Thermodynamic Notes PDF

CHEMICAL THERMODYNAMICS
CHEMICAL THEROMODYNAMICS
1. Thermodynamics
The branch of science which deal with study of different forms of energy and the quantities
relationship between them is known as Thermodynamics.
When the study of thermodynamics is confined to chemical changes and chemical substances
only, it is known as chemical thermodynamics.
Energetics. It is that branch of chemistry which deals with energy changes taking place in a
reaction.
Application in Chemistry :
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Thermodynamics helps in
(a)
Determining feasibility of a particular process i.e., whether or not a particular process
will occur under a given set of conditions.
(b)
Determining the extent to which a reaction would proceed before attainment of
equilibrium.
(c)
Most important laws of physical chemistry such as Raoultss law, vant Hoff law,
distribution law, phase rule, law of equilibrium, laws of thermochemistry and expression
for elevation in boiling point and depression in freezing point are in accordance with
laws of thermodynamics.
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2. Some fundamental Definitions

System : That part of the universe which is chosen for thermodynamics considerations is
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called system.
Surrounding : The remaining portion of the universe which is not chosen for thermodynamic
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consideration is called surrounding.
Boundary : The imaginary line which separates the system from the surrounding is called
boundary.
4.
Types of system :
(i)
(ii)
(iii)
5.
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Open system : A system is said to be an open systems if it can exchange both matter
and energy with the surroundings.
Closed system : If a system can exchange only energy with the surrounding but not
matter is called closed systems.
Isolated system : If a system can neither exchange matter nor energy with the
surrounding it is called an isolated systems.
State of a System and state variable :

(i)
(ii)
The existence of a system under a given set of conditions is called a state of systems.
The properties which change with change in the state of system are called as state
variables e.g., pressure, volume and temperature etc.
The first and last state of a system are called initial state and final state respectively.
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6.
State function and Path Function : A physical quantity is said to be state function if its
value depends only upon the state of the system and does not depend upon the path by which
this state has been attained. For example, a person standing on the roof of a five storeyed
building has a fixed potential energy, irrespective of the fact whether he reached there by
stairs or by lift. Thus the potential energy of the person is a state function. On the other hand,
the work done by the legs of the person to reach the same height, is not same in the two cases
i.e., whether he went by lift or by stairs. Hence work is a path function.
7.
Extensity and Intensive properties : An extensive property of a system is that which

depends upon the amount of the substance or substances present in the system. e.g., mass,
volume, energy etc.
An intensive property of a system is that which is independent of the amount of the substance
present in the system e.g., temperature, pressure, density, velocity etc.
8.
Thermodynamics Processes : The operation by which a thermodynamic system changes
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form one state to another is called is thermodynamic process.

(i)
Isothermal process : A process in which although heat enters or leaves the system
yet temperature of the system remains constant throughout the process is called an
isothermal temperature of the system remains constant throughout the process is
called an isothermal process. For an isothermal process, change in temperature
(T) = 0. Change of state (e.g., freezing, melting, evaporation and condensation) are
all examples of isothermal process.
(ii)
Adiabatic process : A process during which no heats enters or leaves the system
during any step of the process is known as adiabatic process. A reaction carried out
in an isolated system is an example of adiabatic process. For an adiabatic process,
change in heat (q) = 0 or q remain constant.
(iii)
Isobaric process : A process during which pressure of the system remains constant
throughout the reaction is called as isobaric process. For example, heating of water to
its boiling point, and its vaporisation taking place at the same atmospheric pressure.
Expansion of a gas in an open system is an example of isobaric process. For an
isobaric process P = 0.
(iv)
Isochoric process : A process during which volume of the system remains constant
throughout the reaction is known as isochoric processes. The heating of a substance
is a non-expanding chamber or change taking place in a closed system are examples
of isochoric process. For an isochoric process, V = 0.
(v)
Cyclic process : A process during which system comes to its initial state through a
number of different processes is called a cyclic process. For a cyclic process, E = 0,
H = 0.
9.
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Internal energy (E) : The total energy stored in a substance by virtue of tis chemical nature
and state is called its internal energy, i.e., it is the sum of its translation, vibrational, rotational,
chemical bond energy, electronic energy, nuclear energy of constituent atoms and potential
energy due to interaction with neighbouring molecules. It is also called intrinsic energy.
E = Et + Er + Ev + Ee + En + E PE
Internal energy is a state property and its absolute value cant be determined. However,
change in internal energy (difference between the internal energies of the products and that of
reactants) can be determined experimentally using a bomb calorimeter.
Internal energy of a system depends upon :
(a)
the quantity of substance
(b)
its chemical nature
and
(c)
temperature, pressure and volume.
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(i)
(ii)
(iii)
(iv)
(v)
(vi)
10.
For a given system, E is directly proportional to its absolute temperature. (E T)

At constant volume, the quantity of heat supplied to a system (isochoric process) is
equal to the increase in its internal energy, i.e., QV = E
In the adiabatic expansion of a gas, it gets cooled because of decrease in internal energy.
In cyclic process the change in internal energy is zero (E = 0) since E is a state
function.
For exothermic reactions, sign of E is negative (ER > EP).
For endothermic reactions, sign of E is positive (EP > ER).
Work : Work is expressed as the product of two factors, i.e.,

W = Intensity factor capacity factor
where, intensity factor is a measure of force responsible for work and capacity factor is a
measure extent to which the work is done. Thus,
(a)
Mechanical work = Force Displacement = F d
(b)
Electrical work = Potential difference Charge flown = V Q = EnF
(c)
Expansion work = Pressure change in volume = P V
(d)
Gravitation work = Gravitational force Height = mg h
(a)
Pressure volume work (Irreversible)
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It is a kind of mechanical work. The expression for such a work may be derived as follows
Consider a gas enclosed in a cylinder fitter with a weightless and frictionless piston.
Suppose area of cross reaction of a cylinder = a sq. cm.
Pressure on piston = P which is less than the internal pressure such that the gas expands.
Let dl be the distance covered by the piston when the gas expands. work done is given by
dW = Fdl.
= Force distance
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Force
= Pressure Force = Pressure area
Area
dW = P.a.dl
= PdV
Let the volume limits be V1 and V2. Hence integrating the equation dW = PdV
dW = PdV
(b)
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W = PV
W = (V2 V1)
When expansion takes place
V2 > V1 W = ve
Hence work is done by the system
When compression of gas takes place then V 2 < V1, W = +ve. Hence work is done on
the system.
Work during isothermal Reversible Expansion of an Ideal Gas

We know that
dW = PdV
v2
dW = Pdv
v2
For an ideal gas PV = nRT
P =
nRT
V
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w=
v2
v2
nRT
dV
V
W = nRT
v2
v2
dV
V
= nRT n Vv12
v
= nRT ln
V2
V1
V2
V1
W = 2.303 nRT log
1
Also as V P
W = 2.303 nRT log
Work is not a state function because amount of work performed depends upon the
path followed.
Positive value of work signifies that the work has been done on the system by the
surroundings and it leads to an increase in the internal energy of the system. On the
other hand, negative value of work indicates that work has been done by the system
and it leads to decrease in the internal energy of the system.
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3. Zeroth Law
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(ii)
P1
P2
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(i)
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P2
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W = 2.303 nRT log
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It states that Two system in thermal equilibrium separately with the third system are said to
be in thermal equilibrium with each other i.e., If system A and B separately are in thermal
equilibrium with another system, then system A and B are also in thermal equilibrium.
4. I law of Thermodynamics
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The first law of thermodynamics states that Energy can neither be created nor be
destroyed although it can be converted from one form to another.
Let a system be at state I with internal energy E 1,
Let it be change to State II with internal energy E 2
This can be achieve in the ways :
(i)
by heat transfer
(ii)
By doing work (either on system or by system)
Let the heat change taking place during the change of state of system from state I to state II
be q and work done be W.
E2 = E1 + q + W
E2 E1 = q + W
E = q + W
or
E = q PV
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Illustration : 1.
A gas expands by 0.5 litre against a constant pressure one atmosphere. Calculate the
work done in joule and calorie.
Solution :
Work = Pext volume change
= 1 0.5 = 0.5 litre-atm
= 0.5 101.328 J = 50.644 J
0.5 lit-atm = 0.5 24.20 cal = 12.10 cal
Illustration : 2.
One mole of an ideal gas is put through a series of changes as shown in the graph in
which A, B, C mark the three stages of the system. At each stage the variables are
shown in the graph.
(a)
Calculate the pressure at three stages of the system.
(b)
Name the processes during the following changes:
(i) A to B
(ii) B to C
(iii) C to A, and
(iv) overall change.
A
12.0 (L)
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300 K
T
300 K
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At stage A;
V = 24.0 L; T = 300 K; n = 1 ; R = 0.0821 lit-atm K 1 mol1
Substituting these values in the ideal gas eqation,
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Solution :
(a)
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24.0 (L)
1 0.0821 300
= 1.026 atm
24.0
At stage B : Volume remains the same but temperature change from 300 K to 600 K .
Thus, according to pressure law, the pressure will be double at B with respect to A.
Pressure at B = 2 1.026 = 2.052 atm
At stage C : Temperature is 300 K and volume is half that of stage A. Thus, according to
Boyles law, the pressure at C will be double with respect to A.
Pressure at C = 2 1.026 = 2.052 atm
P=
(b)
(i)
(ii)
(iii)
(iv)
During the change from A to B, volume remains constant, the process is

isochoric.
During the change from B to C the pressure remains constant, the process is
isobaric.
During the change from C to A, the temperature remains constant, the process
is isothermal.
Overall, the process is cyclic as it returns to initial state.
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Illustration : 3.
The diagram shows a P-V graph of a thermodynamic behaviour of an ideal gas. Find
out form this graph :
(i)
work done in the process A B, B C, C D and D A
(ii)
work done in the complete cycle A B C D A.
Solution :
P(105Newton/m2)
12
10
8
6
4
2
2 3 4
V (litre)
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Work done in the process AB (the process is expansion, hence work is done by the gas)
= P V = 12 10 5 5 103
= 6000 J
Work done in the process B C is zero as volume remains constant.
Work done in the process C D (The process is contraction, hence work is one on the gas)
= P dV = 2 10 5 5 103
= 1000 J
Work done in teh process D A is zero a volume remains constant.
Net work one in the whole cycle = 6000 + 1000 = 5000 J
i.e. net work is done by the gas.
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Illustration : 4.
Calculate the work done when 1.0 mole of water at 373K vaporizes against an
atmospheric pressure of 1.0 atmosphere. Assume ideal gas behaviour.
Solution :
The volume occupied by water is very small and thus the volume change is equal to the volume
occupied by one gram mole of water vapour.
V=
W =
=
=
=
nRT
1.0 0.821 373
=
= 31.0 litre
P
1 .0
Pext V
(1.0) (31.0) litre-atm
(31.0) 101.3 J
3140.3 J
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Illustration : 5.
Calculate w and E for the conversion of 0.5 mole of water at 100C to steam at
1 atm pressure. Heat of vaporisation of water at 100C is 40670 J mol 1.
Solution :
Volume of 0.5 mole of steam at 1 atm pressure
nRT
0.5 0.0821 373
=
= 15.3 L
P
1 .0
Change in volume = Vol. of steam vol. of water
= 15.3 negligible = 15.3 L
Work done by the system,
w = Pext volume change
= 1 15.3 = 15.3 litre-atm
= 15.3 101.3 J = 1549.89 J
w should be negative as the work has been done by the system on the surroundings.
w = 1549.89 J
Heat required to convert 0.5 mole of water in 100C to steam
= 0.5 40670 J = 20335J
According to first law of thermodynamics,
E = q + w
= 20335 1549.89
= 18785.11 J
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Illustration : 6.
Calculate the work done when 50 g of iron is dissolved in HCl at 25C in :
(i)
a closed vessel
and
(ii)
an open beaker when the atmospheric pressure is 1 atm.
Solution :
(i)
When the reaction is carried in a closed vessel, the change in volume is zero. Hence,
the work done by the system will be zero.
(ii)
When iron dissolves in HCl, hydrogen is produces.
Fe + 2HCl
FeCl2 + H2
56 g
1 mole
1
50 mole
56
Volume of hydrogen produced at 25C
50 g
nRT
50
0.0821 298
=
P
56
1
= 21.84 L
This is equal to volume change when the reaction is carried in open beaker.
Work done by the system = PV = 1.0 21.84
= 21.84 litreatm
= 2212.39 J
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Illustration : 7.
Calculate the amount of work done by 2 mole of an ideal gas at 298 K in reversible
isothermal expansion from 10 litre 20 litre.
Solution :
Amount of work done in reversible isothermal expansion
w = 2.303nRT log
V2
V1
Given n = 2, R = 8.314 JK 1 mol1, T = 298 K, V2 = 20 L and V1 = 10L.

