# Chapter 7 Thermodynamics and Thermochemistry by TEACHING CARE Online coaching and tuition classes

Chapter 7 Thermodynamics and Thermochemistry by TEACHING CARE Online coaching and tuition classes

File name : Chapter-7-Thermodyanamics-and-Thermochemistry.pdf

# Thermodynamics

### Thermodynamics (thermo means heat and dynamics means motion) is the branch of science which deals with the study of different forms of energy and the quantitative relationships between them.

The complete study of thermodynamics is based upon three generalizations celled first, second and third laws of thermodynamics. These laws have been arrived purely on the basis of human experience and there is no theoretical proof for any of these laws.

• System, surroundings and Boundary : A specified part of the universe which is under observation is called the system and the remaining portion of the universe which is not a part of the

system is called the surroundings.

The system and the surroundings are separated by real or imaginary boundaries. The boundary also defines the limits of the system. The system and the surroundings can interact across the boundary.

#### (2)  Types of systems

• Isolated system : This type of system has no interaction with its

The boundary is sealed and insulated. Neither matter nor energy can be exchanged with surrounding. A substance contained in an ideal thermos flask is an example of an isolated system.

• Closed system : This type of system can exchange energy in the form of heat, work or radiations but not matter with its surroundings. The boundary between system and surroundings is sealed but not insulated. For example, liquid in contact with vapour in a sealed tube forms a closed system. Another example of closed system is pressure
• Open system : This type of system can exchange matter as well as energy with its surroundings. The boundary is not sealed and not insulated. Sodium reacting with water in an open beaker is an example of open
• Homogeneous system : A system is said to be homogeneous when it is completely uniform A homogeneous system is made of one phase only. Examples: a pure single solid, liquid or gas, mixture of gases and a true solution.
• Heterogeneous system : A system is said to be heterogeneous when it is not uniform throughout, e., it consist of two or more phases. Examples : ice in contact with water, two or more immiscible liquids, insoluble solid in contact with a liquid, a liquid in contact with vapour, etc.
• Macroscopic system : A macroscopic system is one in which there are a large number of particles (may be molecules, atoms, ions )

Note :®         All physical and chemical processes taking place in open in our daily life are open systems

because these are continuously exchanging matter and energy with the surroundings.

• 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,

The macroscopic properties can be sub – divided into two types

• 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
• 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 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.

• 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 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)
• 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.

• Chemical Equilibrium : A system in which the composition of the system remains fixed and
• Mechanical Equilibrium : No chemical work is done between different parts of the system or between the system and It can be achieved by keeping pressure constant.
• Thermal Equilibrium : Temperature remains constant e. no flow of heat between system and surrounding.
• 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
• Isothermal process : The process is termed isothermal if temperature remains fixed, e., operation is done at constant temperature. This can be achieved by placing the system in a constant temperature bath, i.e., thermostat. For an isothermal process dT = 0, i.e., heat is exchanged with the surroundings and the system is not thermally isolated.
• Adiabatic process : If a process is carried out under such condition that no exchange of heat takes place between the system and

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.

• Isobaric process : The process is known as isobaric in which the pressure remains constant throughout the change e., dP = 0.
• Isochoric process : The process is termed as isochoric in which volume remains constant throughout the change, e., dV = 0.
• 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 =
• 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.
• 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

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

#### Difference between reversible and irreversible process

 Reversible process 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   than   the 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 (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,

E Etranslational + Erotational + Evibrational + Ebonding + Eelectronic + ……

• Characteristics of internal energy
• Internal energy of a system is an extensive
• Internal energy is a state
• The change in the internal energy does not depend on the path by which the final state is
• There is no change in internal energy in a cyclic
• The internal energy of an ideal gas is a function of temperature
• Internal energy of a system depends upon the quantity of substance, its chemical nature, temperature, pressure and
• The units of E are ergs (in CGS) or joules (in SI)

1 Joule = 107 ergs.

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

 DE = Ef  – Ein

system moves from one state to another. Let DE

be the difference of energy of the initial state (Ein ) and the final

state (Ef ) , then,

; DE

is positive if

Ef > Ein and negative if

Ef    < Ein .

• 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.

• 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 = 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

• 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 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.

When w or q is negative, it means that energy has left the system as work or as heat. The internal 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.

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.”

 C
 B
 C
 A

⇌                  ⇌

 B
 A

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.”

• 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 The following observations justify the validity of this law
• The total energy of an isolated system remains constant although it can undergo a change from one form to
• 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
• 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
• Energy is conserved in chemical reactions 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 ¾¾® H

(g) + 1 O

2           2    2

• ; H
• + 1O

2           2    2

(g) ¾¾® H

2O(l) + 286.4 kJ

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

• Mathematical expression for the law : The internal energy of a system can be changed in two ways
• By allowing heat to flow into the system or out of the
• By doing work on the system or by the

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 + or E2 – E1 = q + or DE = q + w

## i.e. (Change in internalenergy) =(Heat added tothe 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, DE = q + (-w) = q w

## i.e. (Change in internalenergy) =(Heat added tothe system) – (Work done by the system)

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

#### (3)

 DE = q + w

#### Some useful conclusions drawn from the law :

• When a system undergoes a change DE = 0 , 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)

• If no work is done, w = 0 and the first law reduces to

DE = 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.

