f
P
6B
IIT–JEE Syllabus
Thermodynamics
2.1. Heat, Energy & Work
2.2. Work
2.3. Properties of a System
2.4. Thermodynamic Process
First law of Thermodynamics
2.6. Differential form of the First
Law
2.7. Work in reversible process
2.8. Work in isothermal process
2.9. Adiabatic Process
2.10. Internal Energy
2.11. Enthalpy
2.12. Heat Capacity
2.13. Limitation of First Law
2.14. Second Law of
Thermodynamics
2.15. Efficiency of Heat Engine
Thermochemistry
Various Types of Enthalpies
Reactions
Laws of Thermochemistry
Lattice Energy of an Ionic
Crystal (Born–Haber Cycle)
Bond Energies
Variation of Heat of Reaction
with Tempeature
Bomb Calorimeter
Solution to Exercises
Solved Problems
11.1. Subjective
11.2. Objective
Assignments
(Subjective Problems)
13. Assignments
(Objective Problems)
Answers to Subjective
Assignments
Answers to Objective
Assignments
Thermochemistry
This lesson introduces some of the basic concepts
of thermodynamics. It concentrates on the
conservation of energy the experimental
observation that it can be neither created nor
destroyed and shows how the principle of
conservation of energy can be used to asess the
energy changes that accompany physical and
chemical processes. Thermodynamics is a subtle
subject, and it is essential to take into account all
the energy transactions that may take place
between the system of interest and its immediate
surroundings. Much of this chapter is based on a
careful development of the means by which a
system can exchange energy with its
surroundings in terms of work it may do or the
heat that it may produce. The target concept of the
lesson is enthalpy, which is a very useful book–
keeping property for keeping track of the heat out
put (or requirements) or physical process and and
chemical reactions under conditions of constant
pressure (which is typical of many laboratory
process in vessels pass to the atmosphere).
Enthalpy will figure in discussion throughout the
rest of the text.
We also discuss the different types of half-cells
and their emf calculations. Certain thermodynamic
properties such as G and G are described in this
lesson.

1. IIT–JEE Syllabus
First and second law of thermodynamics; Internal energy; enthalpy, work and heat; heats
of reaction, fusion, and vaporisation, Hess’s law; pressure – volume work.
2. Thermodynamics
The study of the transformation of energy is called thermodynamics. Actually, it deals with energy
in its various forms, which include thermal, chemical, electrical, and mechanical, with the
restrictions on the transformation of one type of energy in to the other types and with the relation
of energy changes to physical and chemical changes.
Thermodynamic Equilibrium
A system is said to be in thermodynamic equilibrium of its macroscopic properties do not change
with time. The following three types of equilibria exist simultaneously in thermodynamic equilibria.
i) Thermal equilibrium (constancy of temperature)
ii) Mechanical equilibrium (constancy of pressure)
iii) Chemical equilibrium (constancy of composition)
2.1 Heat, Work and the Conservation of Energy
The system is the part of the world in which we have a special interest, and may be a reaction
vessel, an engine, an electric cell, etc. Around the system is its surrounding in which we make
our observations. The two part may be in contact. When matter can be transferred between the
system and its surrounding we call it an open system. A system is said to be closed when it
permits the passage of energy but not mass, across the boundary.
A system is said to be isolated when it can neither exchange energy nor matter with its
surrounding.
2.2 Work
Work has been done by the system if a weight has been raised in the surrounding, work has been
done on the system if a weight has been lowered.
2.3 Types of Systems
Depending on the interactions between system and its surroundings, three kinds of systems are
distinguished namely, isolated system, closed system and open system.
i) Isolated System: A system which neither exchanges energy nor matter with its
surroundings is called an isolated system. For example a liquid in contact with its vapour in an
insulated closed vessel is an isolated system. No exchange of matter and energy is possible with
the surroundings.
However if allowed, it can exchange energy with the surrounding in the form of work.

ii) Closed System: A system which may exchange energy but not matter with its
surroundings is called a closed system. If in the above example the vessel containing liquid in
contact with its vapour is closed one but is not insulated, it is a closed system as it can exchange
energy (gain or lose heat) with the surroundings but it is incapable of exchanging matter with
surroundings.
iii) Open System: A system which may exchange both energy and matter with its
surroundings is called an open system.
One such example is of evaporation of water ad an open beaker. The water in open beaker
absorbs heat for evaporation from the surroundings and escapes into the surroundings as water
vapour.
Exercise 1: A gas present in a cylinder fitted with a frictionless piston expands against at
constant pressure of 1 atm from a volume of 2 litre to a volume of 6 litre. In doing so, it absorbs 800
J heat from surrounds. Determine increase internal energy of process.
Exercise 2: Calculate the maximum work done expanding 16 gm of oxygen at 300 K and
occupying a volume of 5 dm3
isothermally until the volume becomes 25 dm3
.
2.4 Thermodynamic Process
i) Adiabatic Process: When a process is carried out under such conditions that no
exchange of heat takes place between the system and its surrounding, the process is called
adiabatic.
Example: The sudden bursting of a cycle tube. If the process is exothermic the heat evolved will
remain in the system and, therefore, the temperature of the system rises and vice versa.
ii) Isothermal Process: The process which occurs at constant temperature is called
isothermal. In an isothermal change the temperature is kept constant by adding heat or taking it
away from the substance. So in an isothermal process the system exchanges heat with the
surroundings.
iii) Isobaric Process: If the pressure of the system remains constant during each step of the
change in the state of a system, this process is said to be an isobaric process.
iv) Isochoric process: A process is defined to be isochoric if the volume of the system
remains constant during the process.
2.5 First law of Thermodynamics
Energy may be converted from one form to another, but it is impossible to create or destroy it.
There are various ways of enhancing the first law of thermodynamics. Some of the selected
statements are given below.
i) “When work is transformed into heat or heat into work, the quantity of work is mechanically
equivalent to the quantity of heat.
ii) Energy of an isolated system must remain constant, although it may be transformed from
one form to another.

iii) Energy in one form, if it disappears will make its appearance in an exactly equivalent in
another form.
iv) It is never possible to construct a perpetual motion machine that could produce work
without consuming any energy.
2.5.1 Mathematical Formulation of the First Law
Suppose a system absorbs a quantity of heat q and its state
change form A to B (fig – I). This heat is used up.
i) In increasing the internal energy of the system i.e, E =
EB – EA
ii) To do some external work ‘w’ by the system on its
surroundings.
From the first law, we get.
Volume 
Pressure

A
B
I
II
Fig. 1
Heat absorbed by the system = its internal energy + work done by the system.
q = E + w …1
2.6 Differential form of the First Law
For an infinitesimal process, equation (1) takes the form
q = dE + w …2
a) Change in internal energy is independent of the path taken. So in mathematical terms, an
exact differential is always denoted by a notation d, e.g. dE in equation (2).
b) q and w are not state functions because changes in their magnitude is dependent on the
path by which the change is accomplished. Mathematically q & w are not exact differentials and
we always write the inexact – differential by q,  w etc.
c) For a cyclic process, the change in the internal energy of the system is zero because the
system is brought back to the original condition.
dE = 0
or
q =
w i.e. the total work obtained is equal to the net heat supplied.
d) In an isolated system, there is no heat exchange with the surrounding i.e. q = 0
 dE + W = 0
or w = – dE
The sign convention: According to latest S.I. convention, w is taken as negative if work is done
by the system whereas it is taken as positive if work is done on the system. When heat is given

by the system to surrounding it is given as negative sign. When heat is absorbed by the system
from the surrounding then positive sign is given.
2.7 Work in reversible process
a) Expansion of a gas
dv
P
a sq.cm
b) Suppose n moles of a perfect gas is enclosed in a cylinder by a friction less piston.
The whole cylinder is kept in large constant temperature bath at T°K. Any change that would occur
to the system would be isothermal. Suppose area of cross section of cylinder = a sq. cm
Pressure of the piston = P
Distance through which gas expands = dV cm
Then force (F) = P  a
Work done by the gas = F  dl
 P  a  dl
 a  dl = dV
 w = P . dV
Let the gas expand from initial volume V1 to the final volume V2, then the total work done(w) =

2
1
V
V
PdV
2.8 Work done in isothermal reversible expansion of an ideal gas
The small amount of work done, w when the gas expands through, a small volume dV, against
the external pressure, P is given by
w = – PdV
Total work done when the gas expands from initial volume V1 to final volume V2 will be W =

2
1
v
v
PdV
Ideal gas equation PV = nRT

i.e, P =
V
nRT
Hence W =

2
1
v
v
dV
V
nRT [ T = constant]
 W = + nRT ln
1
2
V
V = + 2.303 nRT log
1
2
V
V = + 2.303 nRT log
2
1
P
P
Note: Work in the reversible process is the maximum and is greater than that in the irreversible
process.
Illustration 1 : 70 gms of nitrogen gas was initially at 50 atm and 25°C
a) It was allowed to expand isothermally against a constant external pressure
of one atmosphere. Calculate, U, Q and W assuming the gas to behave ideally.
b) Also find out the maximum work that would be obtained if the gas expanded
reversibly and isothermally to one atmosphere.
Solution: a) Amount of gas n =
28
70 = 2.5 moles
Initial pressure; P1 = 50 atm
Final pressure = 1 atm
Since the gas is ideal, U = 0, as the temperature is constant.
The work obtained w = P2 (V2 – V1)  P2







1
2
P
nRT
P
nRT
=
2
1
2
1
2
P
P
)
P
P
(
nRTP  =
1
50
)
1
50
(
1
298
2
5
.
2




 = 1.5 K cal
b) The maximum work for its isothermal reversible expansion.
W = nRT ln
2
1
P
P = 2.303  2.5  2  298 log
1
50 = 5.8 Kcal
Illustration–2: Isothermally at 27°C, one mole of a vander waal’s gas expands reversibly from 2
litres to 20 litres. Calculate the work done if a =1.42  1012
dynes cm4
per mole and b = 30 c.c
Solution: W =

 