Substituting the values in above equation
20
10
= 2.303 2 8.314 298 0.3010
= 3434.9 J
i.e., work is done by the system.
w = 2.303 2 8.314 298 log
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Illustration : 8.
5 mole of an ideal gas expand isothermally and reversibly from a pressure of 10 atm
to 2 atm at 300 K. What is the largest mass which can be lifted through a height of
1 metre in this expansion ?
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P1
P1
= 2.303 nRT log10 P
2
P2
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= nRT loge
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Solution :
Work done by the system
= 2.303 5 8.314 300 log
10
2
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= 20.075 10 3 J
Let M be the mass which can be lifted through a height of 1m.
Work done in lifting the mass
= M g h = M 9.8 1J
So
M 9.8 = 20.075 103
M = 2048.469 kg
5. Enthalpy
Internally energy changes are usually measured at constant volume. But in actual practice, most
processes are carried out at constant pressure rather than constant volume. Hence volume
changes which occurs cause changes in internal energy. To account for these changes, a new
thermodynamic property is introduced called as Enthalpy.
It is defined as sum of internal energy and product of pressure volume work
It is donated by the letter H.
H = E + PV
Enthalpy of a system is also called heat content of system, because it is the net energy available
in a system which can be converted into heat.
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Characteristics :
(a)
It is a state functions
(b)
It is an extensive property
(c)
Change in its value can be determined by relationships H and E
Let a system at state-I be transformed to state-II at constant pressure condition.
State - I
State-II
Enthalpy
H1
H2
Internal energy
E1
E2
Pressure
P
P
Volume
V1
V2
H1 = E1 + PV1
H2 = E2 + PV2
H2 H1 = H = (E2 + PV2) (E1 + PV1)
= (E2 E1) + P (V2 V1)
H = E + PV
Also for an ideal gas PV = nRT
PV1 = n1 RT and PV2 = n2RT
P(V2 V1) = ng RT
Then H = E + ng RT
qp = qv + ng RT
When ng = 0
H = E
When ng > 1
H > E
When ng < 1
H < E
Also E = qP PV
qP = E + PV
and
H = qP (i.e., Enthalpy change = heat exchange at constant pressure condition)
also E = qV (i.e., Internal energy change = heat exchange at constant volume condition)
H = E + PV
qP qV = PV = ng RT
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Illustration : 9.
The heat of combustion of ethylene at 18C and at constant volume is 335.8 kcal
when water is obtained in liquid state. Calculate the heat of combustion at
constant pressure and at 18C.
Solution :
The chemical equation for the combustion of C2H4 is
C2H4(g) + 3O2(g) = 2CO2(g) + 2H2O (l); E = 335.8 kcal
1mole
3 mole 2 mole
No. of moles of gaseous reactants = (1 + 3 ) = 4
No. of moles of gaseous products = 2
So
n = (2 4) = 2
Given E = 335.8 kcal, n = 2, R = 2 10 3 kcal
and
T = (18 + 273) = 291 K
H = E + nRT
= 335.8 + (2) (210 3) (291) = 336.964 kcal
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Illustration : 10.
The enthalpy of formation of methane at constant pressure and 300 K 75.83 kJ.
What will be the heat of formation at constant volume ? [ R = 8.3 JK 1 mol1 ]
Solution :
The equation for the formation of methane is
C(s) + 2H2(g) = CH4(g) ; H = 75.83 kJ
2 mole 1 mole
n = (1 2) = 1
Given
H = 75.83 kJ, R = 8.3 10 3 kJ K1 mol1
T = 300 K
Applying
H = E + nRT
75.83 = E + (1) (8.3 10 3) (300)
E = 75.83 + 2.49
So
= 73.34 kJ
6. Heat capacity of a system
q
q
=
T2 T1 T
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Then, heat capacity = C =
.in
It is the amount of heat required to raise the temperature of a system through 1C

If q is the amount of heat supplied to a system and as a result let the temperature rise from
T1 to T2 C.
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q
dT
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Specific Heat and Molar Heat Capacity at Constant Volume

Specific Heat : It is the amount of heat required to raise the temperature of 1 gm of a gas
through 1 at constant volume.
Molar heat capacity : It is the amount of heat required to raise the temperature of one mole of
a gas through 1 at constant volume
dq = dE + PdV
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(a)
C=
yS
Then
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When heat capacity varies with temperature then the value of C has to be considered over a
narrow range of temperature.
Molar heat Capacity : C =
dq
dE PdV
=
dT
dT
At constant volume dV = 0
Cv
Hence
E
E
C= =
T V T V
dE
dT
Cv
= SV = Specific heat at constant volume.
Molecular mass
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(b)
Heat Capacity at Constant Pressure :

It is defined as the amount of heat required to raise the temperature of one mole of gas through
1o keeping pressure constant
E
V
C P P
T P
T P
.......(a)
We know that H = E + PV
Differentiating w.r.t. temperature
H
E
H
T P T P
T P
H
CP
T P
From (a) and (b)
CP
= SP = Specific heat at constant pressure
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CP is always greater than CV

When gas is heated at constant volume, the pressure of gas has to increase.
As the gas is not allowed to expand, therefore in case of C V heat is required for raising the
temperature of one mole of a gas through 1o. When gas is heated at constant P. It expands, the
gas has done some work against external pressure. More heat is therefore supplied to raise its
temperature through 1o.
Thus CP is heat required for the purpose of
(i)
Increasing temperatures of one mole of gas through 1o.
(ii)
For increasing the volume of the gas against external pressure.
CP > CV
Relationship between CP and CV.

According to I law of thermodynamics.
dqp = dqV + PdV
CPdT = CVdT + PdV
For one mole of gas

PV = RT
PdV + VdP = RdT
Since the gas is being heated at constant pressure dP = 0
PdV = RdT
CPdT = CVdT + RdT
Dividing both sides by dT
CP = CV + R
CP - CV = R
CP CV R
Calculation of W , E , H , for Isothermal Expansion of Ideal Gas

(A)
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(i)
(ii)
(iii)
E for an ideal gas E depends on temperatures. Since temperature is constant

dE = 0
E = 0
According to law
dE = dq + dW;
since dE = 0
dq = -dW
q = -W
Heat absorbed is equal to work done by the system during isothermal expansion of
ideal gas
Enthalpy change H = E + PV
H = E + PV = E + (nRT)
H = E + nRT
(Because T=0)
H = 0 + 0
Hence, H = 0
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(B)
Adiabatic Reversible Expansion of ideal Gas :

(a)
Calculation of E
= f (T , V)
E
E
dE T TV
T V
T
E
But T = 0
T
Also dE = nCVdT
and dE = CVdT
E = E2 - E1 = nCV(T2-T1)
E
dE = dT
T V
for n moles
for one moles
= E nCV T
.in
E or E2 E1 = nCV (T2 T1)

Also q = 0
E = W = CV(T2 T1) [Because expansion of gas takes place]
W = CV(T2 T1)
W = nCV(T2T1) = nCV T
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Illustration : 11.
A gas expands from a volume of 3.0 dm 3 to 5.0 dm 3 against a constant pressure of
3.0 atm. The work done during expansion is used to heat 10.0 mole of water of
temperature 290.0 K. Calculate the final temperature of water .
[ specific heat of water = 4.184 JK 1 g1 ]
Solution :
Work done = P dV = 3.0 (5.0 3.0) = 6.0 litreatm = 6.0 101.3 J = 607.8 J
Let T be the change in temperature.
Heat absorbed = m S T = 10.0 18 4.184 T
Given
P dV = m S T
St
P dV
607.8
=
= 0.807
mS
10.0 18.0 4.184
Final temperature = 290 + 0.807 = 290.807 K
or
T =
Illustration : 12.
How much heat is required to change 10 g ice 0C to steam at 100C? Latent heat of
fusion and vaporization for H 2O are 80cal/g and 540cal/g respectively. Specific heat
of water is 1 cal/g.
Solution :
Total heat absorbed
= Hfusion + Htemp.rise + Hvap.
= 10 80 + 10 1 100 + 10 + 540 = 7200 cal.
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7. Corollary
(a)
Adiabatic process
(i)
Adiabatic compression
E = W. ,
E = + PV
(ii)
Adiabatic expansion
E = W
E = PV
Hence during adiabatic compression of an ideal gas internal energy of system increases and
during adiabatic energy of system decreases.
Isochoric process E = qv
(i)
Isochoric absorption of heat
E = +q internal energy of system increase
(ii)
Isochoric liberation of heat
E = q internal energy of system decreases
(c)
Heat absorbed by system and work done by system

E = +q W
(d)
Heat liberation by system and work done on the system

E = q + W
(e)
Isobatic process (expansion)

E = +qP PV
(i)
when no gases are involved in reaction
E = +qP
(ii)
When gases are involved but initial and final volumes are not given
E = q ng RT
where ng = nP(g) nR(g)
Q.1
Instructions : From Question (1 to 3) a single statement is made. Write T if the

statement is true & F if the statement is false.
A thermodynamic equilibrium is one when all the three thermal, mechanical and chemical
equilibrium are attained by the system.
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(b)
Q.2
Ice in contact with water constitutes a homogeneous system.
Q.3
The state properties are those which depends on the path followed by a system in bringing a
particular change.
Q.4
The change in entropy in going from one state to another is independent of path.
II.
Instructions : Questions (7 to 18) consist of a problem followed by several alternative

answers, only one of which is correct. Mark the letter corresponding to the correct
answer.
For a hypothetical system, consider these conditions
(i)
heat transferred to the surrounding
(ii)
work done on the system
(iii)
work done by the system.
State whether each of the following will increase or decrease the total energy content of the
system
(A) decreases, increases, decreases
(B) increases, increases, decreases
(C) decreases, increases, increases
(D) decreases, decreases, increases
Q.5
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Q.6
A well stoppered thermos flask contains some ice cubes. This is an example of :
(A) closed system
(B) open system
(C) isolated system
(D) non-thermodynamic system
Q.7
Internal energy change when system goes from state A to state B is 40 kJ/mol. If the system
goes from A to B by reversible path and returns to state A by irreversible path, what would
be net change in internal energy ?
(A) 40 kJ
(B) > 40 kJ
(C) < 40 kJ
(D) zero
Q.8
A system X undergoes following changes

The overall process may be called
(A) reversible process

(C) cyclic as well as reversible
(B) cyclic process

(D) isochoric process
Work done in taking the gas from the state A B

(A) 36 102 J
(B) 36 102 J
(C) 60 102 J
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III. A In the adjoining diagram, the P-V graph of an ideal gas is shown. Answer the following
question (11 to 13) from the graph.
Work done in a complete cycle (1 litre = 10 3 m3)

(A) 36 102 J
(B) 36 102 J
(C) 60 102 J
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Q.11
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Q.10 Work done in taking the gas from B C

(A) zero
(B) 1.5 102 J
(C) 3.0 102 J
(D) 60 102 J
(D) 3.0 102 J
(D) 60 102 J
III. B For 1 mole of an ideal gas following V-T graph is given
Answer the following questions (14 to 18) from the graph

Q.12 The pressure at A and B in atmosphere are respectively :
(A) 0.821 & 1.642
(B) 1.642 & 0.821
(C) 1 & 2
(D) 0.082 & 0.164
Q.13 The pressure at C is :

(A) 3.284 atm
(B) 1.642 atm
(D) 0.821 atm
(C) 0.0821 atm
Q.14 The process which occurs in going from B C is :

(A) isothermal
(B) adiabatic
(C) isobaric
Page 14 of 51
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(D) isochoric
Q.15 The work done in going form C to A is :

(A) zero
(B) 1.15 kJ
(C) 2.3 kJ
Q.16 The process A B refer to :

(A) isoentropic process
(C) isochoric process
(B) reversible process

(D) isobaric process
Q.1
Q.2
Which one is not a state function ?

(A) internal energy (E)
(C) heat
(B) volume
(D) enthalpy
Which is the intensive property ?

(A) temperature
(B) viscosity
(C) density
(D) unpredictable
(D) all
Warming ammonium chloride with sodium hydroxide in a test tube is an example of :

(A) closed system
(B) isolated system (C) open system
(D) none of these
Q.4
The work done by a system in an expansion against a constant external pressure is :

(A) P. V
(B) P. V
(C) Q
(D) V. P
Q.5
An example of extensive property is :

(A) temperature
(B) internal energy
ps
.in
Q.3
(C) viscosity
(D) surface tension
A gas expands isothermally and reversibly. The work done by the gas is :
(A) zero
(B) minimum
(C) maximum
(D) None
Q.7
A system is changed from state A to state B by one path and from B to A by another path. If
E1 and E2 are the corresponding changes in internal energy, then :
(A) E1 + E2 = +ve
(B) E1 + E2 = ve
(C) E1 + E2 = 0
(D) none
Q.8
The maximum work done in expanding 16 g oxygen at 300 K and occupying a volume of
5 dm3 isothermally until the volume becomes 25 dm3 is :
(A) 2.01 103 J
(B) +2.81 103 J
(C) 2.01 10 3 J
(D) +2.01 106 J
Q.9
1 mole of gas occupying 3 litre volume is expanded against a constant external pressure of
1 atm to a volume of 15 litre. The work done by the system is :
(A) 1.215 10 3 J (B) 12.15 10 3 J (C) 121.5 10 3 J (D) none
St
ud
yS
te
Q.6
Q.10 If 50 calorie are added to a system and system does work of 30 calorie on surroundings, the
change in internal energy of system is :
(A) 20 cal
(B) 50 cal
(C) 40 cal
(D) 30 cal
Q.11
Page 15 of 51
One mole of an ideal gas at 300 K is expanded isothermally from an initial volume of 1 litre to
10 litre. The E for this process is : [ R = 2 cal K1 mol1 ]
(A) 163.7 cal
(B) 1381.1 cal
(C) 9 litreatm
(D) zero
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Q.12 A thermodynamic process is shown in the following figure. The pressure and volumes
corresponding to some point in the figure are :
PA = 3 104 Pa , P B = 8 104 Pa , VA = 2 103 m3 , VD = 5 103 m3
In the process AB, 600 J of heat is added to the system and in BC, 200 J of heat is added to
the system. The change in internal energy of the system in the process AC would be :
(A) 560 J
(B) 800 J
(C) 600 J
(D) 640 J
.in
Q.13 The q value and work done in the isothermal reversible expansion of one mole of an ideal
gas from an initial pressure of 1 bar to a final pressure of 0.1 bar at a constant temperature of
273 K are :
(A) 5.22 kJ, 5.22 kJ
(B) 5.22 kJ, 5.22 kJ
(C) 5.22 kJ, 5.22 kJ
(D) 5.22 kJ, 5.22 kJ
ps
Q.14 In a given process of an ideal gas, dW = 0 and dQ < 0. Then for the gas :
(A) the temperature will decrease
(B) the volume will increase
(C) the pressure will remain constant
(D) the temperature will increase
St
ud
yS
te
Q.15 Five moles of a gas is put through a series of changes as shown graphically in a cyclic
process. The process during A B , B C and C A respectively are :
(A) isochoric, isobaric, isothermal

(C) isochoric, isothermal, isobaric
(B) isobaric, isochoric, isothermal

(D) isobaric, isothermal, isochoric
Q.16 20 gm of N2 at 300 K is compressed reversibly and adiabatically from 20 dm 3 to 10 dm3.