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

DE = 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.

• In case of gaseous system, if a gas expands against the constant external pressure P, let the volume change be DV . The mechanical work done by the gas is equal to – P ´ DV .

Substituting this value in DE = q + w ; DE = q PDV

When,

DV = 0 ,

DE = q

or qv

The symbol qv

indicates the heat change at constant volume.

• 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
• It puts no restriction on the direction of flow of heat. But, heat flows spontaneously from a higher to a lower
• It does not tell whether a specified change can occur spontaneously or
• It does not tell that heat energy cannot be completely converted into an equivalent amount of work without producing some changes

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

From first law of thermodynamics, q = E + PV

Heat change at constant pressure can be given as

Dq = DE + PDV

At constant pressure heat can be replaced at enthalpy.

……….(i)

……….(ii)

………..(iii)

Constant pressures are common in chemistry as most of the reactions are carried out in open vessels. At constant volume, D= 0 ; thus equation (ii) can be written as,

Dqv = DE

\DH =

DE =

Heat change or heat of reaction (in chemical process) at constant pressure Heat change or heat of reaction at constant volume.

• In case of solids and liquids participating in a reaction,

DH = DE(PDV » 0)

• Difference between DH

and DE

is significant when gases are involved in chemical reaction.

DH = DE + PDV

Here, Dn =

PDV = DnRT

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 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 ).
• 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 .

• 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

under constant volume conditions, i.e.,

relative to that required

Cp > Cv or Cp  = Cv   + Work done on expansion, PDV(= 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

• Cp Cv  = R = 2 calories = 314 J

• C

= 3 R (for monoatomic gas) and C

v           2                                              v

= 3 + x

2

(for di and polyatomic gas), where x varies from gas to gas.

Cp   = g

Cv

(Ratio of molar capacities)

• For monoatomic gas Cv= 3 calories whereas, Cp = Cv + R = 5calories

• For monoatomic gas, (g ) = Cp

5 R

= 2     = 1.66

Cv             3 R

2

C          7 R

• For diatomic gas (g ) = p

Cv

= 2     = 1.40

5 R

2

• For triatomic gas (g ) = Cp

Cv

= 8R = 1.33 6R

#### Values of Molar heat capacities of some gases,

 Gas Cp Cv Cp– Cv Cp/Cv= g Atomicity He 5 3.01 1.99 1.661 1 N 2 6.95 4.96 1.99 1.4 2 O2 6.82 4.83 1.99 1.4 2 CO2 8.75 6.71 2.04 1.30 3 H2 S 8.62 6.53 2.09 1.32 3

• Isothermal Expansion : For an isothermal expansion, temperature remains fixed e.

ideal gases depends only on temperature) According to first law of thermodynamics,

DE = q + w

\q = –w

DE = 0 ( DE   of

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

Since for isothermal process, DE

and DT

are zero respectively, hence, DH = 0

• Work done in reversible isothermal expansion : Consider an ideal gas enclosed in a cylinder fitted with a weightless and frictionless The cylinder is not insulated.

The external pressure, Let it be P.

Pext

is equal to pressure of the gas,

Pgas .

Pext

= Pgas   = P

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,

 2

V

 V

w = –nRT log e                    or

1

At constant temperature, according to Boyle’s law,

P V  = P V   or

V2 = P1

So,

1   1         2   2

V1       P2

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

 w                = 2.303nRT log V1 = 2.303nRT logcompression                                                        V2 P2 P1
• Work done in irreversible isothermal expansion : Two types of irreversible isothermal expansions are observed, 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

V2

to V2

is given by

 1

w = -òV    Pext   ´ dV  = –Pext (V2   – V1 )

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.

• Adiabatic Expansion : In adiabatic expansion, no heat is allowed to enter or leave the system, hence,

q = 0 .

According to first law of thermodynamics,

DE = q + w

\    DE = w

work is done by the gas during expansion at the expense of internal energy. In expansion, DE decreases

while in compression DE increases.

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

 =

æ dE ö

Cv       ç dT ÷

or dE = Cv .dT

 w = DE = Cv DT

è      øv

and for finite change DE = Cv DT

So,

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

• Reversible adiabatic expansion : The following relationships are followed by an ideal gas under reversible adiabatic

where, P = External pressure, V = Volume

where, volume.

Cp   =

molar specific heat capacity at constant pressure, Cv =

molar specific heat capacity at constant

knowing g ,

P1 , P2

and initial temperature T1 , the final temperature T2

can be evaluated.

• Irreversible adiabatic expansion : In free expansion, the external pressure is zero, e, work done is

zero. Accordingly, DE

which is equal to w is also zero. If DE

is zero, DT

should be zero. Thus, in free expansion

(adiabatically), DT = 0 , DE = 0 , w = 0 and DH = 0 .

In intermediate expansion, the volume changes from V1

to V2

against external pressure,

Pext .