2
1
2
1
V
V
2
V
V
dV
V
a
b
V
RT
PdV
= 2
V
1
V
V
a
)
b
V
ln(
RT 







= RTln
b
V
b
V
1
2

 +







1
2 V
1
V
1
a
= 2.303  8.31107
 300 log
03
.
0
2
03
.
0
20

 + 1.42  1012








2000
2000
2000
2000
= 5778  107
erg

= 5778 joules
2.9 Adiabatic Process (Reversible)
An adiabatic change by definition, is one which does not allow any transfer of heat, i.e., q = 0, it
follows from the 1st
law,
U = – W
dU = – dW
Let only mechanical work of expansion or contraction is involved, dW = PdV. Moreover,
dU = CVdT
CVdT = –PdV
For a system of 1 mole of an ideal gas, expanding adiabatically and reversibly from temp T1 to T2
and volume V1 to V2, we have
CVdT =
dV
V
RT

or
R
C
V
dV V


.
T
dT
or




2
1
2
1
T
T
V
V
V
T
dT
R
C
V
dV
 ln
1
2
V
V =
2
1
V
1
2
V
T
T
ln
R
C
T
T
ln
R
C


or
ln
V
V
ln
1
2

R
/
C
2
1
V
T
T








or R
/
C
2
2
R
/
C
1
1
V
V
T
V
T
V  = K
 Cp – Cv = R
 ln
2
1
v
p
v
1
2
T
T
ln
C
C
C
V
V


or
2
1
1
2
v
v
p
T
T
ln
V
V
ln
C
C
C


or (-1)
2
1
1
2
T
T
ln
V
V
ln 
(  = Cp/Cv)
2
1
1
1
2
T
T
ln
V
V
ln 










or



















2
1
1
1
2
T
T
V
V
or K
T
V
T
V 1
1
1
2
1
2 
 




Substituting T by
R
PV
1
V
R
PV 
 = constant
PV = R  constant = constant
Similarly 1
1
P
RT
T
TV











= constant
TP1- = constant
Adiabatic work:
W = - Cv = –Cv(T2–T1) = Cv (T1 – T2)
Where T1,T2 are initial and final temperatures.
For 1 mole of gas T = PV/R
Hence adiabatic work
W = Cv







R
V
P
R
V
P 2
2
1
1
=
R
Cv
(P1V1 – P1V2)
or w =
1
V
P
V
P 2
2
1
1



Slope of PV curve in adiabatic & isothermal expansion.
For isothermal expansion of the gas, PV = K
The slope of the PV curve will be obtained from
V
P
dV
dP


For the adiabatic expansion of the gas PV = K
 P = K/V ;
dV
dP =
V
P
)
(
V
K
1








V
P

A
isothermal
adiabatic
B
C
In the both the changes the slope is negative, since , is greater than 1, the slope in the adiabatic
P – V curve will be large than that in the isothermal one.
Illustration 3: A sample of a gas initially at 27°C is compressed from 40 litres to 4 litres
adiabatically and reversibly. Calculate the final temperature (Cv = 5 cal/mole)
Solution: Assuming ideal behaviour
Cp = Cv + 2 = 7 cal / mole
  = Cp/Cv = 7/5
If T be the final temperature thus

1
4
40
300
T









T = 300  (10)7/5-1
= 753.6°K
2.10 Internal Energy
It is the energy associated with a system by virtue of its molecular constitution and the motion of
its molecules. The contribution of energy due to molecular constitution is known internal potential
energy and the contribution of energy due to the motion of molecules is called internal energy of
a system is given by the sum of two types of energies.
Determination of E: When a reaction is carried out in such a manner that the temperature and
volume of the reacting system remain constant, then the internal energy change (E) of the
reaction is equal to the heat exchanged with the surrounding.
2.11 Enthalpy
When we deal certain process in open vessels (at constant pressure). It becomes essential to
introduce in place of internal energy, a new themrodynamic function called heat enthalpy. This
new function is denoted by H.
H = E + PV
The change in enthalpy of a given system is given as follows
H = H2 – H1
or H = (E2 + P2V2) – (E1 + P1V1) = (E2 – E1) + (P2V2 – P1V1)
or H = E + PV
If P is maintained constant
H = E + PV
or H = Q
Hence the change in enthalpy of the system H may be defined as the amount of heat absorbed
or released at constant pressure.
Exercise 3: Calculate the change in entropy for the fusion of 1 mole of ice. The melting point of
ice is 273 K and molar enthalpy of fusion for ice = 6.0 kJ/mole.
Exercise 4: 10 gm of Argon gas is compressed isothermally and reversibly at a temperature of
27°C from 10 litre to 5 litre. Calculate q, w. E and H for this process. R = 2.6 kcal/K.
2.12 Heat Capacity
The heat capacity of a system is defined as the quantity of heat required for increasing the
temperature of one mole of a system through 1°C. Heat capacity may be given as follows.

C =
dT
q
 …(1)
i) Heat capacity at constant volume
By first law of thermodynamics,
We have
q = dE + PdV …(2)
on substituting the value of q in equation (2) We get
C =
dT
PdV
dE  …(3)
If the volume is kept constant then
CV =
V
dT
dE





 …(4)
Hence the heat capacity at constant volume of a given system may be defined as the rate of
change of internal energy with temperature.
ii) Heat capacity at constant pressure
If the pressure is held constant, equation (3), becomes as follows.
Cp =
dT
pdV
dE 
or Cp =
V
dT
q





 
Hence the heat capacity at constant pressure of a system may be defined as the rate of change
of enthalpy with temperature.
2.13 Modes of Transference of Energy
A system can lose or gain energy from the surroundings in a variety of ways. The following are
the two important modes of transference of energy between the system and the surroundings.
2.13 Limitations of first Law of thermodyanamics
1. This law fails to tell us under what conditions and to what extent it is possible to bring
about conversion of one form of energy into the other.
2. The first law fails to contradict the existence of a 100% efficient heat engine or a
refrigerator.
State functions: The fundamental 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
function depends upon the initial state and final state of the system, but is independent of the m
anenrin which the change has taken place. It means that the state properties do not depend upon
the path followed. Examples: pressure, volume, temperature, number of moles, internal energy,
enthalpy, entropy, free energy etc.

Note: Work (w) and heat energy (q) are not state functions but the quantity q + w (=E) is a state
function (because E is state function).
Of all the macroscopic properties (state variables), pressure, volume, temperature and
composition assume particular importance in thermodynamics since they can be controlled more
easily.
State of a System: The system is a said to have a definite state when it is in such a condition
where all of its macroscopic properties have definite values.
It follows that under any other condition where the value of any one of its macroscopic properties
changes, the system is said to have a different state. Thus the state of a system is defined when
its macroscopic properties are specified, i.e., the state of a system is fixed by its macroscopic
properties.
System Variables: The macroscopic properties of a system are often called system variables,
since the state of a system changes with the change in any of the macroscopic properties.
Whenever a system changes, from one state (initial), to another state (final), there is always a
change in one or more of its macroscopic properties.
Extensive Properties: Are those properties whose magnitude depend upon the quantity of
matter present in the system. Examples: mass, volume, total energy, enthalpy, internal energy,
entropy, work, mole, heat capacity etc.
Intensive Properties: Are those properties which do not depend upon the quantity of matter
present in the system or size of the system. Examples: pressure, temperature, density, specific
heat, surface tension, refractive index, viscosity, concentration, melting point, boiling point,
volume per mole, molarity, normality, molar enthalpy, mole fraction etc.
2.14 Second law of thermodynamics
It has been stated is several forms as follows.
i) All the spontaneous process are irreversible in nature.
ii) It is impossible to obtain work by cooling a body below it lowest temperature
iii) 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 reservoirs.
iv) Heat cannot of itself pass from a colder body to hotter body without the intervention of
external work.
v) It is impossible to construct a machine functioning in cycle which can convert heat
completely into equivalent amount of work without producing change elsewhere.
vi) The entropy of universe is always increasing in the course of every spontaneous process.
vii) Spontaneous or natural process are always accompanied with an increase in entropy.
2.15 Efficiency of a Heat Engine
The relationship between W, the net work done by the system and q2, the quantity of heat
absorbed at the higher temperature T2, in case of the cyclic process (i.e. the carnot cycle), can
be obtained from the following two equations.

W = R (T2 – T1)
1
2
V
V
ln
q2 = RT2
1
2
V
V
ln
W = q2
2
1
2
T
T
T 
The fraction of the heat absorbed by an engine which it can convert into work gives the
efficiency().
 =
2
1
2
2 T
T
T
q
W 

The net heat absorbed by the system, q is equal to q2 – q1 and according to the first law of
thermodynamics, this must be equivalent to the net work done by the system. Thus,
W = q2 – q1
or
2
1
2
q
q
q  =
2
1
2
T
T
T  = . Since
2
1
2
T
T
T  is invariably less than l, the efficiency of a heat engine is
always less than 1.
3. Thermochemistry
Thermochemistry deals with the transfer of heat between a chemical system and its
surroundings when a change of phase or a chemical reaction takes place within the
system. In general, a chemical reaction can be either exothermic or endothermic. In the former
case, heat is released to the surroundings when the reactants at a given temperature and
pressure are converted to the products at the same temperature and pressure, and in the latter
heat is absorbed by the system, from the surroundings.
Sign conventions: If the heat is absorbed by the system (q>0) then the reaction is said to be
endothermic and ΔE or ΔH value is given a positive sign. If the heat is evolved (q<0) the reaction
is said to be exothermic, and ΔE or ΔH values is given negative sign.
Standard States: In the computation of heat of reactions it is a convention to assume that the
heat of formation of elements in their standard states is zero. The standard state is taken as 1
atm pressure and at a constant temperature. Standard states for various forms of matter are
summarized below:
State Standard State
Gas Ideal gas at 1 atm and the given temperature
Liquid Pure liquid at 1 atm and the given temperature
Solid Stable crsytalline form at 1 atm and given T (e.g. graphite form
of carbon, rhombic form of sulphur)
At standard state the heat of reactions are denoted by o
E
 or o
ΔH
at given temperature.
4. Various Types of Enthalpies of Reactions