Change in internal energy for the process is :
(A) 284.8 J
(B) 142.46 J
(C) 1424.69 J
(D) 3462.89 J
Page 16 of 51
8. Thermochemistry
Thermochemistry is the branch of physical chemistry which deals with the transfer of heat
between a chemical system and its surrounding when a change of phase or a chemical reaction
takes place with in the system. It is also termed as chemical energetics.
Thermochemical equations : A chemical reaction 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 informations.
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(i)
(ii)
(iii)
(iv)
It tells about the physical state of the reactants and products. This is done by inserting
symbols (s), (l) and (g) for solid, liquid and gaseous state respectively with the
chemical formulae.
It tells about the allotropic from (if any) of the reactant.
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, i.e. heat change involved in the system.
Exothermic Reactions :
Heat is evolved in these chemical reactions. It is possible when the bond energy of reactants is
less at constant pressure
H (HP HR) = ve i.e., HP < HR
At constant volume
E E P E R = ve i.e, E P E R
Endothermic Reactions :
i.e.,
At constant volume
E E P E R ve
Sign conventions :
Page 17 of 51
ve
+ve
yS
ve
+ve
ud
Exothermic
Endothermic
E P > ER
te
i.e.,
HP > HR
ps
H H P H R ve
.in
Heat is absorbed in these chemical reactions. It is possible when bond energy of reactants is
greater than the bond energy of products.
At constant pressure
St
9. Heat of Reaction or Enthalpy of Reaction

Enthalpy : Heat content of a system at constant pressure is called enthalpy denoted by H.
From 1st Law of thermodynamics :
Q = E + PV
Heat change at constant pressure can be
Q E PV
Enthalpy of reaction is the difference between the enthalpies of the products and the
reactants when the quantities of the reactants indicated by chemical reaction have completely
reacted. Enthalpy of reaction ( or heat of reaction )
H H P H R
For example, the equation
H2(g) + Cl2 (g) = 2HCl (g) + 44.0 Kcal or H = 44 kcal
C2H4(g) + 3O2(g) = 2CO2 + 2H2O(l) ;
E = 335.8 kcal
This equation indicates that reaction has been carried between 1 mole of C 2H4 and 3 mole
of oxygen at constant volume and 25C. The heat evolved is 335.8 kcal or the internal
energy of the system decreases by 335.8 kcal
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10. Factors which affect the heat of reaction

(a)
Physical State of Reactants and Products : The heat of reaction varies for a given reaction
with the change in physical nature of reactants or products e.g.,
C
CO ; H=-94.3 kcal
For reactant
2
2
diamond
having different
C
O
CO ; H=-97.7 kcal
2
2
Physical state Amorphous
For products having different physical state ;

1
H 2(g ) O 2 g
H 2O l ; H 68.3 Kcal
2
1
H 2 g O 2 g
H 2O g ; H 57.0 Kcal
2
Reaction carried out a constant pressure or constant volume :

From 1st Law of thermodynamics
q = E W
at constant volume
qV = E
and at constant pressure
qp = H
Two values are related as
.in
(b)
H E nRT
yS
Alternatively
: Pressure at which the reaction is carried out

V : Change in volume during the course of reaction.
te
where P
ps
H E PV
n No. of moles of products No. of moles of reactant (only gas phases)
Temperature : Heat of reaction also depends upon the temperature at which reaction is
carried out. The variation in H value with temperature are due to variation in heat capacities
of system with temperature.
Kirchoffs equation
St
ud
(c)
H 2 H1 C P T2 T1
and
E 2 E1 C V T2 T1
where C P C p of products - C P of reactants
C v C V of products - C V of reactants
H 2 , H1 are change in heat enthalpies at temperature T 2 & T1 respectively
E 2 , E are change in heat internal energy at temperature T 2 & T1 respectively

(d)
Page 18 of 51
Enthalpies of Solution : Enthalpies of reaction differ when in one case dry substances react
and in another case when the same substance react in solution.
e.g.,
H 2S g I 2 g
2HI S ;
H 172 Kcal
H 2S g I 2 solution
2HI solution S;
H 21.93 Kcal
CuSO 4 aq
CuSO 4 H 2 O aq ;
H 15.8 Kcal
CuSO 4 .H 2 O aq
CuSO 4 .5H 2 O aq. ;
H 29 Kcal
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11. Terms used for Heat of Reactions

(1)
Heat of Formation or enthalpy of Formation : The amount of heat absorbed or evolved

when 1 mole of the substance is directly obtained from its constituent elements is called Heat
of formation.
Standard Heat Enthalpy : The heat enthalpy of a compound at 25 C and 1 atm pressure is
known as standard heat enthalpy and represented by the symbol H .
Thus the standard heat of formation of 1 mole of CO 2(g) and 1 mole of H 2O(g) from their
respective elements can be represented as below :
C(g) + O2(g) CO2(g) ;
H f 94 Kcal
H2(g) + 12 O2(g) H2O(g) ;
H f 63 Kcal
The compounds which have positive enthalpies of formation are called endothermic compounds
and areless stable than the reactants. The compounds which have negative enthalpies of formation
are known as exothermic compound are more stable than reactants.
Illustration 13 :
The standard enthalpies of formation at 298 K for CCl 4(g), H2O(g), CO2(g) and HCl(g)
.in
are 106.7, 241.8, 393.7 and 92.5 kJ mol 1, respectively. Calculate H 298K for
ps
the reaction, CCl 4(g) + 2H2O(g) CO2(g) + 4HCl(g)

Solution :
The enthalpy change of the given reaction will be given as
te
H H 0f CO2 , g 4H 0f HCl,g H 0f CCl4 , g 2H 0f H 2O,g

Heat of Combustion or Enthalpy of combustion : It is defined as the change in heat
enthalpy when one mole of a substance is completely burnt in oxygen.
C + O2 CO2 ; H 94.3 Kcal
C + 2S CS2 ; H 22.0 Kcal
St
ud
(2)
yS
= ( 393.7 4 92.5 + 106.7 + 2 241.8 ) kJ mol 1 = 173.4 kJ mol 1.
Calorific Value : The amount of heat produced in calorie or joule when one gram of a
substance (food or fuel) is completely burnt or oxidised.
(3)
Page 19 of 51
Enthalpy of Neutralisation : It is defined as the heat evolved or decrease in enthalpy

when 1 gm equivalent of an acid is neutralised by 1 gm equivalent of a base-in solution.
Strong acid + strong Base Salt + Water ;
H 13.7 Kcal
HCl (aq.) + NaOH (aq.) NaCl (aq.) + H2O(l) ;
H 13.75 Kcal
1
1
H 13.7 Kcal
H 2SO 4 aq. NaOH aq.
Na 2SO 4 aq. H 2O l ;
2
2
Thus heat of neutralisation of a strong acid and a strong base is merely the heat of formation
of water from H+ and OH - ions. When strong acid and a weak base or a weak acid and a
strong acid or weak acid and weak base are mixed in equivalent amounts, the heat evolved or
change in enthalpy is less than 13.7 Kcal. eq.
HCl (aq.) + NH 4OH (aq.) NH4Cl (aq.) + H2O (l) ;
H 12.3 Kcal
HCN (aq.) + NaOH (aq.) NaCN (aq.) + H 2O(l) ;
H 2.9 Kcal
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(4)
Enthalpy of Hydration : It is the amount of heat evolved (i.e., change in enthalpy) when
1 mole of anhydrous or a partially hydrated salt combines with required number of moles of
water to form a hydrate. e.g.,
CuSO4 + 5 H2O (l) CuSO4 . 5 H2O ; H = 18.69 Kcal
CaCl2 (l) + 6 H2O (l) CaCl2 . 6 H2O ; H = 18.8 Kcal
(5)
Enthalpy of ionisation : It is defined as the amount of heat absorbed when 1 mole of an

electrolyte completely dissociates into ions.
CH3COOH CH3COO + H+ ;
H = 3 Kcal
HCN H+ + CN ;
H = 10.8 Kcal
12. Law of Thermochemistry

Levoisier and Laplace Law :
Hesss Law of Constant Heat Summation :
Page 20 of 51
.in
According to this enthalpy of decomposition of a compound is numerically equal to

enthalpy of formation of that compound with opposite sign. e.g.,
C (s) + O2 (g) CO2 (g) ;
H = 94.3 Kcal
CO2 (g) C (g) + O2 ;
H = + 94.3 Kcal
H = - x kJ mol
ud
H2(g) + O2(g)
yS
te
ps
For a chemical equation that can be written as the sum of two or more steps, the enthalpy
change for the overall equation is equal to the sum of the enthalpy changes for the individual
steps. Thus, Hesss law enables us to break down a reaction into so many intermediate steps
and passing to each step an individual enthalpy change. The sum of the individual changes
must, of course, equal the overall enthalpy change provided the initial and final states are the
same in each case.
Route A
-1
H1 = - y kJ mol H2 = - z kJ mol
St
1
2
H2O(l) +
-1
O2(g)
-1
H2O2(l)
Route B
For the above example
H H1 H2
An energy level diagram for the above reaction cycle is shown in figure
H2(g) + O2(g)
H1 = - 187.6 kJ mol-1
H = - 285.9 kJ mol-1
H2O2(l)
H2 = - 98.3 kJ mol-1
1
2
H2O(l) + O2(g)
Fig. Energy level diagram to illustrate Hesss Law
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Illustration 14 :
Compute the resonance energy of gaseous benzene from the following data.
(C H) = 416.3 kJ mol 1
(C C) = 331.4 kJ mol 1
(C = C) = 591.1 kJ mol 1
sub
C, graphite = 718.4 kJ m ol -1
H diss
H 2 , g = 435.9 kJ mol -1
H f benzene, g = 82.9 kJ mol -1
Solution :
To compute resonance energy, we compare the calculated value of H f (benzene, g) with the
given one. To calculate H f (benzene, g), we add the following reactions.
H 3CC 3 CC 6 C H
6C(g) 6H(g) C6 H 6 (g)
H 6 718.4 kJ mol 1
6C(graphite) C(g)
3H 2 (g) 6H(g)
H 3 435.9 kJ mol 1
ps
6C(graphite) + 3H2(g) C6H6(g)

The corresponding enthalpy change is
.in
Add
te
f H 3CC 3 CC 6 CH 6 718.4 + 3 435.9 kJ mol 1

= [ ( 3 331.4 + 3 591.1 + 6 718.4 + 3 435.9] kJ mol 1
Page 21 of 51
Experimental Determination of Heat of Reaction
St
ud
yS
The given H 0f is H 0f (benzene, g) = 82.9 kJ mol1

This means benzene becomes more stable by (352.8 82.9) kJ mol 1,
i.e.,269.7 kJ mol1 . This is its resonance energy.
The apparatus used is called calorimeter. There are two types of Calorimeters :
(a)
Bomb Calorimeter
(b)
Water Calorimeter
Bomb Calorimeter : The calorimeter used for determining enthalpies of combustion known
as the bomb calorimeter is shown in figure. 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.
Ignition Wires
O2
Insulating
Container
Water
Sample
Steel Bomb
Bomb Calorimeter
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Bond Energy
When bond is formed between the two free atoms in a gaseous state to form a molecular
product in a gaseous state, some heat is always evolved which is known as the bond
formation energy or the bond energy. The bone energy may be referred to as heat of formation
of the bond.
Alternatively, bond energy may be defined as the average amount of energy required to
dissociate (i.e. break bonds) of that type present in one molecule of the compound. Thus bond
energy of C H in methane (CH4) is the average value of the dissociation energies of the four
C H bonds .
Hreaction =
Bond energy data used for formation Bond energy data used for dissociation
of bond ( to betaken as ve)
+ of bond ( to betaken as ve)
Hreaction = BE(R) - B.E.(P) taking Bond Energies as +ve values.
Bond Enthalpy :
ps
.in
The average energy required to break a bond in gaseous molecule to produce gaseous
species. Enthalpy of Reaction :
H = (bond energy of bonds broken) (bond energy of bonds formed)
Bond dissociation Energy. The energy required to break a particular bond in gaseous molecule
to form gaseous species.
Lattice energy :
Resonance energy :
(observed heat of formation) (calculated heat of formation) .
ud
yS
te
Enthalpy change when one mole of gaseous ions condense to form a solid crystal lattice.
eg. Na+(g) + Cl (g) NaCl (s) .
Born Haber's cycle is useful in determination of lattice energy & related problems .
St
Illustration. 15 :
Calculate the heat of formation of acetic acid from the following data:
(i)
CH3COOH(l) + 2O2(g)
2CO2(g) +2H2O(l) ;
H = 207.9 kcal
(ii)
C(s) + O2(g)
CO2(g)
H = 94.48 kcal
(iii) H2(g) + 12 O2(g)

H = 68.4 kcal
H2O(l)
Solution :
First method : The required eqation is
2C(s) + 2H2(g) + O2(g) = CH3COOH(l);
H = ?
This equation can be obtained by multiplying Eq. (ii) by 2 and also Eq. (iii) by 2 and adding
both and finally substracting Eq.(i)
[2C + 2O2 + 2H2 + O2 CH3COOH(l) 2O2 = 2CO2 + 2H2O 2CO2 2H2O]
Page 22 of 51
H CH3COOH(l ) = 2 (94.48) + 2 (68.4) (207.9)

= 188.96 136.8 + 207.9
= 325.76 + 207.9 = 117.86 kcal
Second method : From eq. (ii) and (iii)
Enthalpy of CO2 = 94.48 kcal
Enthalpy of H2O = 68.4 kcal
Enthalpy of O2 = 0 (by convention)
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H of Eq. (i) = Enthalphies of products = Enthalpies of reactants