æ RT2

RT1 ö

æ TP1 – TP2 ö

w = –Pext (V2V1 ) = –Pext ç P    P    ÷ = –Pext ç       P P           ÷ ´ R

è    2            1   ø             è        1   2        ø

æ TP1 – TP2 ö

or w = Cv (T2 – T1 ) = –RPext ç      P P           ÷

è        1   2        ø

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 20 litres at 25o C is                                                                                                                         [CMC Vellore 1991]

(a)

(c)

2.303 ´ 298 ´ 0.082 log 2

2.303 ´ 298 ´ 0.082 log 0.5

(b)

(d)

– 298 ´ 107 ´ 8.31 ´ 2.303 log 2

2.303 ´ 298 ´ 2 log 2

Solution:(b)

w = -2.303nRT log V2

V1

= -2.303 ´ 1 ´ 8.31 ´ 107 ´ 298 log 20

10

= – 298 ´ 107 ´ 8.31 ´ 2.303 log 2

Example 2.         An ideal gas expands from 10 3 m3 to 10 2 m3 at 300 K against a constant pressure of 105 Nm2 . The work done is

(a)

– 103 kJ

(b)

10 2 kJ

(c)

0.9 kJ

(d)

-900 kJ

Solution:(c)

w = –P(V2 – V1) = -105 (10 2 – 10 3 ) = -105 ´ 10 2 (1 – 0.1) = -0.9 ´ 103 J = -0.9 kJ

Example 3.         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 DE 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

Solution: (a)

w = -2.303nRT log P1  = 2.303 ´ 1 ´ 2 ´ 300 log 2 = 965.84

P2                                                                          10

At constant temperature DE = 0

DE = q + w ; q = –w = -965.84cal

Example 4.         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

Solution: (c)

q = 40J

w = -8 J

(work done by the system)

DE = q + w = 40 – 8 = 32J

Example 5.         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 )

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

Solution:(a)       Work done during adiabatic expansion = Cv (T2 – T1 )

[KCET  2000]

-3000 = 20(T2 – 300) or T2 = 150 K .

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

Joule-Thomson effect.

During this process, enthalpy remains constant. It is, therefore, called isoenthalpic process.

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

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.

• For cooling,

m = +ve

(because dT and dP both will be – ve)

• For heating, m = –ve (because dT = +ve, dP = – ve)

m = 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. m = 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.

• 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

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

• A river flows from a mountain towards the sea is spontaneous
• Heat flows from a conductor at high temperature to another at low temperature is spontaneous
• A ball rolls down the hill is spontaneous
• The diffusion of the solute from a concentrated solution to a dilute solution occurs when these are brought into contact is spontaneous process.
• Mixing of different gases is spontaneous

• Heat flows from a hot reservoir to a cold reservoir is spontaneous
• Electricity flows from high potential to low potential is spontaneous
• Expansion of an ideal gas into vacuum through a pinhole is spontaneous

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

• Spontaneous process and Enthalpy change : A spontaneous process is accompanied by decrease in internal energy or enthalpy, 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 the spontaneous nature of the system. This factor is entropy.

(DH) which explains

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

• 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 ”
• 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 ”
• 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 ”
• 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 ”
• Proof of the law : No rigorous proof is available for the second 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.

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

• Isothermal reversible expansion,

q2 = –w1

= RT2

loge       2

 V

V1

…..(i)

• Adiabatic reversible expansion, DE = –w2= –Cv (T2 – T1)

…..(ii)

• Isothermal reversible compression, –

q1 = w3

= RT1

log e V4

V3

…..(iii)

w4 = Cv (T2 – T1)

…..(iv)

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

q = q

• (-q ) = RT log V2 + RT log V4  = RT

log V2 – RT log V3

……(v)

 V
 V

2             1             2        e                            1        e

1                            3

2        e                            1        e

1                            4

 V
 V

According to the expression governing adiabatic changes,

T        æ V

ög -1

2 = ç   3 ÷

T1      è V2 ø

T       æ V

ög -1

1 = ç   1 ÷

T2      ç V4 ÷

è      ø

V         V             V         V

or    3 =   4 or

2 =   3

V2        V1            V1       V4

V

Substituting the value of  3

V4

in eq. (v),

q = q

• q = RT log  V2  – RT log  V2

2        1            2        e                            1        e

1                            1

 V
 V

q2 = R(T2

• T1 )log e 2
 V
 V

1

……(vi)

Similarly, net work done by the gas is given by

w = –w1 – w2 + w3 + w4

So, w = R(T2

V

 V
• T1 )log e 2

1

……(vii)

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,

wT2 – 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 – TT2

< 1 , it follows that w < q2

. This means that only a part of heat absorbed by the system at the

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 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

Solution: (a)

T2 = 150 + 273 = 423K T1 = 25 + 273 = 298K

q2 = 500J

wT2T1

q2                T2

w = 500é 423 – 298 ù = 147.7 J

ëê     423     úû

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%

• None of these

Solution:(b)       First condition

T2 = 110 o C = 110 + 273K = 383K T1 = 25 o C = 25 + 273K = 298K

h = T2  T1 = 383 – 298 = 0.222 or

h = ?

22.2%

T2                383

Second condition

T2 = 140 o C = 140 + 273K = 413K T1 = 25 o C = 25 + 273K = 298K

h = ?

h = T2 – T1 = 413  298 = 0.278 or  27.8%

T2                          413

• Definition : Entropy is a thermodynamic state quantity which is a measure of randomness or disorder of the molecules of the

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.

If heat is absorbed, then DS = +ve and if heat is evolved, then DS = –ve.

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

degree Kelvin, i.e.,

JK-1 .

