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thermodynamics
B K Sharma
Definition - Thermodynamics
quantitative relationship between
various forms of energies.
Importance of Thermodynamics
Predict the feasibility of a chemical reaction under a
given set of conditions.
Predict the extent to which the reaction is
carried out before attainment of equilibrium.
Basic Terms in Thermodynamics
Universe
system
surroundings
Types of systems
Open system
Closed system
Isolated system
Homogeneous system
e.g. aqueous solution of NaCl, aqueous solution of sugar.
Heterogeneous system
e.g. ice in water, oil in water, etc.
Properties of a system
1. Macroscopic property
shown by large number of particles
of a system.
volume, pressure, temperature,density.
2. Microscopic property
Property of individual atoms and molecules of a system at micro level
atomic mass, molecular mass.
Macroscopic Properties of System
Macroscopic Properties
extensive properties
mass, volume
intensive properties
temperature, pressure, viscosity,
density,
State function Thermodynamic properties which depend
only upon the initial and final states of a system For
example, internal energy, enthalpy, etc. mass, volume,
etc.
State of a system
The macroscopic properties which
causes a change in the state of a
system are called state variables.
For example, pressure, temperature,
etc.
Types of Thermodynamic Process
(i) Isothermal process :-A process which is carried out
at constant temperature.
(ii) Adiabatic process:-A process which is carried out in
such a way that no heat flows from the system to
surroundings and vice versa.
(iii) Isochoric process :-A process which is carried out at
constant volume.
(iv) Isobaric process:-A process which is carried out at
constant pressure.
Types of Thermodynamic Process
(v)
Reversible process :- carried out infinitesimally slow so that all
changes occurring can be reversed and the system remains
almost in a state of equilibrium with the surroundings at every
instant of time.
(vi) Irreversible process:- Occurs rapidly but the system cannot
be reversed to its initial state immediately and the system
does not remain in equilibrium during transition.
(vii)Cyclic process:- When a system undergoes different processes
and finally returns to its initial state, is known as cyclic
process.
Internal energy
Energy stored within a substance of a system
internal energy or intrinsic energy.
E = Ee + Ev + Er
Internal energy
Actual (absolute) value can not be measured
change in internal energy (DE) can be measured
It is an extensive property.
Internal energy is a state function
For a cyclic process, DE = 0
Work
Work done = Force × Displacement
W = P(V2 – V1) (At constant pressure)
W = PDV
Where,
W = work done
W = –ve, work done by the system (expansion)
W = +ve, work done on the system (contraction)
Work
Work done in isothermal reversible expansion
of an ideal gas
Pext–
dP
Pext
dv
Pg
Pg
Wrev  
V2

V1
Wrev
Work
nRT
dV
V
V2
  nRT ln
V1
Wrev = – 2.303 nRT log
V2
V1
1
Since P 
V
\ Wrev = – 2.303 nRT log
P1
P2
First Law of Thermodynamics
Energy can neither be created nor destroyed
E2 = E 1 + q + w
E2 – E1 = q + w
DE = q + w
If work is done by the system, (in case of expansion)
w = - PDv
\ DE = q - PDv
\ q = DE + PDv
First Law of Thermodynamics
a.
During isothermal expansion
DE = 0
\ q = PDv
b. In case of adiabatic process, q = 0
\ DE = - PDv = +w
DE = +w
c. In case, the process is carried out at constant volume
First Law of Thermodynamics
DV = 0
PDV = 0 or w = 0
\ qV = DE
\ Quantity of heat supplied at constant volume, qv is equal to increase in the internal
energy of the system.
Session objectives
1. Enthalpy
2. Various types of enthalpy of reactions
3. Heat capacities of gases
4. Adiabatic process
5. Hess’s law
6. Bond energy
7. Lattice energy
8. Limitation of first law of thermodynamics
Enthalpy
Enthalpy is the total heat contents of the system at constant pressure.
Enthalpy is shown by ‘H’.
H = E + PV
Enthalpy change at constant pressure
DH = DE + PDV
H2 – H1 = E2 – E1 + P(V2 – V1)
Where H1, E1 and V1 are the enthalpy, internal energy
and volume respectively in initial state while H2, E2 and V2
are the enthalpy, internal energy and volume respectively
in final state.
Enthalpy
Enthalpy is a state function
PV1 = n1RT
At constant pressure
(for initial state)
PV2 = n2RT
(for final state)
P(V2 – V1) = RT(n2 – n1)
PDV = Dn g RT
Where Dng = np–nr (gaseous moles only)
\ DH = DE + DngRT
Enthalpy of formation
It is the change in enthalpy when one mole of a
compound is formed from its elements in their
naturally occuring physical states.
2C(s)  2H2(g) 
 C2H4 DHf  52kJ
Enthalpy of combustion
It is the change in enthalpy when one mole of the
substance undergoes complete combustion.
CH4  g  2O2  g  CO2  g  2H2O(g)
; DH   890.3 kJ
Application of heat of combustion
Calorific value
Amount of heat produced per gram of a substance (food or fuel) is completely
burnt.
CH4  g  2O2  CO2  g  2H2O l 
DH = -890.3 kJ/mol
Calorific value of CH4 (g) =
– 890
 – 55.6 kJ / g
16
Hydrogen has the highest calorific value (150 kJ/g)
Enthalpy of solution
Amount of heat evolved or absorbed per
mole of the substance in excess of water,
KCl  s   H2O 
 KCl  aq
DH  – 4.4 Kcal
KOH  s   H2O 
 KOH  aq DH  – 13.3 Kcal
Enthalpy of fusion
One mole of solid substance changes to its liquid
state at its melting point.
Melting
H2O  s   H2O l
Freezing
DH  1.44 Kcal
H2O l 
 H2O  s  DH  – 1.44 Kcal
Enthalpy of vaporization
One mole of the substance changes from liquid state to
gaseous state at its boiling point.
H2 O  l   H2 O  g 
Boiling
DH  10.5 Kcal
H2 O  g  H2 O  l  DH  – 10.5 Kcal
Cooling
Enthalpy of sublimation
Enthalpy change per mole of a solid converts directly
to its vapours
sublimation