207.9 = 2 (94.48) + 2(68.4) H CH3COOH(l )
H CH3COOH(l ) = 188.96 136.8 + 207.9 = 235.76 + 207.9 = 117.86 kcal

Illustration 16. :
) according to
2
the following reaction ?
CaO (s) + 3C(s) CaC2(s) + CO(g)
The heats of formation of CaO (s), CaC 2(s) and CO(g) are 151.6, 14.2 and 26.4
kcal respectively.
Solution :
H = Hf(products) Hf(reactants)
= [Hf(CaC2) + Hf(CO)] [Hf(CaO) + 3Hf(C)]
H
= [14.2 26.4] [151.6 + 3 0]

= 40.6 + 151.6 = 111.0 kcal
For formation of 64 g of CaC 2 111.0 ckal of heat is required.
So, heat required for making 2000 g of
.in
111.0
2000 = 3468.75 kcal
64
ps
CaC2 =
St
ud
yS
te
Example 17. :
Calculate heat of combustion of ethene:
H
H
C=C
+ 3O = O 2O=C=O + 2H OH
H
H
From bond energy data :
C=C CH O=O C=O OH
1
K.E. KJ mol
619 414 499
724
460
Solution :
H = sum of bond energies of reactants Sum of bond energies of products
= [H(C=C) + 4H(CH) + 3 H(O=O)] [4 H(C=O) + 4 H(OH)]
= [619 + 4 414 + 3 499] [ 4 724 + 4 460]
= 964 kJ mol 1
Illustration 18. :
Calculate the lattice energy for the reaction
Li+ (g) + Cl(g) LiCl (s)
From the following data :
Hsub(Li) = 160.67 kJ mol 1;
and
1 (Cl )
2
2
= 122.17 kJ mol 1
I.P. (Li) = 520.07 kJ mol 1; E.A. (Cl) = 365.26 kJ mol 1

Hf (LiCl) = 401.66 kJ mol 1
Solution :
Applying the equation
Page 23 of 51
Q = H +
1D
2
+ I.P. E.A. + U
And substituting the respectrive values,

401.66 = 160.67 + 122.17 + 520.07 365.26 + U
U = 839.31 kJ mol 1
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Illustration 19. :
Bond dissociation enthalpies of H 2(g) & N2(g) are 436.0 kJ mol 1 and 941.8 kJ mol 1
and enthalpy of formation of NH 3(g) is 46 kJ mol 1. What is enthalpy of atomization
of NH3(g)? What is the average bond enthalpy of NH bond?
Solution :
N2(g) + 3H2(g) 2NH3(g) ; H = 2 46 kJ/mol
H = S(B.E.)R S(B.e.)P
= (941.8 + 3 436) (6x) = 2 46 kJ/mol
(here x = B.E. of NH bonds)
x = 380.3 kJ mol 1
NH3 N + 3(H)
Heat of automization = 3 390.3 = 1170.9 kJ mol 1
Which of the following is incorrect about the reaction :

C(Diamond) + O2(g) CO2(g) ; H = 94.3 kcal at 298 K and 1 atm :
(A)
heat of combustion of C D = 94.3 kcal
(B)
heat of formation of CO2 = 94.3 kcal
(C)
H = E
(D)
standard heat of formation of CO 2 = 94.3 kcal
At constant P and T which statement is correct for the reaction CO(g) + 12 O2(g) CO2(g)
(A) H = E
(B) H < E
(C) H > E
(D) H is independent for physical state of reactant
te
yS
Q.2
ps
.in
Q.1
The formation of water from H2(g) and O2(g) is an exothermic reaction because :
(A)
the chemical energy of H2(g) and O2(g) is more than that of water
(B)
the chemical energy of H2(g) and O2(g) is less than that of water
(C)
not dependent on energy
(D)
the temperature of H2(g) and O2(g) is more than that of water
Q.4
Equal volume of C2H2 and H2 are combusted under identical condition. The ratio of their heat
of combustion is :
St
ud
Q.3
H2(g) + 12 O2(g) H2O(g)
H= 241.8 kJ
C2H2(g) + 2 12 O2(g) 2CO2(g) + H2O(g) ;
H = 1300 kJ
(A) 5.37/1
(B) 1/5.37
(C) 1/1
(D) none of these
Q.5
Given N2(g) + 3H2(g) 2NH3(g) ; H =22 kcal. The standard enthalpy of formation of
NH3 gas is :
(A) 11 kcal/mol
(B) 11 kcal/mol
(C) 22 kcal/mol
(D) 22 kcal/mol
Q.6
Given enthalpy of formation of CO 2(g) and CaO(s) are 94.0 kJ and 152 kJ respectively
and then enthalpy of the reaction :
CaCO3(s) CaO(s) + CO2(g) is 42 kJ. The enthalpy of formation of CaCO 3(s) is :
(A) 42 kJ
(B) 202 kJ
(C) +202 kJ
(D) 288 kJ
Page 24 of 51
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Q.7
The enthalpies of formation of N2O and NO are 28 and 90 kJ mol1 respectively. The enthalpy
of the reaction , 2N 2O(g) + O2(g) 4NO(g) is equal to :
(A) 8 kJ
(B) 88 kJ
(C) 16 kJ
(D) 304 kJ
Q.8
Give standard enthalpy of formation of CO (110 kJ mol 1) and CO2(394 kJ mol 1). The
heat of combustion when one mole of graphite burns is :
(A) 110 kJ
(B) 284 kJ
(C) 394 kJ
(D) 504 kJ
Q.9
The H values for the reaction,

C(s) +
1
O (g) CO(g)
2 2
H = 100 kJ
1
O (g) CO2(g) ; H = 200 kJ
2 2
The heat of reaction for C(s) + O 2(g) CO2(g) is :
(A) 50 kJ
(B) 100 kJ
(C) 150 kJ
CO(g) +
SO2 + 12 O2 SO3
SO3 + H2O H2SO4
; H = 298.2 kJ
........ (i)
; H = 98.7kJ
........ (ii)
; H = 130.2 kJ
........ (iii)
.in
Q.10 If, S + O2 SO2
(D) 433.7kJ
te
ps
H2 + 12 O2 H2O
; H = 227.3 kJ
........ (iv)
The enthalpy of formation of H2SO4 at 298 K will be :
(A) 754.4 kJ
(B) +320.5 kJ
(C) 650.3 kJ
Standard heat of formation of CH4(g) , CO2(g) and water at 25 are 17.9 , 94.1 and
68.3 kcal mol1 respectively. The heat change (in kcal) in the following reaction at 25C is :
CH4(g) + 2O2(g) CO2(g) + 2H2O(l)
(A) 144.5
(B) 180.3
(C) 248.6
(D) 212.8
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Q.11
(D) 300 kJ
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Q.12 According to the equation ,

C6H6(l) + 15
2 O2(g) 3H2O(l) + 6CO2(g)
H = 3264.4 kJ/mol, the energy evolved when 7.8 g of benzene is burnt in air will be :
(A) 163.22 kJ/mol
(B) 326.4 kJ/mol
(C) 32.64 kJ/mol
(D) 3.264 kJ/mol
Q.13 When a certain amount of ethylene was burnt 6226 kJ heat was evolved. If heat of
combustion of ethylene is 1411 kJ, the volume of O2 (at NTP) that entered into the reaction is
(A) 296.5 mL
(B) 296.5 litre
(C) 6226 22.4 litre (D) 22.4 litre
Q.14 For the reactions ,
H2(g) + Cl2(g) 2HCl (g) + x 1 kJ
. . . . (i)
2HCl(g) H2(g) + Cl2(g) x 2 kJ
. . . . (ii)
Which of the following statements is correct :
(A) x1 and x2 are numerically equal
(B) x1 and x2 are numerically different
(C) x 1 x 2 > 0
(D) x 1 x 2 < 0
H2(g) + Cl2(g) 2HCl ; H = 44 kcal
2Na(s) + 2HCl(g) 2NaCl (s) + H2(g) ; H = 152 kcal then
Na(s) + 0.5 Cl 2(g) NaCl(s) ; H = ?
(A) 108 kcal
(B) 196 kcal
(C) 98 kcal
(D) 54 kcal
Q.15 If ,
Page 25 of 51
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Q.16 Combustion of carbon forms two oxides CO and CO 2. Heat of formation of CO2 is 94.3 kcal
and that of CO is 26.0 kcal. Heat of combustion of carbon is :
(A) 26.0 kcal
(B) 94.3 kcal
(C) 68.3 kcal
(D) 120.3 kcal
Q.17 S + 3
2 O2 SO3 + 2x kcal
SO2 + 12 O2 SO3 + y kcal
The heat of formation of SO2 is :
(A) y 2x
(B) (2x + y)
(C) x + y
(D) 2x/y
Q.18 H for CaCO3(s) CaO(s) + CO2(g) is 176 kJ mol 1 at 1240 K. The E for the change
is equal to :
(A) 160 kJ
(B) 165.6 kJ
(C) 186.3 kJ
(D) 180.0 kJ
Q.19 From the thermochemical reactions ,
Cgraphite + 12 O2 CO ; H = 110.5 kJ
.in
CO + 12 O2 CO2 ; H = 283.2 kJ
H for the reaction , Cgraphite +O2 CO2 is :
(A) 393.7 kJ
(B) +393.7 kJ
(C) 172.7 kJ
(D) +172.7 kJ
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Q.20 H2(g) + 12 O2(g) H2O (l) ; H298 K = 68.32 kcal. Heat of vapourisation of water at 1 atm
and 25 C is 10.52 kcal. The standard heat of formation (in kcal) of 1 mole of water vapour
at 25 C is :
(A) 78.84
(B) 78.84
(C) +57.80
(D) 57.80
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Q.21 Given that standard heat enthalpy of CH 4, C2H4 and C3H8 are 17.9, 12.5, 24.8 kcal/mol.
The H for CH4 + C2H4 C3H8 is :
(A) 55.2 kcal
(B) 30.2 kcal
(C) 55.2 kcal
(D) 19.4 kcal
St
Q.22 For the reaction ,

C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l)
at constant temperature, H - E is :
(A) +RT
(B) -3RT
(C) +3RT
(D) -RT
Q.23 The standard enthalpy change for a reaction ,

CO2(g) + H2(g) CO(g) + H2O(g) is [ Given that Hf for CO(g) and H2O(g) as -110.5 and -241.8 kJ mol -1 and
heat of combustion of C(s) is -393.5 kJ/mole ]
(A) 41.2 kJ
(B) -41.2 kJ
(C) -393.5 kJ
(D) +393.5 kJ
Q.24 The standard internal energy change (U) for the reaction,
OF2(g) + H2O(g) O2(g) + 2HF(g) is Given that standard enthalpies of formation Hf of OF2(g), H2O(g) and HF(g) as +23.0,
-241.8, -268.6 kJ mol-1 respectively.
(A) -393.5 kJ
(B) -318.4 kJ
(C) -537.2 kJ
(D) -320.8 kJ
Q.25 The standard enthalpies of formation of NO 2(g) and N2O4(g) 8 and 2 kcal mol-1 respectively.
The heat of dimerisation of NO 2 in gaseous state is (A) 10 kcal mol-1
(B) 6.0 kcal mol-1
(C) -14 kcal mol -1
(D) -6.0 kcal mol-1
Page 26 of 51
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Q.26 Given that , C + O2 CO2 ; H = x kJ and 2CO + O 2 2CO2 ; H = y kJ.

The standard enthalpy of formation of carbon monoxide is :
(A) y 2x
(B)
2x y
2
(C)
y 2x
2
(D) 2x y
NH3 + 3Cl2(g) NCl3(g) + 3HCl(g) ; H1 = x 1

N2(g) + 3H2(g) 2NH3(g)
; H2 = x 2
H2(g) + Cl2(g) 2HCl(g)
; H3 = + x 3
The heat of formation of NCl 3(g) from the above data is -
Q.27 Given that ,
(A) -x 1 +
x2 3
x2 3
x3 (B) x 1 +
- x3
3
2
2
2
(C) x 1 -
x2 3
- x3
2
2
(D) -x 1 -
x2 3
- x3
2
2
Q.28 Hf of CO2(g), CO(g), N2O(g) and NO2(g) in kJ/mol are respectively - 393, -110, 81 and
34. H in kJ of the following reaction : 2NO 2(g) + 3CO(g) N2O(g) + 3CO2(g)
(A) 836
(B) 1460
(C) -836
(D) -1460
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Q.29 If E is the heat of reaction,

C2H5OH(l) + 3O2(g) 2CO2(g) + 3H2O(l)
at constant volume, the H (heat of reaction at constant pressure) at constant temperature is (A) H = E - RT
(B) H = E - 2RT
(C) H = E + 2 RT
(D) H = E + RT
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Q.30 Change in enthalpy for the reaction,

2H2O2(l) 2H2O(l) + O2(g)
if heat of formation of H2O2(l) and H2O(l) are -188 kJ mol-1 and -286 kJ mol-1 respectively is
(A) -196 kJ mol -1
(B) +196 kJ mol -1
(C) +948 kJ mol -1
(D) -948 kJ mol -1
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Q.31 Enthalpy of CH4 + 12 O2 CH3OH is negative. If enthalpy of combustion of CH 4 and