• Characteristics of entropy : The important characteristics of entropy are summed up below
• Entropy is an extensive Its value depends upon the amount of the substance present in the system.
• Entropy of a system is a state It depends upon the state variables (T, p, V, n).
• The change in entropy in going from one state to another is independent of the
• The change in entropy for a cyclic process is always

• The total entropy change of an isolated system is equal to the entropy change of system and entropy change of the The sum is called entropy change of universe.

DSuniverse = -DSsys + DSSurr

• In a reversible process, DSuniverse

= 0 and, therefore

DSsys

= -DSSurr

• In an irreversible process, spontaneous

DSuniverse  > 0 . This means that there is increase in entropy of universe is

• Entropy is a measure of unavailable energy for useful

Unavailable energy = Entropy × Temperature

• 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

• Entropy changes in system & surroundings and total entropy change for Exothermic and Endothermic reactions : Heat increases the thermal motion of

the atoms or molecules and increases their disorder and hence their entropy. In case of an exothermic process, the heat escapes into the surroundings and therefore, entropy of the surroundings increases on the other hand in case of endothermic process, the heat enters the system from the surroundings and therefore. The entropy of the surroundings decreases.

In general, there will be an overall increase of the total entropy

(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

increases.

• Entropy change during phase transition : The change of matter from one state (solid, liquid or gas) to another is called phase transition. Such changes occur at definite temperature such as melting point (solid to liquid). boiling point (liquid to vapours) etc, and are accompanied by absorption or evolution of

When a solid changes into a liquid at its fusion temperature, there is

absorption of heat (latent heat). Let The entropy change will be

DHf be the molar heat of fusion.

Similarly, if the latent heat of vaporisation and sublimation are

denoted by DHvap and DHsub , respectively, the entropy of vaporisation and sublimation are given by

and

Since

DHf , DHvap and

DHSub 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.Dn = +ve )

• Entropy change for an ideal gas : In going from initial to final state, the entropy change, ideal gas is given by the following relations,

DS for an

• When T and V are two variables, DS = nC

ln T2 + nR ln V2 . Assuming C is constant

 v
 v

T1                    V1

• When T and p are two variables, DS = nCP

ln T2

T1

• nR ln

p2 . Assuming C

 p

p1

, is constant

• Thus, for an isothermal process (T constant), DS = nR ln V2or = –nR

V1

ln p2

p1

• For isobaric process (p constant), DS = n Cp

ln T2

T1

• For isochoric process (V constant), DS = n Cv

ln T2

T1

• Entropy change during adiabatic expansion : In such process q=0 at all Hence reversible adiabatic processes are called isoentropic process.

DS = 0 . Thus,

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]

 =
 v

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

Solution: (b)

DS         DHv b

30kJ = 30000 J

 T

DHv

= 30000 J ;

DSv

= 75 mole 1 K ;

T   = 30000 = 400K

b          75

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

• 4 J/mole (b) 87.2 J/mole                            (c)  900 J/mole                     (d) Zero

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

DSvap

= DHv

T

; Molecular weight of water = 18

DSvap = 900 ´ 18 = 43.4 JK 1mol1

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)

DS 0 =

DS0(product) – DS0(reactants) = 213.5 -[5.690 + 205]

= 2.81 JK1

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

 G = H – TS
• The free energy of a system is the enthalpy of the system minus the product of absolute temperature and entropy e.,

• Like other state functions E, H and S, it is also expressed as

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

DG . Also

DG = DH TDSsystem where

DS is

• At equilibrium DG = 0
• For a spontaneous process decrease in free energy is noticed e., DG = –ve .

• At absolute zero,

TDS is zero. Therefore if

DG is – ve,

DH should be – ve or only exothermic reactions

proceed spontaneously at absolute zero.

DGsystem   TDSuniverse

• The standard free energy change,

DGo

= -2.303RT log 10 K, where K is equilibrium constant.

• Thus if

K > 1, then

DGo = –ve thus reactions with equilibrium constant K>1 are thermodynamically

• If K<1, then

DGo  = +ve

and thus reactions with equilibrium constant K<1 are thermodynamically

spontaneous in reverse direction.

• Criteria for spontaneity of reaction : For a spontaneous change

DG = DHTDS, provides the following conditions for a change to be spontaneous.

DG = –ve

and therefore use of

DH         DS                              DG

Reaction characteristics                                     Example

–            +        Always negative                             Reaction is spontaneous at all temperatures

+             –         Always positive                             Reaction       is     non      spontaneous      at      all

temperatures

2O3(g) ® 3O2(g)

3O2(g) ® 2O3(g)

–             –         Negative              at             low

temperature but positive at high temperature

+            +        Positive at low temperature

but       negative       at       high temperature

Reaction is spontaneous at low temperature but becomes non spontaneous at high temperature

Reaction is non spontaneous at low temperature but becomes spontaneous at high temperature

CaO(s)  + CO2(g)  ® CaCO3(s)

CaCO3(s)  ® CaO(s)  + CO2(g)

Example 11.      For a reaction at 25°C enthalpy change and entropy change are – 11.7×103 J mol–1 and

