) Effect of change of temperature : According to Le-Chatelier's principle, “If the temperature of the
system at equilibrium is increased (heat is supplied), the equilibrium will shift in the direction in which the added
heat is absorbed. In other words, the equilibrium will shift in the direction of endothermic reaction with increase
in temperature. Alternatively, the decrease in temperature will shift the equilibrium towards the direction in which
heat is produced and, therefore, will favour exothermic reaction.”

Thus,

Increase in temperature Shifts the equilibrium → Endothermic reaction
in the direction of

Decrease in temperature Shifts the equilibrium → Exothermic reaction
in the direction of

(3) Effect of change of pressure : Pressure has hardly effect on the reactions carried in solids and liquids.
However, it does influence the equilibrium state of the reactions that are carried in the gases. The effect of pressure
depends upon the number of moles of the reactants and products involved in a particular reaction. According to Le-
Chatelier's principle, “Increase in pressure shifts the equilibrium in the direction of decreasing gaseous moles.


Alternatively, decrease in pressure shifts the equilibrium in the direction of increasing gaseous moles and pressure
has no effect if the gaseous reactants and products have equal moles.”

Thus,

Increase in pressure Shifts the equilibrium → Decreasing gaseous moles
in the direction of

Decrease in pressure Shifts the equilibrium → Increasing gaseous moles
in the direction of

(4) Effect of volume change : We know that increase in pressure means decrease in volume, so the effect
of change of volume will be exactly reverse to that of pressure. Thus, “decreasing the volume of a mixture of gases
at equilibrium shifts the equilibrium in the direction of decreasing gaseous moles while increasing the volume shifts
the equilibrium in the direction of increasing gaseous moles.”

Thus,
Increase in volume Shifts the equilibrium → Increasing gaseous moles

in the direction of

Decrease in volume Shifts the equilibrium → Decreasing gaseous moles
in the direction of

(5) Effect of catalyst : Catalyst has no effect on equilibrium. This is because, catalyst favours the rate of
forward and backward reactions equally. Therefore, the ratio of the forward to reverse rates remains same and no
net change occurs in the relative amount of reactants and products present at equilibrium. Thus, a catalyst does
not affect the position of the equilibrium. It simply helps to achieve the equilibrium quickly. It may also be
noted that a catalyst has no effect on the equilibrium composition of a reaction mixture.

Thus, Catalyst does not shift the equilibrium in any direction

(6) Effect of addition of inert gas : The addition of an inert gas (like helium, neon, etc.) has the following
effects on the equilibrium depending upon the conditions :

(i) Addition of an inert gas at constant volume : When an inert gas is added to the equilibrium system
at constant volume, then the total pressure will increase. But the concentrations of the reactants and products (ratio
of their moles to the volume of the container) will not change. Hence, under these conditions, there will be no effect
on the equilibrium.

(ii) Addition of an inert gas at constant pressure : When an inert gas is added to the equilibrium
system at constant pressure, then the volume will increase. As a result, the number of moles per unit volume of
various reactants and products will decrease. Hence, the equilibrium will shift in a direction in which there is
increase in number of moles of gases.

Thus,
Addition of an inert gas V =constant → No effect on the equilibrium.

Addition of an inert gas P =constan t → Increasing gaseous moles.
Shifts the equilibrium
in the direction of


Applications of Le-Chatelier's Principle.

The Le-Chateliers principle has a great significance for the chemical, physical systems and in every day life in a
state of equilibrium. Let us discuss in brief a few applications.

(1) Applications to the chemical equilibrium : With the help of this principle, most favourable conditions
for a particular reaction can be predicted.

(i) Synthesis of ammonia (Haber’s process): N 2 + 3H 2 ⇌ 2NH3 + 23kcal (exothermic)
1 vol 3 vol 2 vol

(a) High pressure (∆n < 0) (b) Low temperature (c) Excess of N 2 and H 2 (d) Removal of NH3 favours
forward reaction.

(ii) Formation of sulphur trioxide : 2SO2 + O2 ⇌ 2SO3 + 45 kcal (exothermic)
2 vol 1 vol 2 vol

(a) High pressure (∆n < 0) (b) Low temperature (c) Excess of SO2 and O2 , favours the reaction in forward
direction.

(iii) Synthesis of nitric oxide : N 2 + O2 ⇌ 2NO − 43.2 kcal (endothermic )
1 vol 1 vol 2 vol

(a) High temperature (b) Excess of N 2 and O2 (c) Since reaction takes place without change in volume i.e.,
∆n = 0 , pressure has no effect on equilibrium.