CH3OH are x and y respectively, then which relation is correct :
(A) x > y
(B) x < y
(C) x = y
(D) x y
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Q.32 The heat of combustion of graphite and carbon monoxide respectively are -393.5 kJ mol-1 and
-283 kJ mol-1. Therefore the heat of formation of carbon monoxide in kJ mol -1 is :
(A) +172.5
(B) -110.5
(C) -1070
(D) +110.5
Q.33 In a reaction involving only solids and liquids, which of the following is true
(A) H < E
(B) H = E
(C) H > E
(D) H = E + RT n
Q.34 H for the reaction,
CH3COOC2H5(l) + H2O(l) CH3COOH(l) + C2H5OH(l)
Given heat of formation of CO2(g) H2O(l), C2H5OH(l) is a, b and c kJ/mole respectively and
heat of combustion of CH3COOC2H5(l) and CH3COOH (l) is d and e kJ/mole respectively.
Q.35 H for the reaction,
C3H8(g) + 92 O2(g) CO(g) + 2CO2(g) + 4H2O(g)
Given heat of formation of CO(g), C3H8(g), H2O(l) is 110.5 kJ/mole, 104.16 kJ/mole and
286 kJ/mole respectively. Heat of combustion of CO(g) is 283.2 kJ/mole and heat of
vapourisation of water is 44.2 kJ/mole.
Page 27 of 51
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The amount of heat evolved when one mole of H 2SO4 reacts with two mole of NaOH is :
(A) 13.7 kcal
(B) less than 13.7 kcal(C) more than 13.7 kcal
(D) none
Q.2
Energy required to dissociate 4 g of gaseous hydrogen into free gaseous atom is 208 kcal at
25C. The bond energy of H H bond will be :
(A) 104 kcal
(B) 10.4 kcal
(C) 1040 kcal
(D) 104 cal
Q.3
The heat of neutralisation of HCl by NaOH is 55.9 kJ/mol. If the heat of neutralisation of
HCN by NaOH is 12.1 kJ/mol. The energy of dissociation of HCN is :
(A) 43.8 kJ
(B) 43.8 kJ
(C) 68 kJ
(D) 68 kJ
Q.4
Bond energies of (H H), (O=O) and (OH) are 105, 120 and 220 kcal/mol respectively
then H in the reaction,
2H2(g) + O2(g) 2H2O(l)
(A) 115
(B) 130
(C) 118
(D) 550
Q.5
Heat evolved in the reaction , H 2 + Cl2 2HCl is 182 kJ. Bond energies of H H and
Cl Cl are 430 and 242 kJ/mol respectively. The HCl bond energy is :
(A) 245 kJ mol 1
(B) 427 kJ mol 1
(C) 336 kJ mol 1
(D) 154 kJ mol 1
Q.6
If the enthalpy change for the reaction

CH4(g) + Cl2(g) CH3Cl(g) + HCl(g)
H = 25 kcal. bond energy of C H is 20 kcal mol 1 greater than the bond energy of C Cl
and bond energies of H H and HCl are same in magnitude, then for the reaction :
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Q.1
H = ?
(C) 32.5 kcal/mol
(D) 12.5 kcal/mol
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1 H (g) + 1 Cl (g) HCl(g) ;

2 2
2 2
(A) 22.5 kcal/mol (B) 20.5 kcal/mol
Enthalpy of neutralization of HCl with NaOH is x. The heat evolved when 500 ml of 2 N
HCl are mixed with 250 ml of 4 N NaOH will be
(A) 500 x
(B) 100 x
(C) x
(D) 10 x
Q.8
The bond energies of C C, C = C, H H and C H linkages are 350, 600, 400 and
410 kJ per mole respectively. The heat of hydrogenation of ethylene is :
(A) 170 kJ mol 1
(B) 260 kJ mol 1
(C) 400 kJ mol 1
(D) 450 kJ mol 1
Q.9
The enthalpy change (Hf) of the following reaction,

2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(g) is
Given average bond energies of various bonds, i.e., C H, C C, O = O, C = O, O H
as 414, 814, 499, 724 and 640 kJ mol 1 respectively
(A) 7632 kJ
(B) 186.1 kJ
(C) 2573 kJ
(D) 763.2 kJ
St
Q.7
Q.10 The bond energy of C H bond in methane from the following data :
(a)
C(s) + 2H2(g) CH4(g) ; H = 74.8 kJ
(b)
H2(g) 2H(g)
; H = 435.4 kJ
(c)
C(s) C(g)
;H = 718.4 kJ
1
(A) 416 kJ mol
(B) 1664 kJ mol 1
(C) 217.7 kJ mol 1 (D) 1741 kJ mol 1
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Q.11
For the following reaction,

CDiamond + O2 CO2(g) ; H = 94.3 kcal
CGraphite + O2 CO2(g) ; H = 97.6 kcal
(i)
The heat of transition for C Diamond CGraphite
(A) +3.3 kcal
(B) 3.3 kcal
(C) 19.19 kcal
(ii)
The heat required to change 1g of C diamond Cgraphite is :

(A) 1.59 kcal
(B) 0.1375 kcal
(C) 0.55 kcal
(D) 191.9 kcal

(D) 0.275 kcal
Q.12 For the following reaction,

H2(g) + 12 O2(g) H2O(g) ; H = 57.0 kcal
H2(g) + 12 O2(g) H2O(l) ; H = 68.3 kcal
(i)
The enthalpy of vapourization for water is :
(A) 11.3 kcal
(B) 11.3 kcal
(C) 125.3 kcal
(ii)
The heat required to change 1 g of H 2O(l) H2O(g) is :
(A) 1.56 kcal
(B) 0.313 kcal
(C) 1.25 kcal
(D) 125.3 kcal

(D) 0.628 kcal
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Q.13 The heat of reaction for N 2 + 3H2 2NH3 at 27C is 91.94 kJ. What will be its value
at 50C ? The molar heat capacities at constant P and 27C for N 2, H2 and NH3 are
28.45, 28.32 and 37.07 joule respectively
(A) +45.74 kJ
(B) +92.84 kJ
(C) 45.74 kJ
(D) 92.84 kJ
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Q.14 Enthalpy of neutralization of HCl by NaOH is 57.32 kJ mol 1 and by NH4OH is 51.34 kJ
mol1. The enthalpy of dissociation of NH4OH, is :
(A) 4.98 kJ mol 1
(B) 108.66 kJ mol 1 (C) 108.66 kJ mol 1 (D) 5.98 kJ mol 1
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Q.15 The enthalpy of formation of H 2O (l) is 285.83 kJ mol1 and enthalpy of neutralisation of a
strong acid and a strong base is 55.84 kJ mol 1. The enthalpy of formation of OH ions is :
(A) 341.67 kJ mol 1 (B) 229.99 kJ mol 1 (C) 229.99 kJ mol 1 (D) 341.67 kJ mol 1
St
Q.16 Hf for the reaction,

Ag+(aq) + Cl(aq) AgCl(s)
at 25C. Given Hf (Ag+, aq) = 105.58 kJ mol 1, Hf (Cl, aq) = 167.16 kJ mol 1 and
Hf (AgCl, s) = 127.07 kJ mol 1.
(A) 65.49 kJ mol 1 (B) 65.49 kJ mol 1
(C) 188.65 kJ mol 1 (D) 188.65 kJ mol 1
Q.17 The change in state is
HCl(g) + aq H+(aq) + Cl(aq)
Given Hf (HCl, g) = 92.31 kJ mol 1 and Hf (Cl, aq) = 167.16 kJ mol 1, then
The enthalpy change when one mol of HCl(g) is dissolved in a very large amount of water at
25C is :
(A) 259.47 kJ mol 1 (B) 74.85 kJ mol 1
(C) 259.47 kJ mol 1 (D) 74.85 kJ mol 1
Q.18 Given :
Enthalpy of combustion of methane
Enthalpy of combustion of C (graphite)
Page 29 of 51
rH = 890.36 kJ mol 1
rH = 393.51 kJ mol 1
H2(g) + 12 O2(g) H2O(l)

rH = 285.85 kJ mol 1
Enthalpy of dissociation of H2(g)
rH = 435.93 kJ mol 1
Enthalpy of sublimation of C (graphite)
rH = 716.68 kJ mol 1
The bond enthalpy of C H from the following data at 298 K is :
(A) 415.85 kJ mol 1 (B) 415.85 kJ mol 1 (C) 1663.39 kJmol1 (D) 166.339 kJ mol 1
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Q.19 The enthalpy changes of the following reaction at 25C are

Na(s) + 12 Cl2(g) NaCl(s)
H2(g) + S(s) + 2O2(g) H2SO4(l)
2Na(s) + S(s) + 2O 2(g) Na2SO4(s)
rH = 411.0 kJ mol 1
rH = 811.3 kJ mol 1
rH = 1382.3 kJ mol 1
1 H (g) + 1 Cl (g) HCl(g)

;
rH = 92.3 kJ mol 1
2 2
2 2
From these data, find the heat change of reaction at constant volume at 25C for the process
2NaCl(s) + H2SO4(l) Na2SO4(s) + 2HCl(g)
(A) 57.1 kJ mol 1
(B) 60.92 kJ mol 1 (C) 60.92 kJ mol 1 (D) 57.1 kJ mol 1
Q.20 Energy required to dissociate 4g of H2(g) into free gaseous atoms is x kJ. The value of H
(H H) will be :
(A) x kJ M1
(B) x/2 kJ M1
(C) x/3 kJ M1
(D) 0.25 x kJ M 1
Q.21 For the reaction ,
2Cl(g) Cl2(g). The sign of H and S respectively are :
(A) + and
(B) + and +
(C) and
(D) and +
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Q.22 The enthalpy of atomisation of CH 4 and C2H6 are 360 and 620 k cal mol1 respectively. The
C C bond energy is expected to be :
(A) 210 k cal mol 1 (B) 130 k cal mol 1 (C) 180 k cal mol 1 (D) 80 k cal mol 1
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Q.23 The enthalpy of neutralization of NH4OH and CH3COOH is 10.5 k cal mol1 and enthalpy
of neutralization of CH 3COOH with strong base is 12.5 kcal mol 1. The enthalpy of
ionization of NH4OH will be :
(A) 4.0 k cal mol 1
(B) 3.0 kcal mol1
(C) 2.0 kcal mol 1
(D) 3.2 kcal mol1
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Q.24 Polymerisation of ethene to polyethene is represented by the equation,

n(CH2 = CH2) ( CH2 CH2)n
Given that average enthalpies of C = C & C C bonds at 298 K are 590 and 331 k J mol 1
respectively, predict the enthalpy change when 56 g of ethene changes to polyethylene.
(A) 72 kJ
(B) 72 kJ
(C) 144 kJ
(D) 144 kJ
13. Limitations of First Law of Thermodynamics
(i)
(ii)
(iii)
(iv)
(v)
Page 30 of 51
Though the first law of thermodynamics gives us the exact equivalence of heat and work,
whenever there is a change of heat into work or vice versa, but it suffers from the following two
limitations :
No indication is available as regards the direction in which the change will proceed.
It gives no idea about the extent to which the change takes place.
These limitations can be understood from the following examples :
This law can easily explain the heating of a bullet when it strikes a block due to the
conversion of kinetic energy into heat, but it fails to explain as to why heat in the block cannot
be changed into kinetic energy of bullet and make it fly back from inside of the block.
When a vessel of water is placed over fire, heat flows into the vessel. What prevents the heat
from flowing from water into the fire, and thereby cooling the water and ultimately converting
into ice. Thus direction of (flow) change is not known from first law.
It is practically found that whole of heat can never be converted into work. The first law has
no answer to this observation. Thus, first law fails to tell extent to which the interchange of heat
into work and vice versa is possible.
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14. Spontaneous and Non-Spontaneous Processes

1.
(i)
(ii)
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(iii)
Spontaneous Process : In our daily life, we come across a large number of physical and
chemical processes which occur in a widely varying conditions. For example,
Some processes proceed on their own, e.g. Water always flows down a hill, heat flows from a
body at higher temperature to a body at lower temperature.
Some processes require proper initiation but once properly initiated they continue on their own
e.g. Kerosene oil once ignited continues to burn till whole of it has been consumed or exhausted.
Some processes proceed only so long as the external energy is availably. e.g. Electrolysis of
water continues so long as current is passed and stops as soon as current is cut off. The
process which can take place by itself or after proper initiation, under the given set of conditions,
is called a spontaneous process. the term Spontaneous simply means that given process is
feasible or possible. Therefore, Spontaneous processes are also called as feasible or probable
processes. Spontaneous process may or may not be instantaneous. But all instantaneous process
are spontaneous.
It may be pointed out that the term Spontaneous should not mean that the process occurs
instantaneously. It simply implies that process has an urge to proceed or it is paretically
possible.
e.g.
(i)
Processes which occur on their own without proper initiation.
(a)
HCl(g) + NH3(g) NH4Cl(s)
(b)
H2O(l) H2O(g)
Water keeps on evaporating from ponds and rivers etc.
(c)
Sugar dissolves in water and forms a solution.
(d)
2NO(g) + O2(g) 2NO2(g)
(ii)
Processes which require initiation
(a)
In domestic oven, once coal (carbon) is ignited it keeps on burning
C(s) + O2(g)
CO2(g)
2.
Page 31 of 51
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(b)
CH4(g) + 2O2(g)
CO2(g) + 2H2O(l)
From the above discussion we conclude that spontaneous process is the one which has natural
tendency to occur.
Non-Spontaneous Process : A process which has no tendency to occur or which is made to

occur only if energy from outside is continuously supplied e.g.
(i)
Decomposition of water into H 2 and O2 is non-spontaneous. However, water can be
decomposed by passing an electric current through it, in a process called electrolysis.
Electricity
H2O(l) 2H2(g) + O2(g)
The process will continue as long as electric current is supplied , and as soon as the
supply of electricity is cut off the decomposition stops.
(ii)
Water cannot be made to flow up the hill, without the help of a machine.
(iii)
Gold ornaments do not get tarnished in air even after a number of years. This shows
that gold does not combine with oxygen in the air.
Driving Force for a Spontaneous Process : the natural tendency of various processes to
occur spontaneously indicated that there must be some driving force behind them.
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15. Entropy
It may be defined as the measure of degree of randomness in the molecule. It is represented by
the symbol S.
Characteristic of Entropy
(i)
It is a state of function
(ii)
It is an extensive property
(iii)
The exact value may be determined by applying the III law of thermodynamics
The change in entropy during a process when a system undergoes charge from one state to
another is represented as S.
Thus S = Sfinal - Sinitial
and for chemical reaction S = Sproduct - SReactant
Calculation of Changes in Entropy
Page 32 of 51
Reversible Process at Equilibrium
q Re v
T
In a reversible reaction heat gained in the forward reaction is equal to heat lost in the
reverse reaction.
Hence in a reversible cyclic process the net charge in entropy is zero. This is called
Clausius Theorem
Suniverse = Ssystem + SSurrounding
S
Ssystem =
q Re v (system)
q Re v (surrounding )
te
(ii)
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(i)
Ssurrounding =
ud
yS
T
T
Heat gained by system = heat lost by surrounding
S Rev(system)= Q Rev(surrounding)
Suniverse =
q Re v (system)
T
St
(a)
q Re v (system)
T
Hence Suniverse(Rev)=0
(iii)
(iv)
q Re v (surrounding )
T
q Re v (system)
T
Suniv > 0 for irreversible process