– 105J  mol1K1

 Solution: (b)  Example 12. (a) 15.05 kJ                             (b) 19.59 kJDG = DH – TDS, T = 25 + 273 = 298 KDG = -11.7 ´ 103 – 298 ´(-105) = 19590J = 19.59 kJFor reaction Ag2O(s) ® 2Ag(s) + 1 O2(g) the value of (c) 2.55 kJ   DH = 30.56kJ mol–1 (d) 22.55 kJ   and DS = 0.066kJ mol–1 K–1. Temperature at which free energy change for reaction wil l be zero is [MH CET 1999] Solution: (c) (a) 373 K                                 (b) 413 KDG = DH – TDS (c) 463 K (d) 493 K

respectively, what is the Gibb’s free energy.                                                                                   [BHU 2001]

2

DH = 30.56kJ mol1 ; DS = 0.066 kJmol 1 K 1 ; DG = 0 at equilibrium;

\DH = TDS 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); DGo = -1234kJ

2

C(s) + O2(g) ® CO2(g); DGo = -394kJ

H2(g) + 1 O2(g) ® H2O (l); DGo = -237kJ

2

T = ?

What is the standard free energy change for the reaction H2(g) + 2C(s) ® C2H2(g)

[Kerala PMT 2002]

Solution : (a)

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

DG = DG0(Product) – DG0(Reactants) = (-1234) – [-237 + 2(-394)] = -1234 + 1025 = -209kJ

Example 14.      Calculate the free energy per mole when liquid water boils against 1 atm pressure (DHvap = 2.0723 kJ / g)

[DPMT 1995]

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

Solution: (a)

DH = 2.0723 ´ 18kJ / mole = 37.3kJ / mole ; T = 373K ;

DS = DHvap = 37.3kJ  = 0.1kJK 1mol1

DGvap =?

T              373

DGvap = DHvapTDSvap = 37.3 – 373 ´ 0.1 = 0

Example 15.

DG 0

for the reaction

x + y = z is – 4.606 kcal. The value of equilibrium constant of the reaction at

227o C is

Solution: (a)

(R = 2.0 kcal mol1K1)

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

R = 2.0 kcal mol1K 1 ; T = 500 K ; DG0 = -4.606 kcal ;

DG0 = -2.303RT log K

-4.606 = -2.303 ´ 0.002 ´ 500 log K

log K = 2, K = 100

[Roorkee 1999]

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.

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

• Glassy solids even at 0K has entropy greater than zero.
• Solids having mixtures of isotopes do not have zero entropy at 0K. For example, entropy of solid chlorine is not zero at 0K.
• Crystals of Co, N2O, NO, H2O, 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.”

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.

• It tells about the physical state of the reactants and This is done by inserting symbol (s), (l) and

(g) for solid, liquid and gaseous state respectively along side the chemical formulae.

• It tells about the allotropic form (if any) of the reactant by inserting the respective allotropic name, for example, C (graphite) (s).
• The aqueous solution of the substance is indicated by the word aq.
• It tells whether a reaction proceeds with the evolution of heat or with the absorption of heat, e. heat change involved in the system.

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

• 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 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

 DH = Hp – Hr

; Hp  < Hr ; DH = – ve

Examples : (i) C(s) + O2 (g) ® CO2 (g) + 393.5kJ

(at constant temperature and pressure)

or C(s) + O2 (g) ® CO2 (g);

DH = -393.5kJ

(ii)

H  (g) + 1 O

2            2    2

(g) ® H

2O(l);

DH = -285.8kJ

N 2 (g) + 3H 2 (g) ® 2NH 3 (g);

2SO2 (g) + O2 (g) ® 2SO3 (g);

DH = -92.3kJ

DH = -694.6kJ

(v) CH 4 (g) + 2O2 (g) ® CO2 (g) + 2H 2 O;

DH = -890.3kJ

(vi) Fermentation is also an example of exothermic reaction.

• 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, DH

will be positive (+ve) for the endothermic reactions.

DH = Hp Hr ;

Hp  > Hr ;

DH = +ve

Example : (i)

N 2 (g) + O2 (g) + 180.5 kJ ® 2NO(g)

or N 2 (g) + O2 (g) ® 2NO(g); DH = +180.5 kJ

(ii) C(s) + 2S(s) ® CS2 (l) DH = +92.0kJ

(iii) H 2 (g) + I 2 (g) ® 2HI(g); DH = +53.9kJ

(iv) 2HgO(s) ® 2Hg(l) + O2 (g); DH = +180.4kJ

(v) SnO2 (s) + 2C(s) ® Sn(s) + 2CO2 (g); DH = +360.0kJ

• Fe + S ® FeS
• Preparation of ozone by passing silent electric discharged through oxygen is the example of endothermic
• Evaporation of water is also the example of endothermic
 Reactant DH=Hp–Hr = – ve(Hp

The enthalpy changes for exothermic and endothermic reactions are shown in figure.

Similarly, if we consider heat change at constant volume and temperature, DE is – ve, now it may be concluded that

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) = DH = SHP – SHR

• Factors which influence the heat of reaction : There are a number of factors which affect the magnitude of heat of

• Physical state of reactants and products : Heat energy is involved for changing the physical state of a chemical 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 (g) + 1 O (g) = H

O(g);

DH = – 57.8 kcal

2                 2    2                        2

H (g) + 1 O (g) = H

O(l);

DH = – 68.32 kcal

2                 2    2                         2

It is observed that there is difference in the value of DH if water is obtained in gaseous or liquid state. DH

value in second case is higher because heat is evolved when steam condenses. Hence, physical sate always affects the heat of reaction.