i) Enthalpy of formation : Enthalpy change when one mole of a given compound is formed
from its elements.
H2(g) + 1/2O2(g) ⎯ → 2H2O(l), ΔH = –890.36 kJ / mol
ii) Enthalpy of combustion: Enthalpy change when one mole of a substance is burnt in oxygen
CH4(g) + 2O2(g) ⎯ → CO2(g) + 2H2O(l), ΔH = –890.36 kJ / mol
iii) Enthalpy of Neutralization: Enthalpy change when one equivalent of an acid is neutralized
by a base in dilute solution. This is constant and its values is –13.7 kcal for neutralization of any
strong acid by a base since in dilute solutions they completely dissociate into ions.
H+
(aq) + OH–
(aq) ⎯ → H2O(l) ΔH = –13.7 kcal
For weak acids and bases, heat of neutralization is different because they are not dissociated
completely and during dissociation some heat is absorbed. So total heat evolved during
neutralization will be less.
e.g. HCN + NaOH ⎯ → NaCN + H2O ΔH = –2.9 kcal
Heat of ionization in this reaction is equal to (–2.9 + 13.7) kcal = 10.8 kcal
i) Enthalpy of hydration: Enthalpy of hydration of a given anhydrous or partially hydrated salt
is the enthalpy change when it combines with the requisite no.of mole of water to form a specific
hydrate. For example, the hydration of anhydrous copper sulphate is represented by
CuSO4(s) + 5H2O (l) ⎯ → CuSO45H2O,(aq) ΔH° = –18.69 kcal
ii) Enthalpy of Transition: Enthalpy change when one mole of a substance is transformed from
one allotropic form to another allotropic form.
C (graphite) ⎯ → C(diamond), ΔH° = 1.9 kJ/mol
5. Laws of Thermochemistry
For some reactions it is not convenient to measure the heat change in the laboratory. So
conventional procedure based on the principle of conservation of energy has been suggested
which can be stated as follows:
1. The heat of formation of any compound is equal in magnitude and of opposite sign to the heat
of dissociation of that compound at the given temperature and pressure.
For example, enthalpy of formation of liquid water from its elements hydrogen and oxygen is –
285.830kJ mol–1
and the enthalpy of dissociation is 285.830 kJ mol–1
. Thus the process can be
represented by
H2(g) + 1/2 O2(g)  H2O (l), H(298 K) = –285.830 kJ
H2O(l)  H2(g) +½ O2(g), H (298 K) = + 285.830 kJ
2. The total enthalpy change of a reaction is the same, regardless of whether the reaction is
completed in one step or in several steps. (Hess’s Law of constant heat summation.) It has been
experimentally verified and is also a consequence of the law of conservation of energy. It is of

particular utility in calculation of the heats of reactions which are difficult for practical calorimetric
measurements.
For example: Carbon be converted into CO2 in 1 step
C(s) + O2(g)  CO2(g) H = –94 kcal
Or in two steps
C(s) + 1/2 O2(g)  CO(g) H1 = –26.,4 kcal
CO)(g) + 1/2 O2(g)  CO2(g) H2 = –67.6 kcal
According to Hess’s law: H = H1 + H2 = –26.4 –67.6 = – 94kcal
H
C(s) CO2(g)
H1 H2
CO(g)
6. Lattice Energy of an Ionic Crystal (Born–Haber Cycle)
M(s) +
2
1
X2(g)  MX(s)
Step 1 Step 2
M(g) X(g)
Step5
Step 3 Step 4
M+
(g) + X–
(g)
Born – Haber cycle
The change in enthalpy that occurs when 1 mole of a solid crystalline substance is formed from
its gaseous ions.
Step 1: Conversion of metal to gaseous atoms
M(s) ⎯ → M(g) , ΔH1 = sublimation
Step 2: Dissociation of X2 molecules to X atoms
X2(g) ⎯ → 2X (g), ΔH2 = Dissociation energy
Step 3: Conversion of gaseous metal atom to metal ions by losing electron
M(g) ⎯ → M+
(g) + e–
, ΔH3 = Ionization energy
Step 4: X(g) atoms gain an electron to form M–
ions
X(g) + e–
⎯ → X–
(g) , ΔH4 = Electron affinity

Step 5: M+
(g) and X–
(g) get together and form the crystal lattice
M+
(g) + X–
(g) ⎯ → MX(s) ΔH5 = lattice energy
Applying Hess’s law we get
ΔH1 + 1/2 ΔH2 + ΔH3 + ΔH4 + ΔH5 = ΔHf (MX)
On putting the various known values, we can calculate the lattice energy.
7. Bond Energies
Whenever a chemical bond is formed energy is released. Conversely, energy is required to
rupture a chemical bond. The energy required to break a particular bond in a gaseous molecule
is referred to as bond-dissociation energy.
H–OH(g) ⎯ → H(g) + OH(g) ΔH = 498 kJ
O–H(g) ⎯ →
2
1 H2(g) + ½O2 (g) ΔH = 430 kJ
In such case, bond energy is expressed as the average of the bond dissociation energies of
various similar bonds HO–H = (498 + 430)/2 = 464 kJ mol–1
Hence, bond energy may be defined as the average amount of energy required to break one mole
bonds of that type in gaseous molecules.
Thus, the values given in bond energy data can help us in:
i) Calculating standard enthalpy of reactions
ii) Calculation of bond energies of some specific bond in the molecule
8. Variation Of Heat Of Reaction With Temperature
The heat of reaction depends on the temperature. The relation between the two is known as
Kirchoff’s equation.
i)
1
2
1
2
T
T
H
H



 = CP ii)
1
2
1
2
T
T
E
E



 = CV
CP = molar heat capacity of products – molar heat capacity of reactants (at constant pressure)
C = molar heat capacity of products – molar heat capacity of reactants (at constant volume)
9. Bomb Calorimeter
The calorimeter used for determining enthalpies of combustion known as the bomb calorimeter is
shown as Figure.

–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
+ –
 O2
IGNITION
WIRES
INSULATING
CONTAINER
WATER
STEEL BOMB
SAMPLE
Bomb Calorimeter
This apparatus was devised by Berthelot (1881) to measure the heat of combustion of organic
compounds. A modified form of the apparatus shown in Figure consists of a sealed combustion
chamber, called a bomb, containing a weighed quantity of the substance in a dish along with
oxygen under about 20 atm pressure. The bomb is lowered in water contained in an insulated
copper vessel. This vessel is provided with a stirrer and a thermometer reading up to 1/100th
of
a degree. It is also surrounded by an outer jacket to ensure complete insulation from the
atmosphere. The temperature of water is noted before the substances is ignited by an electric
current. After combustion, the rise in temperature of the system is noted on the thermometer and
heat of combustion can be calculated from the heat gained by water and the calorimeter.
By knowing the heat capacity of calorimeter and also the rise in temperature, the heat of
combustion can be calculated by using the expression
Heat exchange = Z  T
Z–Heat capacity of calorimeter system
T– rise in temp.
Heat changes at constant volumes are expressed in E and Heat changes at constant pressure
are expressed in H.
Also, H = E + nRT
n = gaseous product moles – gaseous reactant moles.
10. Solution to Exercises
Exercise 1: E = 394.8 J
Exercise 2: (– 2.01  103
J)
Exercise 3: (21.8 J/K mole)
Exercise 4: W = – 103.635 cal, E = 0, q = 103.635 cal,  103.635)

11. Solved Problems
11.1 Subjective
Problem 1: A cylinder of gas is assumed to contain 11.2 kg butane. If a normal family needs
20,000 kJ of energy per day for cooking. How long will the cylinder last if the enthalpy of combustion
ΔH = –2658 kJ for butane.
Solution: C4H10(g) +
2
13 O2(g) ⎯ → 4CO2(g) + 5H2O(g)
ΔH = – 2658 kJ
Molecular weight of C4H10 = 58 g/mole
58 g of butane on combustion produces 2658 kJ heat
∴ 11.2 kg of butane on combustion produces
58
10
2
.
11
2658 3


Family needs 20,000 kJ of energy per day
∴ No. of days =
000
,
20
58
10
2
.
11
2658 3


 = 25.66 days
Problem 2: Calculate heat of neutralisation from following data.
200 ml of 1 (M) HCl is mixed with 400 ml 0.5 (M) NaOH the temparature rise in calorimeter was
found to be 4.4°C. Water equivalent of calormeter is 12g and specific heat is 1 cal ml–1
degree–1
for
solution
Solution: Meq. of acid & base = 200
i.e. 200 Meq. of HCl react with 200 Meq. of NaOH to produce heat = ΔH
∴ 1000 Meq. of HCl when react with 1000 Meq. of NaOH will give heat
= 5 × ΔH = heat of neutralisation
Now heat produced during neutralisation of 200 Meq. of acid and base
= heat taken up by calorimeter + solution
= M1 × S1ΔT + M2 + S2ΔT
= 12 × 1 × 4.4 + 600 × 1 × 4.4 = 2692.8 cals
∴ Heat of nuetralisation = 5 × 2692.8 cals
= – 13.464 kcals
Problem 3: 0.16 g of methane was subjected to combustion at 27°in a bomb calorimeter. The
temperature of calorimeter system (including water) was found to rise by 0.5°C. Calculate the heat
of combustion of methane at (i) constant volume (ii) constant pressure. The thermal capacity of the
calorimeter system is
17.7kJ K–1
.
Solution: Heat evolved at constant volume = 0.16 × 17.7 × 0.5 kJ
Heat of combustion of methane (heat evolved per mole)
=
16
.
0
5
.
0
7
.
17
16 
 = 885 kJ / mol
ΔE = – 885 kJ / mol
CH4(g) + 2O2(g) ⎯ → CO2(g) + 2H2O (l)
Δn = 1–3 = – 2
ΔH = ΔE + ΔnRT = – 885 + (–2) × 8.314 × 10–3
× 298
H = – 889.95 kJ / mole

Problem 4: From the following data, calculate the enthalpy change for the combustion of
cyclopropane at 298 K. The enthalpy of formation of CO2(g), H2O(l) and propene (g) are – 393.5, –
285.8 and 20.42 kJ mol–1
respectively. The enthalpy of isomerisation of cyclopropane to propene is
– 33.0 kJ mol–1
.
Solution: C(s) + O2 ⎯ → CO2(g) ΔH1 = – 393.5 kJ ---------- (1)
H2(g) +
2
1 O2(g) ⎯ ⎯ → H2O(l)ΔH2 = – 285.8 kJ -------------- (2)
3C(s) + 3H2(g) ⎯ ⎯ → C3H6 ΔH3 = 20.42 kJ ----------------- (3)
CH2
H2C CH2
 C3H6 H4 = – 33 kJ -------------- (4)
CH2
H2C CH2
+
2
9
O2  3CO2 + 3H2O H = ?
ΔH = 3ΔH1 + 3ΔH2 – ΔH3 + ΔH4
= 3 × – 393.5 + 3 × – 285.3 – 20.42 – 33
= – 2091. 31 kJ
Problem 5: Calculate the resonance energy of N2O from the following data.
o
f
H
 of N2O = 82 kJ mole–1
.
Bond energies N ≡ N 946 kJ mole–1
N = N 418 kJ mole–1
O = O 498 kJ mole–1
N = O 607 kJ mole–1
Solution: N ≡ N +
2
1 O2 ⎯ → 