NH4Cl (s) 
 NH4Cl  g
DH  14.9 Kcal
Heat capacity
Specific heat capacity is the heat required to raise the temperature of
unit mass by one degree.
q
= c × m × DT
m = Mass of the substance
q = Heat required
DT = Temperature difference
c = Specific heat capacity. Specific heat capacity of water is 4.18
J/g K.
Hess’s Law
According to Hess’s law
q = q1 + q 2
q
1
C
q
2
q
A
B
Determination of lattice energy
 D H MX   D Hsub M
1
 (D H
2
)
diss X2
 (IP)M  (  EA)X  (  U)MX
Figure
Determination of bond energies or bond
enthalpies
Energy required to break the bond
or energy released during the bond formation is
called bond energy.
H2S 
 H  g  SH  g
DH  100 kJ / mole
SH  g 
 S  g  H  g
DH  200 kJ / mole
The average of these two bond dissociation energies
gives the value of bond energy of S — H.
Bond energy of S — H bond

100  200
 150 kJ / mole
2
Limitations of first law
1. The first law of thermodynamics
states that one form of energy
disappears, an equivalent amount
of another form of energy is produced.
But it is salient about the extent to which such conversion can
take place.
2. It does not tell about the direction of flow of heat.
3. It does not tell about spontaneity of reaction.
Class exercise 2
When 1 gram of methane (CH4) burns
in O2 the heat evolved (measured under standard
conditions) is 13.3 kcal. What is
the heat of combustion?
(a) –13.3 k cals (b) +213 k cals
(c) – 213 k cals (d) – 416 k cals
Soluti
on
13.3 Kcal/gm evolved
DHo
  13.3  16  Kcal / mol
comb
= – 213 Kcal/mol
Hence, the answer is (c).
Class
exercise
3 is transferred
When
4.184 J of heat
to 1 g of water at 20° C, its temperature
rises to 21° C. The molar heat capacity
at this temperature is
(a) 18 JK–1
(c) 75.4
JK–1
(b)
18
4 .184
JK
-1
(d) 4.184 JK–1
Solution:
n CDT = 4.184
C = (4.184) × 18 = 75.4 J/K
Hence, the answer is (c).
Class
exercise
4 (B.P 80° C)
When
0.532 g of benzene
is burnt in a constant volume system
with an excess of oxygen, 22.3 kJ of
heat is given out. forDthe
H combustion process
is given by
(a) – 21 kJ
(b) – 1234.98 kJ
(c) – 221 kJ
(d) – 3273.26 kJ
Soluti
on
DH   22.3 
Mol. wt C6H6 
 22.3  78

0.532
0.532
= – 3269.5 kJ
Hence, the answer is (d).
Class
exercise
5
Consider
the reaction
1
SO2(g)  O2(g)  SO3(g) DH
2
= – 98.3 kJ. If the enthalpy of formation
of SO3(g) is – 395.4 kJ, then the enthalpy of formation of
SO2(g) is
(a) 297.1 kJ
(b) 493.7 kJ
(c) – 493.7 kJ
(d) – 297.1 kJ
Solution:
DHo
 DHo
  98.3 kJ
f  SO3 
f SO2 
 DHo
 98.3  395.4 kJ  297.1 kJ
f SO2 
Hence, the answer is (d).
Class
exercise
6
Calculate the heat change for the
following reaction:
CH4(g) + 2O2(g) 
 CO2(g) + 2H2O(l)
DH0for CH4 , H2O and CO2 are –17.89, –68.3
f
and–94.05 kcal/mole.
Soluti
on
CH g  2O
4