• 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 DH depends on the allotropic form used in the reaction. For example,

the value of DH is different when carbon in the form of diamond or in amorphous form is used.

C (diamond) + O2 (g) ® CO2 (g);

DH = – 94.3 kcal

C (amorphous) + O2 (g) ® CO2 (g);

DH = – 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.

• Temperature : Heat of reaction has been found to depend upon the temperature at which reaction is The variation of the heat of reaction with temperature can be ascertained by using Kirchhoff’s equation.
 DHT     – DHT2                       1 = DCP T2 – T1

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

 DET     – DET2                      1 = DCn T2 – T1
• 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 But, in reactions which involve gases, the difference in two values is considerable.
 qv + DnRT = qp

DE + DnRT = DH or

DE = qv = heat change at constant volume; DH = qp =

heat change at constant pressure,

#### (2)    Types of heat of reaction

• Heat of formation : It is the quantity of heat evolved or absorbed (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

DHf . When the temperature and pressure are as

25 o C

and 1 atmospheric pressure. The heat of

 f

formation under these conditions is called standard heat of formation. It is usually represented by DH 0 .

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

2 (g) ® NH 3

(g);

DH(g)

= -11kcal

1

H 2

(g) + 1 Cl

2     2

(g) ® HCl ;

DHf

= -22 kcal

It may be calculated by

DH 0 = éSum of standard heats of ù – éSum of standard heats ofù

 û
 ë
 û
 ë

ê Formation of products ú   ê Formation of reactants ú

• Heat of combustion : It is the amount of heat evolved or absorbed (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 2 O(l);

C2 H 6 (g) + 3.5 O2 (g) ® 2CO2 (g) + 3H 2 O(l);

It may be calculated by

DH = – 192 kcal

DH = – 372.8 kcal

DH 0 = éSum of the standard heats of ù – éSum of the standard heats ù

 û
 ë
 û     ë

ê     Combustion of products     ú    ê Combustion of reactants ú

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,

• 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

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 2 O(l);

1 mole (16g)

DHCH

= -890.3 kJ

 4

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.

• 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

• 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, g., Neutralisation reactions are

always exothermic reaction and the value of DH is (-ve) .

HCl(aq.) + NaOH (aq.) ® NaCl(aq.) + H2O;

DH = – 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 +

H + + OH ® H2O; DH = – 13.7 kcal

and OH

ions, i.e.,

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

• 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;

DH = – 2.9 kcal

HCN (aq.)

H+ + CN; DH = 10.8 Kcal

• 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.);

BaCl2(s) + H2O(l) ® BaCl2(aq.);

DH = + 3.90 kcal

DH = – 2.70 kcal

• 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);

DH = – 18.69

• Heat of vapourisation : When a liquid is allowed to evaporate, it absorbs heat from the surroundings and evaporation is accompanied by increase in 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);

H2O(g) ® H2O(l);

DH = + 10.5 kcals

DH = – 10.5 kcals

(+43.93 kJ) (-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.

• 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);

DH = + 1.44 kcals

(+ 6.02 kJ)

HO (liquid) ® HO (ice);

DH = – 1.44 kcals (- 6.02 kJ)

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

• 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

 4

Ba 2+ + SO 2 (aq) ® BaSO4

(s):

DH = – 4.66 kcal

• Heat of sublimation : Sublimation is a process in which a solid on heating changes directly into gaseous state below its melting

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) ;

DH = + 62.39 kJ

Most solids that sublime are molecular in nature e.g. iodine and naphthalene etc.

• 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, 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 system. Different types of calorimeters are used but two of the common

types are,

(i) Water calorimeter and (ii) Bomb calorimeter

Bomb calorimeter : This is commonly used to find the heat of 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 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.

The temperature of the water is noted and the substance is ignited by an electric current. After combustion the rise in temperature of the system is noted. The heat of combustion can be calculated from the heat gained by water and calorimeter.

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

T1 the initial temperature of the

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.

DH can be calculated from the relation, DH = DE + DnRT

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)

combustion of carbon monoxide will be                                                       [MP PET/PMT 1988;  EAMCET 1993]

(a) + 26.4 kcal                        (b) – 67.6 kcal                            (c) –120.6 kcal                  (d) + 52.8 kcal

CO + 1 O2 ® CO2 ; DH = ?

2

DHCO2 = -94.0 kcal , DHCO = – 26.4 kcal

DH = DH(CO2 ) – DH(CO)

= -94.0 – (-26.4) = – 67.6 kcal

Example 17.      Reaction H2(g) + I2(g) ® 2HI(g); DH = -12.40 kcal . According to this, the heat of formation of HI will be

[MP PET 1990]

Solution:(c)

(a) 12.4 kcal                           (b) –12.4 kcal                             (c) –6.20 kcal                    (d) 6.20 kcal

H2(g) + I2(g) ® 2HI(g)

For 2M, DH = -12.40 kcal

1M, -12.40 = -6.20 kcal

2

Example 18.