 N
N
= O
The resonating structures of N2O is as follows
N ≡ N → O ←⎯ → 

 N
N
= O
(i) (ii)
Structure (ii) is more acceptable
The value of ΔH for the above reaction is given by
ΔH = ΣBEreactants – ΣBEproducts
ΔH =







 498
2
1
946
– (418 + 607) = 1195 – 1025 = 170 kJ mole
Resonance Energy = (ΔHf)experimental – (ΔHf)Theoretical = 82 – 170 = – 88 kJ
Problem 6: A 1 L sample of CH4 and O2 measured at 25°C and 740 torr were allowed to react
at constant pressure in calormeter which together will its contents had a heat capacity of 1260
cals/k. The complete combustion of CH4 to CO2 and H2O caused a temperature rise in calormeter
0.687 K. What was the mole percent of CH4 in the original mixture? (ΔH) combustion of CH4(g) = –
210.8 kcals/mole.
Solution: Applying PV = nRT for gaseous mixture
1
760
740






 = n × 0.082 × 298
n = 0.0398

Heat generated = Heat capacity × temperature rise
= 1260 × 0.667 = 840 calos
Mole of CH4 in mixture =
3
10
8
.
210
840

= 3.98 × 10–3
Mole percent of CH4 =
100
398
.
0
10
98
.
3 3

  = 10
Problem 7: A solution of composition CuCl2 + 10H2O is dilute with x moles of water, the heat
released is given by Qv =
24
.
21
x
5023
x

 . The heat of solution of CuCl2 in 600 moles of water is 11080 cals.
Find out the heat of solution of anhydrous CuCl2 in 10 moles of water.
Solution: CuCl2, 10H2O + 590H2O = CuCl2, 600H2O ΔH0 = – Qv
Qv =
24
.
21
x
5023
x

 =
24
.
21
590
5023
590

 = 4850 cals …(1)
CuCl2 + 600H2O = CuCl2⋅ 600H2O ΔH1 = –11080 cals ….(2)
CuCl2 + 10H2O = CuCl2⋅ 10H2O
(2)⎯ (1) gives, ΔH = (–11080 + 4850) cals = – 6230 cals
Heat of solution of CuCl2 in
10 moles of water = – 6230 cals
Problem 8: Calculate the enthapy of combustion of benzene (l) on the basis of the following.
i) Resonance energy of benzene (l) = – 152 kJ mole–1
ii) Enthalpy of hydrogenation of cyclohexane (l) = – 119 kJ mole–1
iii) (ΔHf
0
)C6H12 = – 156 kJ mole–1
iv) (ΔH0
f)H2O = – 285.8 kJ mole–1
v) (ΔHf
0
)CO2 = – 393.5 kJ mole–1
Solution:
+ H2  H = – 119 kJ mole–1
+ H2  H= 3  (–119) kJ mole–1
= -357 kJ mole–1
– 357 = (ΔHf
0
)Product ⎯ (ΔHf
0
)Reactant
= – 156 – (ΔHf
0
)Benzeene
(ΔHf
0
)Benzene = (– 156 + 357)KJ mole–1
=201 KJ mole–1
Resonance energy [(ΔHf
0
)Benzene]Actual – [(ΔHf
0
)Benzene]Theoretical
or, – 152 = [(ΔHf
0
)Benzene]Actual -⎯ 201
or [(ΔHf
0
)Benzene]Actual = 49 kJ mole–1
C6H6(l) +
2
O
2
15 (g) ⎯ → 6CO2(g) + 3H2O(l)
(ΔH0
)reaction = 6(ΔHf
0
)CO2 + 3(ΔHf
0
)H2O – (ΔHf
0
)Benzene
= 6 (– 393.5) + 3(–285.8) – 49
= 3267.4 kJ mole–1
Problem 9: Calculate the lattice energy of AlI3(s) from the following data.
(ΔHf
0
)AlI3 = – 52 kcals mole–1
(ΔHv0
)I = 6Kcals mole–1
(ΔHs
0
)Al = 15 Kcals mole–1
(ΔHb
0
)I2 = 5kcals mole–1
(ΔHf
0
)I2= 10 kcals mole–1
(ΔH0
EA)I2(g) = –28.67 Kcals mole–1
The sum of 1st
two ionisation energies and 3 times the third is 74 kcals mole–1
. The sum of 1st
and
3rd
ionisation energies along with the two times 2nd
ionisation energy is 54 kcals mole–1
while 3

times the 1st
ionisation enerty added to 5 times the 3rd ionisation energy subtracted by 2nd
ionisation energy gives 98 kcals mole–1
.
Solution: Let 1st
, 2nd
& 3rd
ionisation energies of Al be x, y & z kcals/mole respectively.
x + y + 3z = 74 …(1)
x + 2y + z = 54 …(2)
3x – y + 5z = 98 …(3)
Solving we get x = 9 kcals mole–1
Y = 14 kcals mole–1
z = 17 kcals mole–1
)
s
(
AlI
)
s
(
Al
)
s
(
I
2
3
3
1
mole
kcals
52
2 



 




)
g
(
Al
)
g
(
Al 3
e
3
40
17
14
9
z
y
x 












 

3I–
(g)
  kcals
)
67
.
28
(
3
e
3
g
I
3







 

kcals
2
3
10 
)
l
(
I
2
3
2
kcals
2
3
6 
)
g
(
I
2
3
2
kcals
2
3
5 
H2
– 52 = 15 + 40 +
2
3 × 10 + 6 ×
2
3 + 5 ×
2
3 + 3 × (–28.67) + ΔH2
ΔH2 = – 52 – 65 – 15 – 9 – 7.3 + 85.01
= – 148.5 + 85 = – 63.5 kcals mole–1
Problem 10: The commercial production of water gas utilizes the reaction under standard
conditions: C + H2O(g)  H2 + CO. The heat required for this endothermic reaction may be supplied
by adding a limited amount of air and burning some carbon to CO2. How many g of carbon must be
burnt to CO2 to provide enough heat for the water gas conversion of 100g carbon. Neglect all heat
losses to environment. 0
f
H
 of CO, H2O(g) and CO2 are –110.53, –241.81 and –393.51 kJ/mol
respectively.
Solution: C + H2O(g)  H2 + CO 0
f
H
 = ?
( H° for C and H2 are zero)
 H° = 


 0
O
H
0
CO 2
H
H –110.53 – (–241.81) = 131.28 kJ/mol
Thus, H° needed for 100g carbon =
kJ
12
100
28
.
131 
Now 393.51 kJ energy is provided by 12g C
12
100
28
.
131  kJ energy is provided by =
51
.
393
12
100
28
.
131
12


 = 33.36 g
Problem 11: In order to get maximum calorific output a burner should have an optimum fuel to
oxygen ratio which corresponds to 3 times as much oxygen as is required theoretically for complete
combustion of fuel. A burner which has been adjusted for methane as fuel (with x litre/hour of CH4
and 6 x litre/hour of O2) is to be readjusted for butane C4H10. In order to get same calorific output,
what should be the rate of supply of butane and oxygen. Assume that losses due to incomplete

combustion etc., are the same for both fuels and that gases behave ideally. Heats of
combustion:CH4 = 809 kJ/mol; C4H10 = 2878 kJ/mol.
Solution: CH4 + 2O2  CO2 + 2H2O; H = –809 kJ mol–1
Initial volume/hr x 6x x 6x
(in litre)
Let the temperature be T and assume volume of 1 mole of a gas is V litre at this
condition
 V litre or 1 mole CH4 gives energy on combustion = 809 kJ
x litre of CH4 gives energy on combustion =
V
)
x
(
809 kJ
 2878 kJ energy is obtained by 1 mole or V litre C4H10

V
)
x
(
809 kJ energy is obtained by =
2878
V
V
)
x
(
809

 litre C4H10
= 0.281 (x) litre C4H10
Thus butane supplied for same calorific output = 0.281 x litre
 C4H10 +
2
O
2
13  4CO2 + 5H2O; H = –2878 kJ / mole.
Volume of O2 required = 3  Volume of O2 for combustion of C4H10
= 3 
2
13  volume of C4H10 = 3 
2
13  0.281 (x) = 5.48(x) litre O2
Problem 12: An intimate mixture of Fe2O3 & Al is used in solid fuel rocket. Calculate the fuel
value/gm & per ml of mixture. (ΔH)Al2O3 = 399 kcals.
(ΔH)Fe2O3 = 199 kcals. Density of Fe2O3 & Al are 5.2 g//ml & 2.7 g/ml respectively.
Solution: 2Al + Fe2O3 ⎯ → Al2O3 + 2Fe
Given 2Al +
2
3 O2 ⎯ → Al2O3 ΔH = – 399 kcals …(1)
& Fe +
2
3 O2 ⎯ → Fe2o3 ΔH = – 199 kcals …(2)
(1) – (2) gives.
2Al + Fe2O3 ⎯ → 2Fe + Al2O3 ΔH = – 200 kcals
Molecular weight of fuel mixture = 2 × 27 + 2 × 56 + 48 = 214g
214g mixture produces = 200 kcals heat
1g mixture produces =
214
200 = 0.9346 kcals/g
Also volume of fusion mixture = volume of Al + volume of Fe2O3
=
2
.
5
160
7
.
2
54

= 20 + 30.76 = 50.76 ml
50.76 ml mixture produces 200 kcals
1 ml mixture produces =
76
.
50
200 = 3.94 kcals/ml
Problem 13: An athlete is given 100 g glucose of energy equivalent to 1560 kJ. He utilises
50% of this gained energy in an event. In order to avoid storage of energy in body calculate the
weight of water he need to perspire. Enthalpy of H2O for evaporation is 44kJ mole–1

Solution: ∴ Gained energy by athlete =







100
50
1560
= 780 kJ
∴ amount of water evaporated =
g
18
44
780







Problem 14: Calculate the proton affinity of NH3(g) from following data (in kJ mole–1
)
ΔH0
(atomisation) : H : 218 Ionisation energy of H : 1310
ΔH0
(atomisation) : Cl : 124 Electron affinity of Cl : 348
ΔH0
(formation) : NH3 : – 46
ΔH0
(formation) : NH4Cl - 314
Lattice energy of NH4Cl – 683
Solution:
2
1
N2(g) + 2H2(g) +
2
1
Cl2(g)
3
NH
0
f
H