(iv) Formation of nitrogen dioxide : 2NO + O2 ⇌ 2NO2 + 27.8 Kcal
2 vol 1 vol 2 vol

(a) High pressure (b) Low temperature (c) Excess of NO and O2 favours the reaction in forward direction.

(v) Dissociation of phosphours pentachloride : PCl5 ⇌ PCl3 + Cl 2 − 15 kcal
1 vol 1 vol 1 vol

(a) Low pressure or high volume of the container, ∆n > 0 (b) High temperature (c) Excess of PCl5 .
(2) Applications to the physical equilibrium : Le-Chatelier's principle is applicable to the physical
equilibrium in the following manner;

(i) Melting of ice (Ice – water system) : Ice ⇌ Water − x kcal

(Greater Volume) (Lesser Volume)

(In this reaction volume is decreased from 1.09 c.c. to 1.01 c.c. per gm.)

(a) At high temperature more water is formed as it absorbs heat.(b) At high pressure more water is formed as it
is accompanied by decrease in volume.(c) At higher pressure, melting point of ice is lowered, while boiling point of
water is increased.

(ii) Melting of sulphur : S(s) ⇌ S(l) − x kcal

(This reaction accompanies with increase in volume.)
(a) At high temperature, more liquid sulphur is formed. (b) At higher pressure, less sulphur will melt as
melting increases volume.(c) At higher pressure, melting point of sulphur is increased.

(iii) Boiling of water (water- water vapour system) : Water ⇌ Water Vapours− x kcal

(Low volume) (Higher volume)

(It is accompanied by absorption of heat and increase in volume.)

(a) At high temperature more vapours are formed.(b) At higher pressure, vapours will be converted to liquid as
it decreases volume.(c) At higher pressure, boiling point of water is increased (principle of pressure cooker).


(iv) Solubility of salts : If solubility of a salt is accompanied by absorption of heat, its solubility increases

with rise in temperature; e.g., NH 4 Cl, K2SO4 , KNO3 etc. KNO3(s) + (aq) → KNO3(aq) − x kcal

On the other hand if it is accompanied by evolution of heat, solubility decreases with increase in temperature;

e.g., CaCl 2 , Ca(OH)2 , NaOH, KOH etc. Ca(OH)2( s) + (aq) → Ca(OH)2 (aq) + x kcal

(3) Application in every day life : We have studied the application of the Le-Chatelier's principle to some
equilibria involved in the physical and chemical systems. In addition to these, the principle is also useful to explain
certain observations which we come across in every day life. A few out of them are discussed below,

(i) Clothes dry quicker in a windy day : When wet clothes are spread on a stand, the water evaporates

and the surrounding air tends to get saturated thus hampering the process of drying. On a windy day when breeze

blows, the nearby wet air is replaced by dry air which helps the process of evaporation further. Thus, clothes dry

quicker when there is a breeze.

(ii) We sweat more on a humid day : The explanation is the same as given above. In a humid day, the air
is already saturated with water vapours. This means that the water that comes out of the pores of the body as sweat
does not vaporise. This will result in greater sweating in a humid day.

(iii) Carriage of oxygen by haemoglobin in blood : The haemoglobin (Hb) in red corpuscles of our blood

carries oxygen to the tissues. This involves the equilibrium, Hb(s) + O2 (g) ⇌ HbO2 (s)

The blood that is in equilibrium with the oxygen of the air in the lungs finds a situation in the tissues where the
partial pressure of oxygen is low. According to Le-Chatelier's principle, the equilibrium shifts towards the left so that
some of the oxyhaemoglobin changes to haemoglobin giving up the oxygen. When the blood returns to the lungs,
the partial pressure of the oxygen is higher and the equilibrium favours the formation of more oxyhaemoglobin.

(iv) Removal of carbon dioxide from the tissues by blood : Blood removes CO2 from the tissues. The

equilibrium is, CO2(g) + H2O(l)⇌ H2CO3(aq) ⇌ H + (aq) + HCO3− (aq)

Carbon dioxide dissolves in the blood in the tissues since the partial pressure of CO2 is high. However in the
lungs, where the partial pressure of CO2 is low, it is released from the blood.

(v) Sweet substances cause tooth decay : Tooth enamel consists of an insoluble substance called
hydroxyapatite, Ca5 (PO4 )3 OH. The dissolution of this substance from the teeth is called demineralization and its
formation is called remineralization. Even with the healthy teeth, there is an equilibrium in the mouth as

exothermic 5Ca 2+ (aq) + 3PO43− (aq) + OH − (aq)

Ca5 (PO4 )3 OH(s)
endothermic

When sugar substances are taken, sugar is absorbed on teeth and gets fermented to give H+ ions. The H+ ions

produced distrub the equilibrium by combining with OH– to form water and with PO43− to form HPO42− . Removal of
products cause the equilibrium to shift towards right and therefore, Ca5 (PO4 )3 OH dissolves causing tooth decay.