Entropy change for an isothermal process
E = q + w
E = 0 (Isothermal process)
V
But w = 2.303 nRT log 2
V1
q = -w
or for a reversible process
qrev = 2.303 nRT log
S =
q Re v
1
V
(2.303 nRT log 2 ),
T
T
V1
Also V
1
P
q=2.303 nRT log
V2
V1
V2
V1
S = 2.303 nR log
V2
V1
S = 2.303 nR log
P1
P2
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Page 33 of 51
(v)
Entropy change in an isobaric process

dS =
dq Re v ( P )
also dqRev(P) = dH
But dH = CpdT
Integrating both sides
dS = CpdT/T
S2
2
dT
C
dS = p
T ,
T
T2
T
= Cp n T T12 = Cp ln T
1
T2
S = 2.303 Cp log T
1
Entropy change for Isochoric process

dS =
dq Re v ( V )
Also dq (Rev)(v) = dE
But dE = CvdT
Integrating both sides one get
dqv = CvdT
T2
T
S = V n T T12 = Cvln T
1
Entropy change during mixing of ideal gas
yS
Smix = R n n x
te
(vii)
ps
T2
S = 2.303 Cv log T
1
.in
(vi)
= R [n1 ln x 1+ n2ln x 2 + n3 ln x 3 ......]
n i n x i
ud
Smix = R
St
ni = m n i n x i no. of moles of the gas

ni = mole fraction of the gas
(viii) Entropy changes during phase transformation
q Re v
T
Entropy of fusion : The entropy changes taking place when 1 mole of a solid
substance change into liquid form, at the melting temperature.
S =
(a)
Sfusion =
(b)
H fusion
Tfusion
Entropy of vaporization is the entropy change when one mole of a liquid changes into
vapours at boiling point.
Svap = Svap Sliquid =
(c)
H vapourisation
Tboiling po int
Entropy of sublimation is the entropy change when one mole of a solid changes into
vapours at sublimation temperature.
Ssub =
H sub
Tsub
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Page 34 of 51
(d)
Hence for any physical transformation

Stransition =
(ix)
H transition
Ttransition
Let a given mass of a liquid be heated from temperature T 1 to T2. Assuming specific
heat of liquid to be constant between T 1 and T2 and that no change occurs the
amount of heat required to raise the temperature by dT is given by
dq = mCdT
T2
S = mC ln T
1
16. Second Law of Thermodynamics
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17. Gibbs Free energy
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It can be defined in number of ways as follows

1.
All spontaneous processes (or naturally occurring processes) are thermodynamically
irreversible.
2.
Without the help of the an external agency, a spontaneous process cannot be
reversed e.g., heat cannot by itself flow from a older to hotter body.
3.
The complete conversion of heat into work is impossible without leaving some effect
elsewhere.
4.
All spontaneous processes are accompanied by a net increase of entropy.
5.
The entropy of the universe is increasing.
6.
The entropy is a time arrow.
St
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It is defined as, the energy available in the system for conversion into useful work.
It is that thermodynamic quantity of a system, the decrease in whose value during a process is
equal to the useful work done by the system.
G = H TS
where H is the that content, T is the absolute temperature and S is the entropy of the system.
As before, for the isothermal processes, we have
G = H TS (Gibbs Helmotz equation)
H = H2 H1 is the enthalpy change of the system
Deriving the criteria for spontaneity from Gibbs-Helmholtz equation. According to GibbsHelmholtz equation
G = H TS
The equation combines in itself both the factor which decide the spontaneity of a process,
namely
(i)
the energy factor, H
(ii)
the entropy factor, TS
Depending upon the signs of H and TS and their relative magnitudes, the following different
possibilities arise.
1.
When both H and TS are negative i.e., energy factor favours the process but
randomness factor opposes it, Then
(i)
If H > TS , the process is spontaneous and G is negative.
(ii)
If H < TS , the process is non-spontaneous and G is positive.
(iii)
If H = TS , the process is in equilibrium and G is zero.
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2.
3.
When both H and TS are positive i.e., energy factor opposes the process but
randomness factor favours it. Then
(i)
If H > TS , the process is non-spontaneous and G is positive.
(ii)
If H < TS, the process is spontaneous and G is negative.
(iii)
IF H = TS, the process is in equilibrium and G is zero.
When H is negative but TS is positive i.e., energy factor as well as the randomness
factor favour the process. The process will be highly non-spontaneous and G will be
highly positive.
An important advantage of free energy criteria over the entropy criteria lies in the fact
that the former requires free energy change of the system only whereas the latter requires
the total entropy change for the system and the surroundings.
Physical significance of Gibbs free energy :

S = q + w
E = q + wexpansion + wnonexpansion
E = q PV = + wnonexpansion
E + PV = q + wnonexpansion
H = qRev + wnonexpanison
(because wexpansion = PV)
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(For a reversible process taking place at constant temperature. S =
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As qRev = TS and E + PV = H we get

H = TS + wuseful [wnonexpansion = wuseful]
H TS = wuseful
G = wuseful
When useful work is done by the system
wuseful = ve value
G = ve value and G = wuseful
Gproduct Greactant = ve
Gproduct Greaction
Capacity to do useful work by product is less than the capacity to do useful

work by reactant
Product more stable than reactant.

Hence according to II law of thermodynamics the process is a spontaneous process as
every substance wants to be in the state of maximum stability.
When work is done on the system
wuseful = +ve
G = +ve
GP > GR
Capacity to do useful work by product is more than the capacity to do useful

work by reactant
Reactant more stable than product
Process non-spontaneous according to II law of thermodynamics
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(i)
(ii)
Relationship between G and Equilibrium constant

G = 2.303 RT log K
where K = equilibrium constant
Relationship between G and standard cell potential
G = nFE
Page 35 of 51
q Re v
qRev = TS)
T
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18. Third Law of Thermodynamics

It states that, Entropy of all perfectly crystalline solids may be taken as zero as the absolute
zero of temperature. Third law of thermodynamics may also be defined as
(i)
The entropy of a solid is zero at the absolute zero of temperature.
(ii)
It is impossible to reduce the temperature of any system to absolute zero by any process.
(iii)
At any pressure, the entropy of every crystalline solid in thermodynamic equilibrium at
absolute zero is zero.
(iv)
At the absolute zero, increment in entropy for isothermal process in crystalline
approaches zero as the limit.
(v)
At absolute zero, every crystal becomes ideal crystal.
In which case, a spontaneous reaction is possible at any temperature

(A) H < 0, S > 0
(B) H < 0, S < 0
(C) H > 0, S > 0
(D) in none of the cases
Q.2
In a chemical reaction H = 150 kJ and S = 100 JK1 at 300 K. The G for the reaction is
(A) zero
(B) 300 kJ
(C) 330 kJ
(D) 120 kJ
Q.3
For reaction at 25 C enthalpy change (H) & entropy change (S) are 11.7 10 3 J mol-1
and 105 J mol1 K1 respectively. The reaction is :
(A) spontaneous
(B) nonspontaneous
(C) instantaneous
(D) none
Q.4
The temperature at which the reaction,

Ag2O(s) 2Ag(s) + 1/2 O2(g)
is at equilibrium is ............... ; Given H = 30.5 kJ mol 1 and S = 0.066 kJ K1 mol1
(A) 462.12 K
(B) 362.12 K
(C) 262.12 K
(D) 562.12 K
Q.5
The enthalpy and entropy change for a chemical reaction are 2.5 10 3 cal and 7.4 cal deg1
respectively. Predict that nature of reaction at 298 K is
(A) spontaneous
(B) reversible
(C) irreversible
(D) nonspontaneous
Q.6
Which is not correct ?

(A)
in an exothermic reaction, the enthalpy of products is less than that of reactants
(B)
Hfusion = Hsublimaiton Hvaporisation
(C)
a reaction for which H < 0 and S > 0 is possible at all temperature
(D)
H is less than E for the reaction C(s) + (1/2) O 2(g) CO2(g)
Q.7
For the reaction :

2NO(g) + O2(g) 2NO2(g)
the enthalpy and entropy changes are -113.0 kJ mol -1 and -145 J K-1 mol-1 respectively. Find
the temperature above which the reaction is spontaneous
(A) 432.3 K
(B) 570.5 K
(C) 1035.7 K
(D) 779.3 K
Q.8
The enthalpy change for a given reaction at 298 K is -x cal/mol. If the reaction occurs
spontaneously at 298 K, the entropy change at that temperature (A)
can be negative but numerically larger than x/298 cal K-1
(B)
can be negative, but numerically smaller than x/298 cal K-1
(C)
cannot be negative
(D)
cannot be positive
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Q.1
Page 36 of 51
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Q.9
The favorable condition for a process to be spontaneous is :

(A) TS > H, H = +ive, S = -ive
(B) TS > H, H = +ive, S = +ive
(C) TS = H, H = -ive, S = -ive
(D) TS = H, H = +ive, S = +ive
Q.10 At 300 K, the reaction which have the following values of thermodynamic parameters, occur
spontaneously
(A) G = -400 kJ mol -1
(B) H = 200 kJ mol -1, S = -4 JK-1 mol-1
-1
-1
-1
(C) H = -200 kJ mol , S = 4 JK mol (D) H = 200 kJ mol -1, S = 40 JK-1 mol-1
Q.11
Which of the following statement(s) is/are correct ?

(A)
the system of constant entropy and constant volume will attain the equilibrium in a state
of minimum energy
(B)
the entropy of the universe is on the increase
(C)
the process would be spontaneous when (S)E,V < 0, (E)S, V > 0
(D)
the process would be spontaneous when (S)E,V > 0, (E)S, V < 0
Q.12 For melting of ice at 25C the enthalpy of fusion is 6.97 kJ, mol -1, entropy of fusion is 25.4 J
K-1 mol-1 and free energy change is -0.6 kJ mol -1. Predict whether the melting of ice is
(A) non spontaneous (B) spontaneous
(C) at equilibrium
(D) not predicted
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Q.13 For a process at H and TS both are positive in what conditio, the process is spontaneous
(A) H > TS
(B) H < TS
(C) H = TS
(D) not predicted
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Q.14 At 0C, ice and water are in equilibrium and H = 6.0 kJ mol-1 for the process
H2O(s)
H2O(l)
The value of S and G for the conversion of ice into liquid water are
(A) -21.8 J K-1 mol-1 and 0
(B) 0.219 J K-1 mol-1 and 0
(C) 21.9 J K-1 mol-1 and 0
(D) 0.0219 J K-1 mol-1 and 0
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Q.15 The entropy change at 373 K for the transformation

H2O(l) H2O(g) is :
[ Given : vapH = 40.668 kJ mol -1. Pressure = 1 bar ]
(A) 128.17 J K-1 mol-1
(B) 109.03 J K-1 mol-1
-1
-1
(C) 19.15 J K mol
(D) 98.35 J K-1 mol-1
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Q.16 The entropy change for the conversion of one gram of ice to water at 273 K and one
atmospheric pressure is :
[Hfusion = 6.025 kJ mol -1 ]
(A) 7.30 J K-1 mol-1 (B) 1.226 J K-1 g-1
(C) 1.226 J K-1 mol-1 (D) 7.30 J K-1 g-1
Q.17 In a reaction A+ + B A + B+, there is no entropy change. If enthalpy change is 22 kJ
for the reaction G for the reaction is :
(A) 22 kJ mol-1
(B) 11 kJ mol-1
(C) 33 kJ mol -1
(D) 44 kJ mol-1
Q.18 H and S for Br2(l) + Cl2(g) 2BrCl(g) and 29.37 kJ and 104.0 J K -1 respectively.
Above what temperature will this reaction become spontaneous ?
(A) T > 177.8 K
(B) T > 354.1 K
(C) T > 282.4 K
(D) T > 141.2 K
Q.19 H and S for the system H2O(l)
H2O(g) at 1 atmospheric pressure are 40.63 kJ
-1
-1
-1
mol and 108.8 J K mol respectively. The temperature at which the rates of forward and
backward reactions will be same, is :
(A) 373.4 K
(B) 256.2 K
(C) 316.8 K
(D) 278.5 K
Q.20 The equilibrium constant Kc for the following reaction at 400 K
2NOCl(g)
2NO(g) + Cl2(g) is :
[ Given H = 77.2 kJ and S = 122 J K-1 at 400 K ]
(A) 2.577 10-4
(B) 1.958 10-4
(C) 28.4 10 -3
(D) 1.466 10-2
Page 37 of 51
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Problem 1 :
C(s) + O2(g) CO2(g) + 394 kJ

C(s) + 1/2 O2(g) CO + 111 kJ
(a)
In an oven using coal (assume the coal is 80% carbon in weight), insufficient
oxygen is supplied such that 60% of carbon is converted to O 2 and 40% carbon
is converted to CO.
Find out the heat generated when 10 kg of coal is burnt in this fashion.
(b)
Calculate the heat generated if a more efficient oven is used such that only
CO2 is formed.
(c)
Calculate the percentage loss in heating value for the inefficient oven.
.in
C(s) + O2(g) CO2(g) + 394 kJ ............ (i)