 






 
NH3(g) +
2
1
H2(g) +
2
1
Cl2(g)
NH4Cl(s)
(Hf
0
)NH4Cl
Cl–
(g)
H(g)
H+
(g
)
I
AH
P.A.
NH4
+
(g)
ACl
Cl(g)
U EA.
Putting the known values of different heat terms with proper sign we have
+ (ΔHf
0
)NH4Cl = + (ΔHf
0
)NH3 + AH + ACl + I + E.A. + P.A. + U
or, –314 = – 46 + 218 + 124 + 1310 + P.A. – 348 – 683
or, P.A.= – 889 kJ mole–1
Problem 15: The standard molar ΔH of formation of cyclohexane (I) and benzene (l) at 25°C are –
156 & +49 kJ/mol respectively. The standard enthalpy of hydrogenation of cyclo hexane (l) at 25°C
is – 119 kJ/mole. Use these data to estimate the magnitude of resonance energy of benzene.
Solution:
+ H2  H
0
= – 119 kJ/mol
This implies that the generation of one double bond is cyclohexane requires 119
kJ/mol of enthalpy. To calculate the resonance energy we may proceed as follows:

 

0
1
H
Δ

 

0
2
H
Δ

 

0
3
H
Δ

 

0
3
H
Δ
H5
0
ΔH1
0
= ΔH2
0
= ΔH3
0
= 119 kJ mol
ΔH4
0
= resonance energy
ΔH5
0
= ΔHf
0
(benzene) – ΔHf
0
(Cyclohexane)
= 49 – (–156)
= 205 kJ/mol
According to Hess’s Law
ΔH4
0
= ΔH5
0
– (ΔH1
0
+ ΔH2
0
+ ΔH3
0
)
= 205 – 3 × 119
= – 152 kJ/mole
11.2 Objective
Problem 1: Smelting of iron ore takes place through this reaction

2Fe2O3(s) + 3C(s) ⎯ → 4Fe(s) + 3CO2(g)
ΔHf
0
of Fe2O3 and CO2 are – 8242 kJ mole–1
& –393.7 kJ mole–1
The reaction is
(A) Endothermic (B) Exothermic
(C) ΔH = 0 (D) None of these
Solution: ΔH = (ΔHf
0
)Products – (ΔHf
0
)Reactants
= 3 × (– 393.7) – 2 × (–824.2)
= 476.3 kJ/mole ∴ (A)
Problem 2: (1) For the given heat of reaction,
(i) C(s) + O2(g) = CO2(g) + 97 kcal
(ii) CO2(g) + C(s) = 2CO(g) – 39 kcal
the heat of combustion of CO(g) is:
(A) 68 kcal (B) – 68 kcal
(C) +48 kcal (D) None
Solution: Substracting equation (ii) from equation (i), we get
C(s) + O2(g) = CO2(g) + 97 kcal
CO2(g) + C(s) = 2CO(g) – 39 kcal
————————————————————————
or, –CO2(g) + O2(g) = CO2(g) – 2CO(g) + 136 kcal
or, 2CO(g) + O2 = 2CO2(g) + 136 kcal
or, CO(g) + 1/2 O2(g) = CO2(g) + 68 kcal
Required value = 68 kcal
∴ (A)
Problem 3: What is the amount of heat to be supplied to prepare 128 g of CaC2 by using
CaCO3 & followed by reduction with carbon. Reactions are
CaCO3(s) = CaO(s) + CO2(g) ΔH0
= 42.8 kcals
CaO(s) + 3C(s) = CaC2 + CO(g) ΔH0
= 111 kcals
(A) 102.6 kcals (B) 221.78 kcals
(C) 307.6 kcals (D) 453.46 kcals
Solution: CaCO3(s) = CaO(s) + CO2(g) ΔH0
= 42.8 kcals
CaO(s) + 3C(s) = CaC2(s) + CO(g) ΔH0
= 111 kcals
–––––––––––––––––––––––––––––––––––––––––––
CaCO3(s) + 3C(s) = CaC2(s) + CO(g) + CO2(g) ΔH = 153.8 kcals
Thus heat required to prepare 1 mole of CaC2 from CaCO3 = 153.8 kcals
Molecular weight of CaC2 = 40 + 24 = 64
64 gm of CaC2 requires 153.8 kcals of heat
128 gm of CaC2 requires 307.6 kcals of heat
∴ (C)
Problem 4: Ionisation energy of Al = 5137 kJ mole–1
(ΔH) hydration of Al3+
= – 4665 kJ mole–
1
. (ΔH)hydration for Cl–
= – 381 kJ mole–1
. Which of the following statement is correct
(A) AlCl3 would remain covalent in aqueous solution
(B) Only at infinite dilution AlCl3 undergoes ionisation
(C) In aqueous solution AlCl3 becomes ionic
(D) None of these
Solution: If AlCl3 is present in ionic state in aqueous solution, therefore it has Al3+
& 3Cl–
ions
Standard heat of hydration of Al3+
& 3Cl-
ions = – 4665 + 3 × (–381) kJ mole–1

= -5808 kJ/mole
Required energy of ionisation of Al = 5137 kJ mole–1
∴ Hydration energy overcomes ionisation energy
∴ AlCl3 would be ionic in aqueous solution
∴ (C)
Problem 5: At 0°C, ice and water are in equilibrium & ΔS & ΔG for the conversion of ice to
liquid water is
(A) 0, 21.98 JK–1
mole–1
(B) –ve, zero
(C) +ve, 21.98 JK–1
mole–1
(D) Zero, zero
Solution: Since the given process is in equilibrium i.e. ΔG = 0
ΔG = ΔH – T ΔS
or, O = ΔH – TΔS or ΔH = TΔS or, ΔS =
T
H
 = 3
10
273
6

JK–1
mole–1
= 21.98 JK–1
mole–1
∴ (A)
Problem 6: Calculate the maximum efficiency of an engine operating between 110°C & 25°C.
(A) 11.1% (B) 22.2%
(C) 33.3% (D) 44.4%
Solution: Maximum efficiency of an engine working between temperature T2 & T1 is given
by
η =
383
293
.
383
T
T
T
2
1
2

 = 0.222 η = 22.2%
∴ (B)
Problem 7: 5 mole of an ideal gas expand reversibly from a volume of 8 dm3
sp 80 dm3
at a
temperature of 27°C. Calculate the change in entropy.
(A) 70.26 JK–1
(B) 82.55 JK–1
(C) 95.73 JK–1
(D) 107.11 JK–1
Solution: Entropy change (ΔS) = 2.303 nRlog
1
2
V
V = 2.303 × 5 × 8.314 ×
8
80
log
= 95.73 JK–1
∴ (C)
Problem 8: Entropy change involved in conversion of one mole of liquid water at 373 K to
vapour at the same temperature (latent heat of vaporisation of water = 2.257 kJ g–1
).
(A) 30.7 JK–1
mole–1
(B) 60.3 JK–1
mole–1
(C) 90.8 JK–1
mole–1
(D) 108.9 JK–1
mole–1
Solution: ΔHvap = 2.257 kJ/g = (2.257 × 18) kJ mole–1
= 40.626 kJ mole–1
ΔSvap=
373
626
.
40 = 0.1089 kJ k–1
mole–1
∴ (D)
Problem 9: What is value of ΔE (heat change at constant volume) for reversible isothermal
evaporation of 90g water at 100°C. Assuming water vapour behaves as an ideal gas & (ΔHvap)water =
540 cals g–1
(A) 9 × 103
cals (B) 6 × 103
cals
(C) 4.49 cals (D) none of these
Solution: Heat of system (ΔH)v = m × Lv
= 90 × 540 = 486000 cal

ΔH = ΔE + P (VV – VL) = ΔE + nRT
or, 48600 = ΔE +
18
90 × 2 × 373
or, ΔE = 48600 – 3730 = 44870 cals ∴ (D)
Problem 10: Which of the statement is correct?
(A) Slope of adiabatic P – V curve will be same as that of isothermal one.
(B) Slope of adiabatic P – V curve is smaller than that in isothermal one.
(C) Slope of adiabatic P – V curve is larger than that in isothermal one.
(D) Slope of adiabatic P – V curve may be smaller or larger depending on the value
V.
Solution:
isothermal
adiabatic
(C)
(B)
(A) Slope of PV curve, for isothmermal process
V
P
dV
dP


Slope of PV curve for adiabatic process
V
P
dV
dP



As γ > 1
isothermal
adiabatic dv
dp
dV
dP













∴ (C)
Problem 11: The specific heats of iodine vapour and solid are 0.031 and 0.055 cals/g
respectively. If hets of iodine is 24 cals/g at 200°C,m its value of 250 C is
(A) 5.7 cals gm–1
(B) 11.4 cal g–1
(C) 22.8 cal g–1
(D) 45,6 cal g–1
Solution: Solid I2 is converted in I2 vapour
I2(s) = I2(g) ΔH = 24 cal/g at 200°C
ΔCp = (CP)2(g) = (Cp)I2(s) = 0.031 – 0.055 = – 0.24 cal g–1
From Kirchoff’s equation
 
C
200
C
250
p
C
200
C
250
T
T
C
H
H 


 





C
250
H 
 = 24 – 0.024 (523 – 473)
= 24 – 1.2 = 22.8 cals g–1
∴ (C)
Problem 12: Bond dissociation energy of CH3 – H bond is 103 kcals mole–1
and heat of
formation of CH4(g) as – 17.88 kcals mole–1
. The heat of formation of methyl radical is
(A) 3.61 kcals mole–1
(B) – 3.61 kcals mole–1
(C) 33.61 kcals mole–1
(D) – 33.61 kcals/mole–1
Solution: CH4(g) = CH3(g) + H(atom, g)
  H
3
CH
f
H 
 ) CH3 – H = 103 = ΔHf (CH3, g) + ΔHf (H atom, g) – ΔHf (CH4, g)
  g
,
3
CH
f
H
 = 103 –
 
889
.
17
2
103


= 33.611 kcals mole–1
∴ Heat of formation of methyl radical is ΔH = 33.611 kcals mole–1
∴ (C)
Problem 13: Given, Hs(g) = 2H(g) DH – H = 103 kcals mole–1
CH4(g) = CH3(g) + H(g) H
3
CH
D 
= 103 kcals mole–1
The heat of reaction of
CH4(g) = CH3(g) + H(g)