Relation between vapour density and degree of dissociation.

In the following reversible chemical equation.

Initial mol A ⇌ yB
10

At equilibrium (1–x) yx x = degree of dissociation

Number of moles of A and B at equilibrium = 1 − x + yx = 1 + x(y − 1)

If initial volume of 1 mole of A is V, then volume of equilibrium mixture of A and B is, = [1 + x(y − 1)]V

Molar density before dissociation, D= molecular weight = m
volume V


Molar density after dissociation, d m ; D = [1 + x(y − 1)] ; x = D−d
= [1 + x(y − 1)]V d d(y − 1)

y is the number of moles of products from one mole of reactant. D is also called Van’t Hoff factor.
d
In terms of molecular mass,

x = M −m ; Where M = Initial molecular mass, m= molecular mass at equilibrium
(y − 1)m

Thus for the equilibria

(I) PCl 5(g) ⇌ PCl 3(g) + Cl 2(g) , y = 2 (II) N 2O4(g) ⇌ 2NO2(g) , y = 2 (III) 2NO2 ⇌ N 2O4 , y = 1
2

∴ x = D−d (for I and II) and x = 2(d − D) (for III)
d d
Also D × 2 = Molecular weight (theoretical value)

d × 2 = Molecular weight (abnormal value) of the mixture

***


Redox reactions

Chemical reactions involve transfer of electrons from one chemical substance to another. These
electron – transfer reactions are termed as oxidation-reduction or redox-reactions.

Redox reactions play an important role in our daily life. These reactions are accompanied by energy
changes in the form of heat, light, electricity etc. Generation of electricity in batteries and many industrial
processes such as production of caustic soda, KMnO4 , extraction of metals like sodium, iron and aluminium
are common examples of redox reactions.

Molecular and Ionic equations.

(1) Molecular equations : When the reactants and products involved in a chemical change are written in
molecular forms in the chemical equation, it is termed as molecular equation.

Examples : (i) MnO2 + 4 HCl → MnCl2 + 2H2O + Cl2

(ii) 2FeCl3 + SnCl2 → 2FeCl2 + SnCl4

In above examples, the reactants and products have been written in molecular forms, thus the equation is
termed as molecular equation.

(2) Ionic equations : When the reactants and products involved in a chemical change are ionic compounds,
these will be present in the form of ions in the solution. The chemical change is written in ionic forms in chemical
equation, it is termed as ionic equation.

Examples : (i) MnO2 + 4 H + + 4 Cl − → Mn2+ + 2Cl − + 2H 2O + Cl2

(ii) 2Fe 3+ + 6Cl − + Sn2+ + 2Cl − → 2Fe 2+ + 4Cl − + Sn4+ + 4Cl −

In above examples, the reactants and products have been written in ionic forms, thus the equation is termed
as ionic equation.

(3) Spectator ions : In ionic equations, the ions which do not undergo any change and equal in number in
both reactants and products are termed spectator ions and are not included in the final balanced equations.

Example : Zn + 2H + + 2Cl− → Zn2+ + H2 + 2Cl − (Ionic equation)

Zn + 2H + → Zn2+ + H2 (Final ionic equation)

In above example, the Cl − ions are the spectator ions and hence are not included in the final ionic
balanced equation.

(4) Rules for writing ionic equations
(i) All soluble ionic compounds involved in a chemical change are expressed in ionic symbols and covalent
substances are written in molecular form. H2O, NH3 , NO2 , NO, SO2, CO, CO2, etc., are expressed in molecular
form.

(ii) The ionic compound which is highly insoluble is expressed in molecular form.

(iii) The ions which are common and equal in number on both sides, i.e., spectator ions, are cancelled.

(iv) Besides the atoms, the ionic charges must also be balanced on both the sides.

The rules can be explained by following examples,

Example : Write the ionic equation for the reaction of sodium bicarbonate with sulphuric acid, The molecular
equation for the chemical change is,

NaHCO3 + H2SO4 → Na2SO4 + H2O + CO2
NaHCO3, H2SO4 and Na2SO4 are ionic compounds, so these are written in ionic forms.

Na+ + HCO3− + 2H + + SO42− → 2Na + + SO42− + H 2O + CO2


Click to View FlipBook Version
Previous Book
Ajánló 2019. aug-okt
Next Book
FRMA Toluna News Août