C(diamond) + O2(g) ® CO2(g); C(graphite) + O2(g) ® CO2(g);

DH = – 395 kJ

DH = – 393.5 kJ

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

Solution:(b)

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

C(graphite) ® C(diamond)

DH(diamond) = -395 kJ ; DH(graphite) = -393.5 kJ,

DH = DHgraphite –

DHdiamond =393.5 – (-395.0)

DH = +1.5 kJ

Example 19.      The enthalpy of combustion of benzene from the following data will be (i) 6C(s) + 3H 2 (g) ® C6 H6 (l); DH = + 45.9 kJ

(ii)

H 2 (g) + 1 O2 (g) ® H 2O(l);

2

DH = -285.9 kJ

(iii) C(s) + O2(g) ® CO2(g);

DH = – 393.5 kJ

Solution:(d)

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

C6 H6 + 15 O2 ® 6CO2 + 3H2O; DH = ?

2

DH = DH(Product) – DH(Reactant) = [(6 ´ -393.5) + (3 ´ -285.9)] -(45.9) = -3264.6 kJ

Example 20.      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

Solution:(b)

H +(aq) + OH(aq) ® H2O(l);

DH = -56.07kJ

DH  = DHf (H2O) – [DHf (H +) + DHf (OH)]

-56.07 = -285.77 – (0 + x)

[DH f (H +) = 0]

\ x = -285.77 + 56.07 = -229.70 kJ

Example 21.      The heat of combustion of glucose is given by C6 H12O6 + 6O2 ® 6CO2 + 6H2O ; DH = -2840 kJ . Which of the following energy is required for the production of .18 gm of glucose by the reverse reaction.

(a) 28.40 kJ                             (b) 2.84 kJ                                    (c)  5.68 kJ                            (d) 56.8 kJ

Solution:(b)

DHcomb = -2840 kJ

for 180g

of C6 H12O6

\ DHcomb

for 0.18gm glucose

= 2840 ´ 0.18 = -2.84kJ

180

For the reverse reaction DHcomb

of glucose = 2.84 kJ

Example 22.      One gram sample of

NHNO3

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

Solution:(a)       Heat evolved = 1.23× 6.12

\ Molar heat capacity = 1.23 ´ 6.12 = 7.5276

Molar heat of decomposition = -7.53kJ / mole

(c)

-16.1 kJ / mole

(d)

-602 kJ / mole

Example 23.      If

DH0 for H O    and H O are – 188kJ/mole and -286kJ / mole.

What will be the enthalpy change of the

f                      2   2                2

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 ;

DH0 = -188 kJ / mole

 f

H  + 1 O  ® H O                DH0 = -286kJ / mole

2     2    2           2                       f

DH = DH0(product) – DH0(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

(a) + 94 kcal                          (b) –94 kcal                                (c) + 47 kcal                      (d) –47 kcal

Solution:(b)

C(s) + O2 (g) ® CO2(g); DH = -94 kcal

DH = DE + Dng RT ; DE = ?

D ng = 1 – 1 = 0 ; DH = DE ; DE = -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 DE

(a)

1430.03 cal

(b)

1500.0 cal

(c)

1450.0 cal

(d)

1440.03 cal

Solution:(d)

q = 1440cal

\DH = 1440 cal

(absorbed)

given H2O(s) ⇌ H2O(l)

PDV = 1´ (0.0180–0.0196) =– 0.0016 L atm

 (1L atm = 24.206 cal)

\- 0.0016Latm = -0.039cal

DH = DE + PDV Þ DE = DH PDV = 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 2 O(l) at 25 o C in kJ is                               [IIT 1991; AFMC 1994]

Solution : (a)

(a) –7.43                         (b)

Dng = 12 – 15 = -3

+3.72

(c)

-3.72

(d)

+7.43

qP qv = Dng RT = -3 ´ 8.314 ´ 298 = -7.43kJ.

1000

• 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); DH = -94.3 kcal ; CO2 (g) ® C(s) + O2 (g); DH = +94.3kcal

• Hess’s law (the law of constant heat summation) : This law was presented by Hess in 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,

DH1 calorie, from B to

C, DH 2

calorie and from C to

D, DH3

calorie. Total enthalpy change from A to D will be equal to the sum of

enthalpies involved in various steps,

Total enthalpy change DH steps = DH1 + DH 2 + DH 3

Now if D is directly converted into A, let the enthalpy change be

DH direct .

According to Hess’s law

DH steps + DH direct = 0, i.e.

DH steps

must be equal to

DH direct

numerically but with opposite sign. In case it is not so,

say

DH steps (which is negative) is more that

DH 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, numerically.

• 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); DH = -94.0 kcal

DH steps

must be equal to

DH direct

Second method : Carbon is first converted into been carried in two steps,

CO(g)

and then

CO(g)

into

CO2 (g), i.e. conversion has

C(s) + 1 O

2    2

= CO(g)

; DH = -26.0 kcal

CO(g) + 1 O

2    2

= CO2

(g);

DH = – 68.0 kcal

Total enthalpy change C(s) to CO2 (g);

DH = -94.0 kcal

(b) Formation of ammonium chloride from ammonia and hydrochloric acid:

First method :

NH 3 (g) + HCl = NH 4 Cl(g); DH = – 42.2 kcal

NH4Cl(g)+aq= NH4Cl(aq); DH = + 4.0 kcal

NH3 (g) + HCl(g) + aq = NH4 Cl(aq);

DH = -38.2kcal

Second method :