(A) 103 kcals mole–1
(B) 206 kcals mole–1
(C) 51.5 kcals mole–1
(D) zero
Solution: CH4(g) = CH3(g) + H(g) …1 H
3
CH
D 
= 103 kcals mole–1
H(g) + H(g) = H2(g) …2 H
H
D 
= – 103 kcals mole–1
–––––––––––––––––––––––
CH4(g) + H(g) = CH4(g) + H2(g) ΔH = 103 – 103 = 0 ∴ (D)
Problem 14: For the combustion of n - octane
C8H18 + O2  CO2 + H2O at 25°C (ignoring resonance in CO2)
(A) H = E – 5.5  8.31  0.298 in kJ/mol
(B) H = E + 4.5  8.31  0.298 in kJ/mol
(C) H = E – 4.5  8.31  0.298 in kJ/mol
(D) H = E – 4.5 + 8.31  0.298 in kJ/mol
Solution: C8H18(l) + 12.5 O2(g)  8CO2(g) + 9H2O(l)
H = E + (8 – 12.5)  8.31  0.298 in kJ/mol
(C)
Problem 15: The enthalpies of neutralisation of HClO4 and Cl3C – COOH are
– 13.5 kcal/g – equivalent and
– 13.5 kcal/g – equivalent respectively
when 40 gm of solid NaOH is added to a mix of 1 gm mol. HclO4 and 1 g mol. Cl3C –
COOH, sodium perchlorate and sodium trichloroacetate are formed in molar ratio of 3:1. Then.
(A) H for the reaction of NaOH with the mix. Is 6.45 kcal.
(B) H for the reaction of NaOH with the mix. Is 13.8 kcal.
(C) After reaction the total no. of moles of acid remained are 0.5 (HclO4 & Cl3C –
COOH)
(D) After reaction the total wt. of acid remained is 147.75 gm. (HclO4 + Cl3Cl – COOH)
and H for the reaction of NaOH with mix. is –13.8 kcal.
Solution: HclO4 + NaOH  NaClO4 + H2O
Cl3C – COOH + NaOH  Cl3C – COONa + H2O
Let x mol of NaOH are used in 1st
reaction and (1 – x) mol of NaOH are used in 2nd
reaction.
Given
4
3
x
3
x
1
x




H = –13.5 
4
3 + (–14.7) 
4
1
=
4
7
.
14
5
.
40 
 = –13.8 kcal
Weak acids =
4
1 (1 + 35.5 + 64) + 3
4
3 (35.5  3 + 24 + 32 + 1)
= 147.75 gm
(D)
12. Assignment (Subjective Problems)
LEVEL – I

1. A Carnot’s engine working between two temperatures has an efficiency of 40%. When
the temperature of the sink is reduced by 60°C, the efficiency increases to 55%. Calculate the
two temperatures in the first case.
2. The bond enthalpy of H2(g) is 436 kJ mol–1
and that of N2(g) is 941.3 kJ
mol–1
. Calculate the average bond enthalpy of an N – H bond in ammonia 0
f
H
 (NH3) = –46.0 kJ
mol–1
.
3. 0.50 g of benzoic acid was subjected to combustion in a bomb calorimeter at 15°C
when the temperature of the calorimeter system (including water) was found to rise by 0.55°C.
Calculate the heat of combustion of benzoic acid: (i) at constant volume, (ii) at constant pressure.
The thermal capacity of the calorimeter system including water was found to be 23.85 kJ.
4. The heats of formation of carbon dioxide and water are 395 kJ and 285 kJ respectively,
and the heat of combustion of acetic acid is 869 kJ (all exothermic). Calculate the heat of formation
of acetic acid.
5. The heats of formation of PCl3 and PH3 are 306 kJ mole–1
and +8kJ mole–1
respectively, and the heats of atomization of phosphorus, chlorine and hydrogen are given by:
Ps = Pg ΔH = 314 kJ mole–1
Cl2(.g) = 2Clg ΔH = 242 kJ mole–1
H2,(g) = 2Hg ΔH = 433 kJ mole–1
Calculate EP – Cl and EP – H
6. A Carnot’s engine works between 120°C and 30°C. Calculate the efficiency. If the
power produced by the engine is 400 watts, calculate the heat absorbed from the source and
rejected to the sink every second.
7. Heats of neutralisation (H) of NH4OH and HF are –51.5 and 68.6 kJ respectively.
Calculate their heats of dissociation.
8. If EC – C is 344 kJ mole–1
and EC – H is 415 kJ mole–1
, calculate the heat of formation of
propane. The heats of atomization of carbon and hydrogen are 716 kJ mole–1
and 433 kJ mole–1
respectively.
9. Calculate the amount of heat evolved during the complete combustion of 100 ml of
liquid benzene from the following data:
i) 18 gm graphite on complete combustion evolves 590 KJ of heat
iii) 15889 kJ of heat is required to dissociate all the molecules of 1 letre water into H2 &
O2.
iii) The heat of formation of liquid benzene is +50 kJ/mole.
iv) Density of C6H6(l) = 0.87 gm ml–1
10. From the given data calculate the heat of hydration of CuSO4.
i) CuSO4 (anmhydrous)+ 800 H2O(l) CuSO4 (800 H2O); H = –66.50 kJ
iii) CuSO45H2O + 795 H2O(l)  CuSO4 (800 H2O); H = +11.71 kJ
11. The heat of reaction (H) for the formation of NH3 according to the reaction
N2 + 3H2 2NH3 at 27°C was found to be –91.94 kJ kJ. What will be the heat of reaction H
at 50°C. R [= 8.31 J/mol – 0
k].

12. For the reaction A2(g) + 3B(g)  AB(g) + AB2(g)
.
kJ
100
H K
3000 

 Calculate .
H K
6000
 Given molar heat capacity of monoatomic gas
= ( + T) J/mol
diatomic gas =
mol
/
J
T
2
5
7









polyatomic gas =
mol
/
J
T
4
5
8









(,  are constatns)
13. Calculate the enthalpy of vaporization for water from the following
H2(g) + 1/2 O2(g) ⎯ → H2O (g) ΔH = – 57.0 kcal
H2(g) + 1/2 O2(g) ⎯ → H2O(l) ΔH = – 68.3 kcal
Also calculate the heat required to change 1 gm H2O (l) to H2O (g)
14. The standard enthalpy of combustion of H2, C6H10 and Cyclohexane (C6H12) are – 241,
– 3800, – 3920 kJ mole–1
at 25°C respectively. Calculate the heat of hydrogeneation of
cyclohexane.
15. Calculate the heat of formation of KOH (s) using the following equations.
K(s) + H2O(l) + aq ⎯ → KOH (aq) + 1/2 H2(g) ΔH = – 48 kcal …(1)
H2(g) + 1/2 O2 (g) ⎯ → H2O(l) ΔH = – 68.4 kcal …(2)
KOH (s) + aq ⎯ → KOH (aq) ΔH = –14.0 kcal …(3)
LEVEL – II
1. A reversible carnot engine produces 3.36×104
J. ,of work by taking up 10 Kcal heat. If
the temperature of the source is 527°C. at what temperature the sink be placed ?
2. 10 gm of Helium at 127°C is expanded isothermally from 100 atm to 1 atm
Calculate the work done when the expansion is carried out (i) in single step (ii) in three
steps the intermediate pressure being 60 and 30 atm respectively and (iii) reversibly.

3. The temperature and pressure of one mole of an ideal monoatomic gas are changes
from 25°C and 1 atm pressure to 250°C and 25 atm. Calculate the change of entropy for the
process
4. For the reaction
2Hg(l) + O2(g) ⎯ → 2HgO(s)
the entropies are
S0
(Hg) = 18.5 cal/K mole
S0
(O2) = 49.0 cal/K mole
S0
(HgO) = 17.2 cal/K mole
The heat of formation, ΔHf
0
of HgO is – 21.68 kcal/mole. What is the standard free energy of
formation of HgO(s)? Is the reaction at 25°C, for the production of HgO in its standard state from
Hg and O2 in their standard state spontaneous.
5. A natural gas may be assumed to be a mixture of CH4 and C2H6 only. On complete
combustion of 10 L of the gas at STP, the heat evolved was 474.6 kJ. Assuming ΔHc(CH4(g)) = –
894 kJ/mol and ΔHC(C2H6(g)) = – 1560 kJ/mole. Calculate the % of volume of each gas in the
mixture.
6. The enthalpy of evaporation of water at 373 K is 40.67 kJ mol–1
. What will be the
enthalpy of evaporation at 353 K and 393 K if average molar heats at constant pressure in this
range for water in liquid and vapour states are 75.312 and 33.89 JK–1
mol–1
respectively?
7. The heat of formation of one mole of HI from hydrogen and iodine vapour at 25°C is
8000 cal.
2
1 H2(g) +
2
1 I2(g) ⎯ → HI(g), ΔH = – 8000 cal
Calculate the heat of formation at 10°C and also the total change in the heat capacity at
constant pressure. It is given that the molar heat capacities of hydrogen, iodine and HI vapours
are given by the equations.
CP = 6.5 + 0.0017 T for Hydrogen (g)
CP = 6.5 + 0.0013 T for Iodine (g)
Cp = 6.5 + 0.0016 T for HI (g)
Where T is the absolute temperature
8. Calculate the boiling point benzene under a pressure of 80 cm of Hg. The normal
boiling point is 80°C. Latent heat of vaporization is 380 Joule/gm density of vapour at the boiling
point is 4 gm/litre and that of the liquid 0.9 gm/ml.
9. Calculate o
f
H
 for chloride ion from the following data
1/2 H2 (g) + 1/2 Cl2(g) → HCl (g) o
f
H
 = – 92.4 kJ
HCl (g) + nH2O → H+
(aq) + Cl–
(aq) 298
H
 = –74.8 kJ
o
f
)
aq
(
H
H 
 = 0.0 kJ
10. Determine the enthalpy of formation of anhyd. Al2Cl6(s)
Given
2Al(s) + 6HCl(aq.) ⎯ → Al2Cl6(aq.) + 3H2(g) + 239760 cal.
ΔHf (HCl(g)) = –22 kcal/mol
Maximum amount of heat of solution of HCl(g) in water = –17315 cal/mol