C(s) + 1/2 O2(g) CO + 111 kJ ............. (ii)
Weight of C in 10 kg coal = 10000 0.8 = 8000 g
Weight of C converted into CO 2 = 8000 0.6 = 4800 g
weight of C converted into CO = 8000 0.4 = 3200 g
12 g (1 mole) C on conversion into CO 2 liberates = 394 kJ
4800 g of c on conversion into CO 2 liberates = 153,600 kJ
12 g(1 mole) C on conversion into CO liberates = 111 kJ
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Solution :
(a)
111 3200
= 29600 kJ
12
= 183200 kJ
Total heat liberated = 153600 + 29600 kJ

C(s) + O2(g) CO2(g) + 294 kJ
12 g carbon liberates heat = 394 kJ
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(b)
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3200 g of c on conversion into CO liberates =
294 8000
= 262666.67 kJ
12
Heat lost by oven = 26266.67 183200 = 79466.67 kJ
8000 g of carbon liberates heat =

(c)
% lost of heat =
79466 .67 100

= 32.25%
262666 .67
Problem 2 :
Calculate the standard internal energy change for the following reaction at 25C.
2H2O2(l) 2H2O (l) + O2(g)
Hf at 25C for H 2O2(l) = 188 kJ mol 1, H2O (l) = 286 kJ mol 1
Solution :
H = H (product) H(reactants)
= 2(286) + 0 2 (188)
= 572 + 376 = 196 kJ
n(g) = 1 0 = 1
H = E + n(g)RT
= 196 1 8.314 10 3 298 = 198.4775 kJ
Page 38 of 51
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Problem 3 :
A sample of argon gas at 1 atm pressure and 27 oC expands reversibly and

adiabatically from 125 dm 3 to 250 dm 3. Calculate the enthalpy change in this
process. C V. for argon is 12.48 JK -1 mol-1.
Solution :
For adiabatic expansion, we have
T2
V2
Cv
ln
= ln
,
T1
V1
R
V2
=2
V1
12.48
T2
ln
= ln2
8.314
300
12.48
T2
ln
= 0.3010
8.314
300
T2
0.3010 8.314
T2
=
In
= 0.200
300
12.48
300
T2 = 300 1.586 =475.8 K or 202.8 oC
T = T2-T1 = 475.8 -300 =175.8K
CP = CV + R
CP = 12.48 + 8.314 = 20.794 JK -1
Now, PV = nRT
1 1.25 = n 0.0821 300
n=0.05
We know
H = nCp T = 0.05 20.794 175.8 = 182.77 J
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Problem 4 :
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Calculate the heat of neutralization from the following data 200 ml of 1 M HCl is
mixed with 400 ml of 0.5 M NaOH. The temperature rise in calorimeter was found to
be 4.4oC. Water equivalent of calorimeter is 12 g and specific heat is 1 cal/ml/degree
for solution.
Solution :
The heat produced ( H1) during neutralization of 200 Meq. of NaOH and HCl each
(Meq. = N V) is taken up by calorimeter and solution in it.
H1 = Heat taken up by calorimeter + solution
H 1 = m 1S 1 T + m 2S 2 T
= 12
[ total solution = (200 + 400) ml.] =2692.8 cal
Neutralization of 200 Meq. gives heat =-2692.8 cal
Neutralization of 1000 Meq. gives heat = -2692 5 = -13464 cal
= -13.464 k cal
Problem 5 :
Page 39 of 51
The thermochemical equation for the combustion of ethylene gas, C 2H4, is

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(1) ; H = -337 KCAL
Assuming 70% efficiency, calculate the weight of water at 20 oC that can be
converted into steam at 100 oC by burning 1 m 3 of C2H4 gas measured at S.T.P. Heat of
vaporization of water at 20 oC and 100oC are 1.00 kcal/kg and 540 kcal/kg respectively.
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Solution :
No. of moles in 1 m3 of ethylene = 44.6 mol
H for 1 m3 of ethylene (44.6 mol of ethylene) = n(C 2H4) H (1 mole)
= 1.50 104 kcal
The useful heat = 1.05 104 cal
For the overall process, consider two stages :
H = (1.00 kcal/kg, K) (80 K) =80 kcal/kg
H2O (1) 20oC H2O(l) 100oC ;
o
o
H2O(1) 100 C H2O(g) 100 C ; H = 540 kcal/kg
H (total ) =620 kcal/kg
Wt. of water converted into steam =
Amount of heat available 1.05 10 4

=
= 16.9 kg
Heatrequired / kg
620
Problem 6 :
.in
Calculate the heat of formation of anhydrous aluminium chloride, Al 2Cl6, from the
following data.
(i)
2Al(s) + 6HCl(aq.) Al2Cl6(g); 3H2(g) ;
H =-240 kcal
H=-44.0 kcal
(ii)
H2(g) + Cl2(g) 2HCl(g)
;
(iii) HCl(g) + aq . HCl(aq.)
;
H=-17.5 kcal
(iv)
Al2Cl6(s) + aq. Al2Cl6 (aq)
;
H=-153.7 kcal
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Solution :
The required equation is
H=?
2Al(s) + 3Cl 2(g) Al2Cl6(s) ;
For obtaining this ,
Multiply (ii) by 3, (iii) by 6 and add the resulting equations to (i)
(i)
2Al (g) + 6HCl(aq.) Al2Cl6(aq.) + 3H2(g) ; H = 240.0 kcal
Subtract (iv) from (v) and rearrange the product
2Al (s) + 3Cl 2(g) + aq. A2Cl6(aq.) ; H = -477.0 kcal
-Al2Cl6(s) - aq. -Al2Cl6(aq);
H = +153.7 kcal
2Al (s) + 3Cl 2(g) - Al2Cl6(s) ; H = -323.3 kcal
or
2Al(s) + 3Cl 2(g) Al2Cl6 ; H=-323.3 kcal
Hence the heat of formation of anhydrous aluminium chloride = -323.3 kcal
Problem 7 :
10 g of argon gas is compressed isothermally and reversibly at a temperature of 27 oC

from 10L to 5 L. Calculate q, W and E for this process. (At wt. of Ar =40)
Solution :
Page 40 of 51
No. of moles of argon =
10
= 0.25 mole
40
For isothermal reversible compression

= 2.302 2.5 2 300 log
and
V1 = 10 litre, V2 = 5 litres, T = 300 K
V2
W = 2.303 nRT log V
1
10
= 103.6 cal
5
Amount of heat absorbed =103.6 cal

Now we know that during isothermal reversible process, internal energy remains constant
throughout the process, hence the change in energy ( E) will be zero.
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Problem 8 :
Calculate the maximum work done when pressure on 10 g of hydrogen is reduced

from 20 to 1 atm at a constant temperature of 273 K. The gas behaves ideally. Will
there be any change in internal energy. Also calculate Q.
Solution :
10
20
P1
= 2.303
= 8180 calories
2 273 log
2
1
P2
Since the change is taking place constant temperature, internal energy will not change, i.e.
E=0
Q = E + W = 0 + 8180 = 8180 calories
W = 2.303 nRT log
Problem 9 :
The heat of combustion of glycogen is about 476 kJ/mol of carbon. Assume that
average heat loss by an adult male is 150 watt. 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 ?
Solution :
Total energy required in the day
Units of glycogen required =
= 12960 kJ
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24 60 60
kJ (1 watt = J sec1)
1000
12960
27.22 units.
476
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Problem 10 :
The heat of total cracking of hydrocarbons H TC is defined as H at 298.15 K and
101.325 kPa for the process below :
ud
C n H m + 2n - H 2(g)
nCH 4 g
2
Given that H TC is 65.2 kJ for C 2H6 and 87.4 kJ for C 3H8, calculate H for
Solution :
St
CH 4 g + C 3 H 8 g
2C 2 H 6 g
H TC of CH4 = 0
CH4(g) + C3H8(g) 2C2H6(g)
H 2H T.C. C2 H 6 H T.C. C3H8
= 2 ( 65.2 ) ( 87.4 ) = 43 kJ
Problem 11 :
A constant pressure calorimeter consists of an insulated beaker of mass 92 g made
up of glass with heat capacity 0.75 J K 1 g1. The beaker contains 100 mL of 1 M
HCl of 22.6C to which 100 mL1 M NaOH at 23.4C is added. The final temperature
after the reaction is complete is 29.3C. What is H per mole for this neutralization
reaction ? Assume that the heat capacities of all solutions are equal to that of same
volumes of water.
Page 41 of 51
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Solution :
Initial average temperature of the acid and base
22.6 23.4
23.0C
2
Rise in temperature = (29.3 23.0) = 63C
Total heat produced = ( 92 0.75 + 200 4.184) 6.3
= (905.8) 6.3 = 5706.54 J
Enthalpy of neutralisation =
57065.54
1000 1
100
= 57065.4 J = 57 kJ
Problem 12 :
Find bond enthalpy of S S bond from the following data :
C 2H 5 S C 2H 5
H f = -147.2 kJ mol -1
C 2H 5 S S C 2H 5
H f = -201.9 kJ mol -1
H f = 222.8 kJ mol -1
S(g)
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Solution :
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H H
H
H
H CC SCC
4C(s) + 5H2 + S
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H H
H
H
H CC SSCC
4C(s) + 5H2 + 2S
(i)
(ii)
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H B.E.R B.E.P
147.2 = Heat of atomization of 4C, 10H, 1S

B.E. of 10(C H), 2(C S), 2(C C)
201.9 = Heat of atomization of 4C, 10H, 2S
B.E. of 10(C H), 2(C S), 2(C C), (S S)
Subtracting (i) from (ii)
201.9 + 147.2 = Heat of formation of 1S B.E. of (S S)
= 222.8 kJ B.E. of (S S)
B.E. of (S S) = 277.5 kJ
Problem 13 :
From the data at 25C :
Page 42 of 51
Fe2O3(s) + 3C(graphite) 2Fe(s) + 3CO(g) ,

H = 492.6 kJ/mol
FeO(s) + C(graphite)
CO2(g)
,
H = 155.8 kJ/mol
C(graphite) + O 2(g)
CO2(g)
,
H = -282.98 kJ/mol
Calculate standard heat of formation of FeO(s) and Fe 2O3(s).
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Solution :
Fe(s) + CO(g) FeO(s) + C(graphite)
H 158.88 kJ / mol
C(graphite) + O2(g) CO2(g)
H 393.5 kJ / mol
CO2(g) CO(g) + 1/2 O 2(g)
H 282.98 kJ / mol
On adding Fe(s) + 1/2 O 2(g) FeO(s) ;
H 393.5 kJ / mol
Similarly we may calculate heat of formation of Fe2O3.
Problem 14 :
Show that the reaction, CO(g) + ( 1/2 ) O 2(g) CO2(g)
at 300 K is spontaneous and exothermic, when the standard entropy change is
0,094 kJ mol 1 K1. The standard Gibbs free energies of formation of CO 2 and CO
are 394.4 and 137.2 kJ mol 1 respectively.
Solution :
The given reaction is ,
CO(g) + (1/2) O 2(g) CO2(g)
1
2
= 394.4 ( 137.2 ) 0
= 257.2 kJ mol 1
G = H TS
257.2 = H 298 (0.094)
or
H = 288.2 kJ
G is ve, hence the process is spontaneous, and H is also ve, hence the process is
also exothermic.
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G (for reaction) = G CO2 G CO G O2
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Problem 15 :
Assume that for a domestic hot water supply 150 kg of water per day must be heated
from 10C to 65C and gaseous fuel propane C 3H8 is used for this purpose. What moles
& volume of propane (in litre at STP) would have to be used for heating domestic
water. H for combustion of propane is 2050 kJ mol 1 & specific heat of water is
4.184 103 kJ/g.
Solution :
Heat taken up by water = m S T
= 150 103 4.184 10 3 55 = 34518 kJ
2050 kJ heat is provided by 1 mole C 3H8
34158 kJ heat is provided by = 34518/2050

= 16.838 mole of C 3H8
Volume of C 3H8 at NTP = 16.838 22.4 litre
= 3.77 102 litre
Problem 16 :
Using the data ( all values in k cal mol 1 at 25 C ) given below, calculate bond energy
of C C & C H bonds.
C(s) C(g)
;
H = 172 kcal
H2 2H
;
H = 104 kcal
Page 43 of 51
H2 +
1
2
O2 H2O(l)
C(s) + O2 CO2
H = -68.0 kcal
H = -94.0 kcal
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For C3H8 : 3C + 4H2 C3H8 ; H = ?