NH3 (g) + aq = NH3 (aq); DH = -8.4 kcal

HCl(g) + aq = HCl(aq); DH = -17.3 kcal

NH3 (aq) + HCl(aq)= NH4 Cl(aq);

DH = -12.3 kcal

#### Conclusions

NH3 (g) + HCl(g) + aq = NH4 Cl(aq);

DH = -38.0 kcal

• The heat of formation of compounds is independent of the manner of its
• The heat of reaction is independent of the time consumed in the
• The heat of reaction depends on the sum of enthalpies of products minus sum of the enthalpies of
• Thermochemical equations can be added, subtracted or multiplied like algebraic
• Applications of Hess’s law
• For the determination of enthalpies of formation of those compounds which cannot be prepared directly from the elements easily using enthalpies of combustion of
• For the determination of enthalpies of extremely slow
• For the determination of enthalpies of transformation of one allotropic form into
• For the determination of bond

DH reaction

= S Bond energies of reactants – S Bond energies of products.

• For the determination of resonance
• For the determination of lattice energy.

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);

Cl Cl(g) ® 2Cl (g);

DH = + 433 kJ mol -1

DH = + 242.5 kJ mol -1

H Cl(g) ® H(g) + Cl(g); DH = + 431kJ mol -1

I I(g) ® 2I(g);

H I(g) ® H(g) + I(g);

DH = + 15.1 kJ mol -1

DH = + 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); DH = – 431kJ mol -1

Bond energy of

HCl = –

(enthalpy of formation) = -(-431) = + 431kJ mol -1

Consider the dissociation of water molecule which consists of two two stages.

O H

bonds. The dissociation occurs in

H 2 O(g) ® H(g) + OH(g);

DH = 497.89 kJ mol -1

OH(g) ® H(g) + O(g); DH = 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

 4

[CH (g) ® C(g) + 4 H(g); DH = 1664 kJ mol -1 ]

#### Applications of bond energy

• Heat of a reaction = S Bond energy of reactants – S Bond energy of

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
• 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

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

bond is                                                                                                                             [UPSEAT   2003]

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

Solution:(d)

CH4 ® 4C H;

1EC– H = 320 = 80

4

DH = 320

calories then 6EC– H = 480 cal.

C2H6 ® ECC + 6EC– H ; DH = 360 cal. 360 = ECC + 480

ECC = 360 – 480 = -120cal

Example 28.      Given that C(g) + 4 H(g) ® CH4(g);

DH = -166 kJ . The bond energy C H

will be                   [AMU 2002]

(a) 208 kJ/mol                        (b) – 41.5 kJ/mol                       (c) 832 kJ/mol                    (d) None of these

Solution:(b)

C(g) + 4 H(g) ® CH4 (g);

Bond energy for C H

DH = -166kJ

= – 166 = -41.5 kJ / mole

4

Example 29.      Calculate resonance energy of N O from the following data. observed DHo(N O) = 82kJ mol 1

2

B.E. of N º N Þ 946 kJ mol1; B.E. of

f         2

N = N Þ 418 kJ mol1

B.E. of O = O Þ 498 kJ mol 1 ;

B.E. of

N = 0 Þ 607 kJ mol 1

[Roorkee 1991]

Solution:(a)

(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 º N + 1 O = O ® N = N = O

2

Calculated DHo(N O) = [B.E.         + 1 B.E.(O = O)] – [B.E.

+ B.E.      ]

f         2                   (N º N)      2

(N=N)

N =O

= é946 + 498 ù – [418 + 607] = +170 kJ / mole

ëê              2 úû

Resonance energy = observed DH 0 – calculated  DH 0 = 82 – 170 = -88kJ mol1

f                                                     f

Example 30.      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

Solution: (d)

will be                                                                                                                                 [AIEEE   2003]

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

CH2 = CH2(g) + H2(g) ¾¾® H3C CH3(g)

4 EC– H Þ 414 ´ 4 = 1656

ECH

Þ 414 ´ 6 = 2484

1 EC=C

Þ 615 ´ 1 = 615

1ECC Þ 347 ´ 1 = 347

1EH – H Þ 435 ´ 1 = 435

4DHC– H + DHC=+ DHH – H  = 2706 ¾¾® 6DHC– H  + 1DHCC  = 2831

DH = 2706 – 2831 = -125kJ

Example 31.      The bond dissociation energies of gaseous

H 2 ,

Cl2

and HCl are 104, 58 and 103 kcal respectively. The

Solution:(c)

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

• H2 + 1Cl2 ® HCl

2         2

DH = SB.E.(Reactants) – SB.E.(Products)

DH = é 1 B.E H2  + 1 BE(Cl2)ù – B E

 ú

êë 2

.(    )    2

.   .(HCl)

û

= 1 (104) + 1 (58) – 103

2            2

= 81 – 103 = -22 kcal

DH = -22 kcal.

Example 32.      Given the bond energies

N º N, H H

and NH bonds are 945, 436 and 391 kJ mole–1 respectively the

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

H

Solution:(a)

N º N + 3H H

9–45+3 –´ 4–3–6

energy absorbed

¾¾®

|

• N H

|

H

2´–(3–´–391–) =–2–346

energy released

Net energy (enthalpy) = BEreactants – BEproducts = 2253 – 2346 =– 93

DH = -93kJ

***