Maximum amount of heat of solution of Al2Cl6(s) in water = –153.69 kcal/mol.
*11. The heat of neutralization of (i) CHCl2 – COOH by NaOH is 12830 cazl. (ii) HCl by NaOH
is 13680 cal. (iii) NH4OH by HCl is 12270 cal. What is the heat of neutralization of dichlioro acetic
acid by NH4OH? Calculate also the heats of ionization of dichloro acetic acid and NH4OH.
12. Calculate the value of 0
f
H
 for OH–
(aq.) if
0
f
H
 (H2O(l)) = –68 kcal/mol
0
f
H
 (H+
(aq.)) = 0
*13. A constant pressure calorimeter consists of an insulated beaker of mass 92 g made up of
glass with heat capacity 0.75JK–1
g–1
. The beaker contains 100 ml of 1 M HCl at 22.6°C to which
100 ml 1 M NaOH at 23.4°C is added. The final temperature after the reaction is complete is
29.3°C. What is ΔH per mole for this neutralization reaction. Assume that the heat capacities of
all solutions are equal to that of same volumes of water.
14. The heat of combustion of glycogen is about 476 kJ/mole of carbon. Assume that
average rate of heat loss by an adult male is 150 watts. If we were to assume that all the heat
comes from oxidation of glycogen, how many units of glycogen (1 mole carbon per unit) must be
oxidised per day to provide for this heat loss.
*15. At 25°C, the following heat of formation are given:
Compound SO2(g) H2O(l)
0
f
H
 kJ/.mole –296.81 –285.83
For the reactions at 25°C
2H2S(g) + Fe(s) → FeS2(s) + 2H2(g) ΔH° = –137 kJ /mole
H2S(g) + 3/2O2(g) → H2O(l) + SO2(g) ΔH° = –562 kJ/mole
Calculate heat of formation of H2S(g) and FeS2(s) at 25°C
LEVEL - III
1. Calculate the depression of melting point of ice by one atmosphere increase of
pressure, given latent heat of ice = 3.35 × 105
J/kg and the specific volumes of 1 kg of ice and
water at 0°C are 1.090 × 10–3
m3
and 10–3
m3
respectively.
2. 12.5 gm of ice at – 24°C is converted into steam at 100°C. Calculate the change in
entropy. It is given that latent heat of steam = 536 cal/g, latent heat of ice = 80 cal/gm. Specific
heat of ice = 0.5 cal/gm K.
3. For the standard cell
Ag(s) | AgCl(s) | Cl–
(aqs)| Hg2Cl2(s) } Hg(I) | Pt
Ecell is 0.0454V and the cell reaction is
2Ag(s) + Hg2Cl2(s) ⎯ → 2Hg(I) + 2AgCl(s)
the heat of formation of AgCl(s) is – 30.36 kcal/mol and of Hg2Cl2 is – 63.32 kcal/mole. Calculate
(a) ΔH (b) Δ?G and (c) ΔS for the cell reaction at 25°C.

4. An aero plane weighing 70,000 Kg files up from sea level to a height of 7000 meter,
Its engine run with pure normal octane [C8H18] has 30% efficiency. Calculate the fuel cost of the
flight, if octane sells at Rs 5/- per litre.
Given density of octane = 0.705 gm/ml, heat of combustion of octane=1300 Kcal/mol (g=981
cm/sec2
)
5. 5 mole of ice at 0°C and 7.6 mm Hg pressure is converted to water vapour at a constant
temperature and pressure. Find ΔH and ΔE if the latent heat of fusion of ice is 80 cal/gm and
latent heat of vaporization of liquid water at 0°C is 596 cal/gm and the volume of ice in comparison
to that of water(vapour) is neglected.
6. H2 gas is mixed with air at 25°C under a pressure of 1 atmosphere and exploded in a
closed vessel. The heat of the reaction.
H2(g) +
2
1 O2(g) ⎯ → H2O (g)
At constant volume, ΔE298 = –240.60 kJ mol–1
and Cp values for H2O vapour and N2 in the
temperature range 298 K and 3,200 K are 39.06 JK–1
mol–1
and 26.40 JK–1
mol–1
respectively.
Calculate the explosion temperature under adiabatic conditions.
7. Estimate the adiabatic flame temperature of a spirit lamp, provided the following data.
C2H5OH(l) + 3O2(g) ⎯ → 2CO2(g) + 3H2O(g)
ΔH298K = 1370 kJ mol–1
CPm for (i) CO2(g) = 36.43 JK–1
mol–1
, (ii) H2O(g) = 33.70 JK–1
mol–1
and (iii) N2(g)
= 294.4 JK–1
mol–1
If the above reaction is carried out in air consisting of 80% N2 and 20% of O2 by volume,
what would be the final temperature?
8. Use Born-Haber cycle to determine the lattice energy of KCl(s) at 298.15 K. The
following data are given at the same temperature,
Enthalpy of formation of KCl = –435.9 kJ mol–1
Ionization energy of K = 418.9 kJ mol–1
Electron affinity of Cl = 348.6 kJ mol–1
Dissociation enthalpy of Cl2 = 242.0 kJ mol–1
Sublimation enthalpy of K = 89.8 kJ mol–1
9. Compute the heat of reaction at 10000
C for
2
1 H2(g) +
2
1 Cl2(g) ⎯ → HCl(g) ΔH0
298K = 92.236 kJ mol–1
0
P
C (H2, g = 29.0284 – 0.8355 × 10–3
T + 2.0097 × 10–6
T2
0
P
C (Cl2, g) = 31.6555 + 1.0134 × 10–2
T – 4.0337 × 10–6
T2
0
P
C (HCl, g) = 28.1359 + 1.8078 × 10–3
T + 1.5453 × 10–6
T2
10. The combustion of 1 g of benzene in a bomb calorimeter evolves 41.746 kJ of heat
at 25°C, (i) What is ΔU0
for combustion of benzene? (ii) Calculate heat of formation of benzene.
11. From the following data, calculate ΔHf at 298.15 K for (i) OH–
(aq1) and (ii) Na+
(aq)
0
f
H
 values in kJ mol–1
at 298.15 K for
H2O (l) = – 285.8 HCl (aq) = –167.4
NaOH (aq) = –469.6 NaCl (aq) = –407.0

12. From the following data:
ΔHC – H = 416 kJ/mol ΔHO – O = 494 kJ/mol
ΔHC = O – 711 kJ/mol ΔHO – H = 464 kJ/mol
mol
/
kJ
40
H )
g
(
2
)
l
(
2 O
H
O
H 
 
ΔHH – H = 435 kJ/mol
ΔHCl – Cl = 243 kJ/mol RE (CO2) = 181 kJ/mol
Calculate ΔH0
comb (CH4),
0
comb
H
 (CH3Cl(g), 0
f
H
 (H2O(l))
and then calculate ΔH° for CH4(g) + Cl2(g) ⎯ → CH3Cl(g) + HCl(g)
Also calculate ΔH° for the monochlorination of methane without finding heats of
combustion of CH4(g), CH3Cl(g) and heat of formation of H2O(l) and HCl(g)
13. An intimate mixture of Fe2O3 and Al is used in solid fuel rocket. Calculate the fuel value
per gram and per ml of mixture.
3
2O
Al
H
 = 399.0 Kcal.
3
2O
Fe
H
 = 199.0 kcal density of Fe2O3 and
Al are 5.2 g / ml and 2.7 g / ml respectively.
14. Compute the heat of formation of liquid methyl alcohol in kilojoules per mole, using the
following data. Heat of vaporization of liquid methyl alcohol = 38 kJ / mole . Heat of formation of
gaseous atoms from their elements in their standard states : H = 218 kJ / mole, C2 = 715 kJ / mole
, O = 219 kJ . mole . Average bond energies:
C–H = 415 kJ / mole, C–O = 356 kJ / mole , O – H =463 kJ / mole
15. If the heat released in the combustion of 1 mol of methane at constant volume and
298K is transferred to 5 L of water, how much temperature of the latter is raised? Given
(g)
4
CH
o
f
H
 = – 749. kJ mol–1
)
g
(
CO
o
f 2
H
 = – 393.5 kJ / mol and o
f )
(
O
2
H
H l
 = – 285 kJ mol–1
.
13. Assignment (Objective Problems)
LEVEL – I
1. Which one is not a state function’
(A) Internal Energy (B) Volume
(C) Heat (q) (D) Enthalpy
2. The spontaneous nature of reaction is impossible if
(A) ΔH is +ve, ΔS is also +ve (B) ΔH is +ve, ΔS is –ve
(C) ΔH is –ve ΔS is +ve (D) ΔH is +ve, ΔS is –ve
3. An ideal gas undergoing expansion in vacuum shows
(A) ΔE =0 (B) W = 0
(C) q = 0 (D) All
4. Human body is an example
(A) Open system (B) Closed system
(C) Isolated system (D) None
5. In a reversible process, the values of ΔSsystem + ΔSsurr. Is
(A) > 0 (B) < 0
(C) = 0 (D) All

6. A system is provided 50 joule of heat and work done on the system is 10 J. The change
in internal energy during the process is
(A) 40 J (B) 60 J
(C) 80 J (D) 50 J
7. Molar heat capacity for a gas at constant temperature and pressure is
(A) 3/2R (B) 5/2R
(C) depends on atomicity of gas (D) infinity (∞)
8. The work done in an open vessel at 300K, when 112g iron reacts with dil. HCl is
(A) 1200 cal (B) 600 cal
(C) 300 cal (D) 200 cal
9. Which reaction proceeds with increase entropy
(A) CaCO3 ⎯ → CaO + CO2 (B) Fe + S ⎯ → FeS
(C) NaOH + HCl ⎯ → NaCl + H2O (D) CH4 + 2O2+ ⎯ → CO2 + H2O
10. What is enthalpy change of the following reaction
CH2 = CH2 + H2(g) ⎯ → CH3 – CH3
Given, C – H, C – C, C = C, H – H are 414, 347, 615 & 435 kJ mole–1
respectively.
(A) – 100 kJ (B) – 125 kJ
(C) – 150 kJ (D) – 175 kJ
11. What is average S-F bond energy in SF6 ? given ΔHf of SF6(g)
S(g) & F(g) are – 1100, 275 & 80 kJ mole–1
respectively
(A) 520 kJ (B) 401 kJ
(C) 309 kJ (D) 254 kJ
12. The heat of reaction for N2 + 3H2 ⎯ → 2NH3 at 27°C is – 91.94 kJ. What will be its
value at 50°C? The molar heat capacities at constant P & 27°C for N2, H2 and NH3 are 28.45,
28.32 & 37.07s J respectively.
(A) – 92.843kJ (B) – 47.7723 kJ
(C) –132.5 kJ (D) –176.114 kJ
13. Given ΔHionisation (HCN) = 45.2 kJ mole–1
& ΔHionization (CH3CO2H) = 2.1 kJ mole–1
which
of the following statement is correct?
(A) pKa(HCN) = pKa(CH3CO2H) (B) pKa(HCN) > pKa(CH3CO2H)
(C) pKa(HCN) < pKa(CH3CO2H) (D) pKa(HCN) =
07
.
2
17
.
45 pKa(CH3CO2H)
14. If H2(g) 2H(g) ΔH = 104 kcals the
Then heat of atomisation of H2 is
(A) 52 kcals (B) 104kcals
(C) 20 kcals (D) none of these
15. At 298 k in a constant volume calorimeter, 0.01 mole of TNT was detonated when
8180 cals of heat was released. Each mole of TNT gives 6 moles of gaseous products on
detonation. What is ΔH/mole of TNT exploded?
(A) – 714 kcals mole–1
(B) – 814 kcals mole–1
(C) –914 kcals mole–1
(D) none of these
16. The latent heat of vaporisation of Br2 at 59°C at 1 atm is 29.2 kJ mole–1
. ΔE (energy
change at constant volume) of this process is