For C2H6 : 2C + 3H2 C2H6 ; H = ?
Solution :
H1 2 C C 8 C H 3Csg 4 H H
H 2 1 C C 6 C H 2Csg 3 H H
Also given C + O2 CO2 ; H 94.0 k cal
H2 +
1
2
O2 H2O; H = 68.0 k cal
. . . . . (1)
. . . . . (2)
. . . . . (5)
. . . . . (6)
C 2 H 6 7 / 2 O 2 2CO 2 3H 2O ; H 530 kcal

C3H8 + 5O2 3CO2 + 3H2O ; H = 530 kcal
By inspection method : 2 (5) + 3 (6) (7) gives
. . . . . (8)
. . . . . (9)
....
(10)
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2C + 3H2 C2H6
; H 2 20 k cal
and 3 (5) + 4 (6) (8) gives
3C + 4H2 C3H8
; 24 k cal
By equation (3), (4), (9) and (10)

a + 6b = 676
2a + 8b = 956
a = 82 k cal and b = 99 k cal

Bond energy of C C bond = 82 k cal
and
Bond energy of C H bond = 99 k cal
. . . . . (7)
Page 44 of 51
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te
Problem 17 :
The standard enthalpy of combustion at 25C of hydrogen, cyclohexene (C 6H10) and
cyclohexane (C 6H12) are 241, 3800 and 3920 kJ/mole respectively. Calculate the
heat of hydrogenation of cyclohexane.
Solution :
The required reaction is
C6 H10 H 2
C6 H12 , H1 ?
Cyclohexane
St
Cyclohexene
. . . . . (1)
Let us write the given facts
H2 + 12 O2 6 CO2 + 5 H2O ; H = - 241 kJ/mole
17
O 2
6CO 2 5H 2O, H 3 = 3800 kJ/mole
2
C6H12 + 9O2
6CO2 + 6H2O, H 4 = 3920 kJ/mole
C6 H10
. . . . . (2)
. . . . . (3)
. . . . . (4)
The required reaction (1) can be obtained by adding equations (2) and (3) and subtracting
(4) from the sum of (2) and (3).
C6H10 + H2 C6H12.
H1 = ( H 2 + H 3 ) H 4
= [241 + ( 3800)] (3920)
= (241 3800) ( 3920)
= 4041 + 3920 = 121 kJ/mole
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Problem 1 :
If 1.00 kcal of heat is added to 1.2 L of oxygen in a cylinder at constant pressure of

1.000 atm, the volume increases to 1.5 L, Hence E for this process is:
(A) 0.993 kcal
(B) 1.0073 kcl
(C) 0.0993 kcal
(D) 1.00073 kcal
Solution :
(A)
H = E + PV
1(1.5 1.2)L atm
2 103 kcal
0.082 L atm
E = 0.993 kcal
1.00 = E +
Problem 2 :
te
ps
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Using only the following data:

(I)
Fe2O3(s) + 3CO (g)
2Fe(s) + 3CO2 (g) ; H = 26.8 kJ
(II)
Fe(s) + CO(g)
FeO(s) + CO(g) ;
H = + 16.5 kJ
the H value, in kilojoules, for the reaction Fe 2O3(s) + CO(g)
2FeO(s) +
CO2(g) is calculated to be:
(A) 43.3
(B) 10.3
(C) +6.2
(D) +10.3
Solution :
(C)
from equation (I) + (2 II) ;
H = 6.2 kJ
Problem 3 :
Problem 4 :
ud
1.435
kcal
18
= 0.0797 kcal g
St
H (per g) =
yS
Enthalpy change when 1.00 g water is frozen at 0C, is : (Hfus = 1.435 kcal mol 1)
(A) 0.0797 kcal
(B) 0.0797 kcal
(C) 1.435 kcal
(D) 1.435 kcal
Solution :
(B)
Heat of neutralisation of CsOH with all strong acids is 13.4 kcal mol 1. the heat
released on neutralization of CsOH with HF (weak acid) is 16.4 kcal mol 1
H of ionisation of HF is:
(A) 3.0 kcal
(B) 3.0 kcal
(C) 6.0 kcal
(D) 0.3 kcal
Solution :
(B)
CsOH + H+ = Cs + H2O
H = 13.4 kcal
Heat of ionisation of CsOH = 13.7 13.4 = + 0.3 kcal
CsOH + HF
CsF + H2O
H = 16.4 kcal
Heat of ionisation of HF = x kcal
Heat of ionisation of CsOH = 0.3 kcal
Heat of neutralization = 13.7
(of H+ and OH )
13.9 + x + 0.3 = 1.64
x = 3.0 kcal
Page 45 of 51
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Problem 5 :
The CCl bond energy can be calculated from:

(A)
Hf (CCl4, l) only
(B)
f (CCl4, l) and D(Cl 2)
(C)
Hf (CCl4, l) D(Cl2)
(D)
Hf (CCl4, l) D(Cl2), Hf (C, g) and Hvap (CCl4)
Solution :
(D)
C(s) + 2Cl2(g) CCl4(l)
Hf (CCl4, l) = H[C(s) C(g)] + 2(BE)ClCl [Hvap (CCl4) + 4(BE)ClCl]
Problem 6 :
Enthalpy of fusion of a liquid is 1.435 kcal mol 1 and molar entropy change is
5.26 cal mol 1 K1. Hence melting point of liquid is :
(A) 100C
(B) 0C
(C) 373 K
(D) 273
Solution :
(B)
S =
H
T
T=
H
S
1435cal
= 273 K 0C
5.26
Problem 7 :
yS
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For the reaction

X2O4 (l)
2XO2(g)
E = 2.1 kcal. S = 20 cal/K at 300 K
Hence G is :
(A) 2.7 kcal
(B) 2.7 kcal
(C) 9.3 kcal
(D) 9.3 kcal
Solution :
(B)
H = E + ng RT = 2.1 + 2 0.002 300 = 3.03 kcal
G = H TS
= 3.3 300 (0.02)
= 2.7 kcal
ud
Problem 8 :
H
nF
Solution :
(C)
(A)
St
d ( G )
then variation of EMF of a cell E,
If G = H TS and G = H + T
dT P
with temperature T, is given by :
(B)
G
nF
(C)
S
nF
d ( G )
dE
d ( nFE )
On comparison :S =
S =
= nF
dT
dT
dT
(D)
S
nF
dE S
=
nF
dT
Problem 9 :
1 g H2 gas at S.T.P. is expanded so that volume is doubled. Hence work done is:
(A) 22.4 L atm
(B) 5.6 L atm
(C) 11.2 L atm
(D) 44.8 L atm
Solution :
(C)
V1 (volume of 1 g H2) = 11.2 L at NTP
V2 (volume of 1 g H2) = 22.4 L
W = PV = 11.2 L atm
Page 46 of 51
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Problem 10 :
Following reaction occurs at 25C :

2NO(g, 1 10 5 atm) + Cl 2(g, l 102 atm)
2NOCl(g, l 10 2 atm)
G is:
(A) 45.65 kJ
(B) 28.53 k J
(C) 22.82 kJ
(D) 57.06 kJ
Solution :
(A)
G = 2.303 RT log K eq
Keq =
2
PNOCl
2
PNO
PCl2
= 108
Hence G = 45.65 kJ
Problem 11 :
1 mol of NH 3 gas at 27C is expanded under adiabatic condition to make volume 8

times ( = 1.33). Final temperature and work done respectively are:
(A) 160 K, 900 cal (B) 150K, 400 cal (c) 250 K, 1000 cal (D) 200 K, 800 cal
Solution :
(A)
1
1.331
V1
1
T2 = T1
= 300
8
V2
= CvT = Cv(T2 T1)
= 150 K
Cv R
= 1.33 CV = 3 R)
Cv
te
Problem 12 :
w = q)
ps
= 3 2 (150 300) = 900 cal
.in
( for adabatic process
(A) R
ud
(A)
W = P (V)
PV = RT
P(V + V) = R(T + 1)
PV = R
St
Solution :
(B) 2R
yS
Temperature of 1 mol of gas is increased by 1 at constant pressure. Work done is:
(C) R
2
(D) 3R
PV + PV = RT + R
Problem 13 :
The gas absorbs 100 J heat and is simultaneously compressed by a constant external
pressure of 1.50 atm from 8 L to 2L in volume. Hence E will be:
(A) 812 J
(B) 812 J
(C) 1011 J
(D) 911 J
Solution :
(C)
H = E + PV
Page 47 of 51
1.5 6
8.314
0.0821
E = 1011.4 J
100 = E +
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Problem 14 :
The standard heat of combustion of solid boron is equal to:

(A) Hf (B2O3)
Solution :
(B) 12 Hf (B2O3)
(C) 2Hf (B2O3)
(D) 12 Hf (B2OP3)
(B)
Problem 15 :
CP Cv = R . This R is :
(A)
Change in K.E.
(B)
Change in rotational energy
(C)
work done which system can do on expanding the gas per mol per degree
increase in temperature
(D)
All correct
Solution :
(C)
PV = RT at temp T for one mol
P(V + V) = R(T + 1) at temp. (T + 1) for one mol
PV = R
Problem 16 :
TS > H, T >
H
4000
T >
T > 400 K
S
10
yS
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For A B , H = 4 kcal mol 1 ,

S = 10 cal mol 1K1.
Reaction is spontaneous when temperature can be :
(A) 400K
(B) 300K
(C) 500K
(D) none is correct
Solution :
(C)
G = ve for spontaneous charge
G = H TS
ud
Problem 17 :
(D) G > 0
St
If a process is both endothermic and spontaneous , then :

(A) S > 0
(B) S < 0
(C) H < 0
Solution :
(A)
As
G = H TS
For spontaneous process, G = negative
For endothermic process, H = positive
Therefore S > 0
Problem 18 :
For which change H E
(A) H2 + I2
2HI
(C) C(s) + O2(g)
CO2(g)
Solution :
(D)
H E nRT
Page 48 of 51
NaCl + H O
(B) HCl + NaOH
2
2NH
(D) N2 + 3H2
3
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Problem 19 :.
If a chemical change is brought about by one or more methods in one or more
steps, then the amount of heat absorbed or evolved during the complete course of
reaction is same, which ever method was followed. This law is known as
(A) Le Chateliers principle
(B) Hesss law
(C) Joule Thomson effect
(D) Troutons law
Solution :
(B)
The statement is definition of Hesss law
Problem 20 :.
The Kirchhoffs equation gives the effect of ..... on heat of reaction.
(A) Pressure
(B) Temperature
(C) Volume
(D) Molecularity
Solution :
(B)
Kirchhoffs equation is : H2 H1 = CP (T2 T1)
ps
.in
Problem 21 :.
The heats of neutralisation of four acids A , B , C , D are 13.7, 9.4, 11.2 and
12.4 kcal respectively when they are neutralised 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
Solution :
(B)
Lower is heat of neutralisation , more is dissociation energy, weaker is acid
Problem 22 :.
St
ud
yS
te
The Hf for CO 2(g), CO(g), and H 2O(g) are 393.5, 110.5 & 241.8 kJ mol 1
respectively. The standard enthalpy change (in kJ) for the reaction ,
CO2(g) + H2(g) H2O(g) is :
(A) 524.21
(B) 41.2
(C) 262.5
(D) 41.2
Solution :
(B)
Given
C + O2 CO2
;
H = 393.5 kJ
. . . . (i)
C + (1/2)O2 CO
;
H = 110.5 kJ
. . . . (ii)
H2 + (1/2)O2 H2O
;
H = 241.8 kJ
. . . . (iii)
By (ii) + (iii) (i),
CO2 + H2 CO + H2O ;
H = + 41.2
1.
Daily Practice Problem Sheet
1. T
2. F
3. F
4. T
5. A
6. C
7. D
8. C
9. D
10. A
11. B
12. A
13. B
14. C
15. B
16. C
Page 49 of 51
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2.
1. C
2. D
3. C
4. B
5. B
6. C
7. C
8. A
9. A
10. A
11. D
12. A
13. A
14. A
15. A
16. C
3.
1. D
2. B
3. A
4. A
5. A
6. D
7. D
8. C
9. D
10. A
11. D
12. B
13. B
14. A
15. C
16. B
17. A
18. B
19. A
20. D
21. D
22. B
23. A
24. D
25. C
26. C
27. D
28. C
29. A
30. A
31. B
32. B
33. B
7. C
34. d + c - e - 2a - 3b
2. A
3. B
4. D
8. A
9. C
10. A
11. A
14. A
15. C
16. A
17. D
21. C
22. D
23. C
24. B
6. A
12. (i) A , (ii) (D)
13. D
18. B
19. B
20. B
4. A
5. A
6. D
7. D
11. A B D
12. B
13. B
14. C
18. C
19. A
20. B
3. B
8. B
9. B
10. A C
15. B
16. B
17. A
St
ud
2. D
yS
1. A
Page 50 of 51
5. B
te
1. C
5.
.in
ps
4.
35. - 1539.94 kJ
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Page 51 of 51
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ud
yS
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.in
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CHEMICAL THERMODYNAMICS

2. Some fundamental Definitions

consideration is called surrounding.

State of a System and state variable :

Extensity and Intensive properties : An extensive property of a system is that which

Thermodynamics Processes : The operation by which a thermodynamic system changes

form one state to another is called is thermodynamic process.

For a given system, E is directly proportional to its absolute temperature. (E T)

Work : Work is expressed as the product of two factors, i.e.,

Pressure volume work (Irreversible)

Work during isothermal Reversible Expansion of an Ideal Gas

For an ideal gas PV = nRT

W = 2.303 nRT log

W = 2.303 nRT log

W = 2.303 nRT log

During the change from A to B, volume remains constant, the process is

Given n = 2, R = 8.314 JK 1 mol1, T = 298 K, V2 = 20 L and V1 = 10L.

w = 2.303 2 8.314 298 log

= 2.303 5 8.314 300 log

PV1 = n1 RT and PV2 = n2RT

6. Heat capacity of a system

Then, heat capacity = C =

It is the amount of heat required to raise the temperature of a system through 1C

Specific Heat and Molar Heat Capacity at Constant Volume

Molar heat Capacity : C =

Heat Capacity at Constant Pressure :

From (a) and (b)

CP is always greater than CV

Relationship between CP and CV.

For one mole of gas

Dividing both sides by dT

Calculation of W , E , H , for Isothermal Expansion of Ideal Gas

E for an ideal gas E depends on temperatures. Since temperature is constant

Adiabatic Reversible Expansion of ideal Gas :

E or E2 E1 = nCV (T2 T1)

Heat absorbed by system and work done by system

Heat liberation by system and work done on the system

Isobatic process (expansion)

Instructions : From Question (1 to 3) a single statement is made. Write T if the

Ice in contact with water constitutes a homogeneous system.

Instructions : Questions (7 to 18) consist of a problem followed by several alternative

A system X undergoes following changes

(A) reversible process

(B) cyclic process

Work done in taking the gas from the state A B

Work done in a complete cycle (1 litre = 10 3 m3)

Q.10 Work done in taking the gas from B C

III. B For 1 mole of an ideal gas following V-T graph is given

Answer the following questions (14 to 18) from the graph

(D) 0.082 & 0.164

Q.13 The pressure at C is :

(D) 0.821 atm

(C) 0.0821 atm

Q.14 The process which occurs in going from B C is :

Q.15 The work done in going form C to A is :

Q.16 The process A B refer to :

(B) reversible process

Which one is not a state function ?

Which is the intensive property ?