 CO2  g  2H2O l
2  g 
o
o
o
o
DHreaction
 2DHo

D
H

D
H

2
D
H
f H2O
f CO2
f CH4
f O2
= – 2 × 68.3 – 94.05 + 17.89 – 0
= – 212.76 kcal/mol
Class
exercise
Calculate
the heat7of combustion of
benzene from the following data:
6C (s) + 3H2(g) 
 C6H6(l)
DH = 11720 cal
1
H2(g) + O2(g) 
 H2O(l) DH = -68320 cal
2
C(s) + O2(g) 
 CO2(g) DH =-93050 cal
Soluti
onrequired reaction is
The
C6H6 
15
O2 
 3H2O  6CO2    (1)
2
6C + 3H2  C6H6    (2)
C + O2  CO2    (3)
1
H2  O2 
 H2O
2
3 × (4) + 6 × (3) – (2)
o
o
o
 DH1
 3DHo

6
D
H

D
H
4
3
2
= 3(– 68320) + 6(– 93050) – (1720 Cal)
= – 774.980 kcal/mol
Class
exercise
8
H the following
Calculate Dfor
reaction at 27o C.
C2H4 (g) + 3O2(g)  2CO2 (g) + 2H2O(l)
Given D=H– 337 kcal R= 1.987 cal deg–1 mole–1
Soluti
on
DH = DE + PDV
= DE + DnRT
{Dn = -2}
 DE =DH - DnRT
=– 337 – (–2)(1.987)(300) × 103
= – 335.8078 kcal/mol
Class
exercise
Calculate
the heat9of combustion of
acetic acid at 25o C if the heat of
formation of CH3COOH(l),CO2(g) and H2O(l) are
–116.4, –94.0
and –68.3 kcal mole–1 respectively.
Solution:
CH3COOH l  3O2 g 
 2CO2 g  2H2O l
DHreactions = 2 . DHf (CO2) + 2 . DHf (H2O)
-DHf (CH3COOH) - 3DHf (O2 )
= 2(– 94.0) + 2(– 68.3) – (–116.4)
= – 208.2 Kcal/mol
Spontaneous process
A process which takes place by itself is
called a spontaneous process.
For example,
1.
Dissolution of common salt in water
2.
Evaporation of water in open vessel
3.
Flow of heat from hot end to cold end of a
metal rod
Non-spontaneous process
A process which cannot take place by itself is called a
non-spontaneous process.
For example,
1. Flow of water up a hill
2. Flow of heat from a cold body to a hot body
Entrop
y
Entropy is measurement of randomness.It is
denoted by S
Unit of entropy is JK–1
q
Entropy Change DS 
T
Entropy  Disorderness
Crystalline solid – Lowest entropy
Gaseous state – Highest entropy
Entropy
Entropy is state function
DS  SFinal  SInitial
standard entropy change
DSo   So (Products)  So (Reactants)
Entropy changes during phase transformations
Entropy of fusion
One mole of a solid changes into liquid at its melting point.
DSfusion  Sm (liquid)  Sm (solid)
DHfusion

T
ΔSfusion is entropy change of fusion per mole
Entropy of vaporization
One mole of a liquid changes into vapours at its boiling point.
DSvap  Sm (vapour)  Sm (liquid)
DHvap

T
DSvap is entropy of vaporization per mole
Entropy of sublimation
One mole of solid changes into vapours
DSsub  Sm (vapours)  Sm (solid)
DHsub

T
DSsub is entropy of sublimation per mole
Second law of thermodynamics
Various statements of second law of thermodynamics
Total entropy change of system and surroundings is positive for all
spontaneous processes.
The entropy of the universe is continuously
increasing.
Total heat absorbed by a system cannot be converted
completely into work
Gibbs free energy
Amount of energy available for useful work. It is denoted by G.
G = H – TS … (i)
Gibbs free energy for isothermal process,
For initial state
G1  H1  TS1
For final state
G2  H2  TS2
\ G2  G1  H2  H1   T  S2  S1 
DG  DH  TDS
This is Gibbs-Helmholtz equation.
Gibbs-Helmholtz equation and spontaneity
DG  DH  TDS
DH
(a) If
is positive
is non-spontaneous.
DH  TDS,
then process
DG
(b) If
negative
DH  isTD
S, DGand thus,
the process is spontaneous.
(c) If
is D
zero
DH  T
S, and
DGthus, the process is in
equilibrium.
Such a process would be spontaneous at high temperatures
DS
Physical significance of Gibb’s free energy
Calculation of electrical work done
If work involved is the electrical work
(like in Galvanic cells),
DGo   nFEo
n is number of electrons involved in the cell reaction
Eo is electromotive force (EMF) of the cell
F is Faraday’s constant
Calculation of equilibrium constant (Kc)
DGo   2.303RTlogK
(where R is the gas constant)
Third Law of Thermodynamics
Nernst in 1906 gave third law of thermodynamics.
It states that the entropy of all pure and perfectly
crystalline solids may be taken as zero at absolute
zero temperature.
Class exercise 1
Which of the following has the highest entropy?
(a) Normal egg
(b) Half boiled egg
(c) Hard boiled egg
(d) (b) and (c) have same entropy
Solution:
There is maximum randomness in normal boiled egg
due to denaturation of proteins in the egg which involves
conversion of helical form into more random coil form.
Hence, the answer is (a).
Thank you