(A) 25 kJ (B) 50 kJ
(C) 75 kJ (D) none of these
17. The H – H bond energy is 430 kJ mole–1
& Cl – Cl bond energy is 240 kJ mole–1
. (ΔH)f
for HCl is – 90kJ. The H – Cl bond energy is about
(A) 425 kJ mole–1
(B) 213 kJ mole–1
(C) 360 kJ mole–1
(D) 180 kJ mole–1
18. ΔH for combustion of ethane are – 341.1 & – 310 kcals respectively. Which is better
gas welder
(A) ethane
(B) ethyne
(C) both a having same gas welding capacity
(D) None of these
19. What is ΔE for the reversible isothermal evaporation of 90g of water at 100°C?
(A) 44870 cals (B) 48600 cals
(C) 49320 cals(D) none of these
20. What is ΔH0
for 2Al + Fe2O3 ⎯ → 2Fe + Al2O3.? Standard heat enthalpy of Fe2O3 &
Al2O3 are –196.5 & – 399.1 kcals
(A) – 302.6 kals (B) – 202.6 kcals
(C) – 102.6 kcals (D) +20.2 kcals
LEVEL – II
1. Which plot represents for an exothermic reaction
(A)
H
R
P
Time
(B)
H
R
P
Time
(C)
H
R P
Time
(D)
H
R
P
Time
2. The word ‘standard’ in standard molar enthalpy change implies
(A) temperature 298 K
(B) pressure 1 atim
(C) temperature 298 K and pressure 1 atm
(D) All temperatures and all pressures
3. The heats of neutralization of four acids A, B, C, D are – 13.7, – 9.4, – 11.2 and – 12.4
kcal respectively when they are neutralized by a common base. The acidic character obeys the
order.

(A) A > B > C > D (B) A > D > C > B
(C) D > C > B > A (D) D > B > C > A
4. Which fuel provides the highest calorific value
(A) Charcoal (B) Kerosene
(C) Wood (D) Dung
5. Given at
A(s) ⎯ → A(l); ΔH = x A(l) ⎯ → A(g); ΔH = y
The heat of sublimation of A will be
(A) x – y (B) x + y
(C) x or y (D) – (x + y)
6. A person requires 2870 kcal of energy to lead normal daily life. If heat of combustion
of cane sugar is – 1349 kcal, then his daily consumption of sugar is
(A) 728 g (B) 0.728 g
(C) 342g (D) 0.342 g
7. If combustion of 4 g of CH4 liberates 2.5 kcal of heat, the heat of combustion of CH4 is
(A) – 20 kcal (B) – 10 kcal
(C) 2.5 kcal (D) – 5 kcal
8. AB, A2 and B2 are diatomic molecules. If the bond enthalpies of A2, AB and B2 are in
the ratio 1:1:0.5 and the enthalpy of formation of AB from A2 and B2 – 100 kJ mol–1
, what is the
bond enthalpy of A2
(A) 400 kJ mol–1
(B) 200 kJ mol–1
(C) 100 kJ mol–1
(D) 300 kJ mol–1
9. ΔCp for the change N2(g) + 3H2(g) = 2NH3(g) is
(A) 2
pN
3
pNH
C
C  (B) )
C
3
C
(
C
2 2
pH
2
pN
3
pNH


(C) )
C
(
C 2
pH
3
pNH
 (D) )
C
3
C
(
C
2 2
pH
2
pN
3
pNH


10. The heats of oxidation of Mg & Fe are given
a) Mg +
2
O
2
1 ⎯ → MgO ΔH = – 145700 cals
b) 2Fe +
2
O
2
3 ⎯ → Fe2O3 ΔH = – 193500 cals
The heat produced in the reaction 3Mg + Fe2O3 ⎯ → 3MgO + 2Fe
(A) – 243600 cals (B) – 257800 cals
(C) – 271400 cals (D) – 222800 cals
11. The heats produced in oxidation of cane sugar, carbon & hydrogen are given as
a) C12H22O11 + 12O2(g) = 12CO2(g) + 11H2O(l) ΔH = – 1350 kcals
b) C + O2(g) = CO2(g) ΔH = – 94.05 kcals
c) H2(g) +
2
1 O2(g) = H2O(l) ΔH = – 68.3 kcals
The heat produceds of canesugar is produced directly from the elements is
(A) – 527.8 kcals (B) – 525.6 kcals
(C) – 529.9 kcals (D) none of above
12. H+
(aq) + NaOH(aq) = Na+
+ H2O(l) ΔH1 = – 1390 cals

HCN (aq) + NaOH(aq) ⎯ →Na+
+ C–
N + H2O (l) ΔH2 = –2900 cals
What is ΔH value for HCN (aq) = H+
(aq) + CN–
(aq)
(A) 11400 cals (B) 10790 cals
(C) 12500 cals(D) 9800 cals
13. The latent heat of vaporisation of Br2 at 59°C at 1 atim is 29.2 kJ/mole. ΔE (energy
change at constant volume) of this process is
(A) 25 kJ (B) 50 kJ
(C) 75 kJ (D) none of these
14. Isothermally at 27°C, 1 mole of Vander Waals’ gas expands reversibly from 2 lits to 20
lits. The work done if
a = 1.42 × 1012
dynes cm4
/mole & b = 30 cc/mole
(A) 5.677 ×103
kJ (B) 2.3 × 103
kJ
(C) 9.2 × 103
kJ (D) 7.689 × 103
kJ
15. At 298°K in a constant volume calorimeter, 0.01 mole of TNT detonates when 8180
cals of heat was released. Each mole of TNT gives 6 moles of gaseous proudcts on detonation.
What is ΔH/moles of TNT exploder.
(A) – 714 kcals/mole (B) – 814 kcals/mole
(C) – 914 kcals/mole (D) none of these
16. Identify the intensive quantities from the following
(A) Enthalpy (B) entropy
(C) volume(D) refractive index
17. If the enthalpy of vaporisation for water is 186.5 kJ mole–1
, its entropy of vaporisation
will be
(A) 0.5 kJ mole–1
k–1
(B) 1 kJ mole–1
(C) 1.5 kJ k–1
mole–1
(D) 2 kJ k–1
mole–1
18. During isothermal expansion of an ideal gas its
(A) internal energy increase (B) enthalpy decreases
(C) enthalpy remains unaffected (D) Enthalpy reduces to zero
19. The heat of combustion of C is 394 kJ. The heat evolveds in combustion of 6.023 ×
1022
atoms of carbon is
(A) 3940 kJ (B) 394 kJ
(C) 39.4 kJ(D) 0.394 kJ
20. The temperature at which the reaction,
Ag2O(s) ⎯ → 2Ag(s) + 1/2O2(g)
Is at equilibrium is …; Given ΔH = 30.5 kJ mol–1
and ΔS = 0.066 kJ K–1
mol–1
(A) 462.12 K (B) 362.12 K
(C) 262.12 K (D) 562.12 K

14. Answers to Subjective Assignments
LEVEL – I
1. 400 K 2. 390.2 kJ/mol
3. – 3201.9 kJ 4. – 491 kJ
5. 319.33 kJ 6. 1347 watts
7. – 11.3 kJ 8. – 128 kJ /mol3
9. 3645 kJ 10. – 78.21 kJ
11. –93.087 kJ 12. (–110 + 135 β – 0.42α)kJ
13. 0.628 KCal 14. – 121 kJ
15. –102.4 Kcal
LEVEL – II
1. 157 K 2. i) 3230.365; (ii) 0.5519.45J; (iii) 38.294 kJ
3. – 3.6226 cal 4. – 14.0
5. % of CH4 = 74.5%, % of C2H6 = 25.5% 6. – 41.498 kJ, 39.841 kJ
7. – 7.995 Kcal, – 0.00115 Tcal deg–1
8. 81.233°C
9. 310.56 kJ mol–1
10. – 321.96 Kcal mol–1
11. –11.42 Kcal, 1410 Cal / mol 12. – 55 Kcal mol–1
13. –57 kJ 14. 27.22 units
15. –20.64 kJ mol–1
, –178.28 kJ
LEVEL – III
1. – 0.0074°C 2. 109.58 J/K

3. 2.6 kcal, – 2.10 kcal, 15.8 cal/K 4. 2375.88 Rs
5. 58.125 kcal 6. 2917 K
7. 2898 K 8. – 475.0 kJ/mol
9. – 94.827 kJ/mol 10. 44.35 kJ/mol
11. – 239.6 kJ/mol 12. – 887kJ/mol, – 683 kJ/mol, – 102 kJ/mo
13. – 167.2 kJ 14. – 266 kJ/mol
15. 42.44 K
15. Answers to Objective Assignments
LEVEL – I
1. C 2. B
3. D 4. A
5. C 6. A
7. C 8. A
9. A 10. B
11. C 12. A
13. B 14. A
15. B 16. D
17. A 18. B
19. A 20. B
LEVEL – II
1. D 2. C
3. B 4. B
5. B 6. A
7. B 8. B
9. B 10. A
11. C 12. B
13. D 14. A
15. B 16. D
17. A 18. C

19. C 20. A
6