Chemistry : Thermodynamics (Class 12)

1. 1

2. 2

3.  Does not give information on rate of
physical & chemical process
 Does not describe the mechanism of
process
 Does not deals with microscopic
entities
 Independent of path followed to bring
the change.
3

4. SYSTEM
SURROUNDING
UNIVERSE
4
BASIC CONCEPTS

5. 


SYSTEM + SURROUNDING = UNIVERSE
5

6. Open system
Closed system
isolated system
TYPES OF SYSTEM
6

7. LET’S STUDY WITH AN EXAMPLE
7

8. 8

9. THERMOS FLASK :-
EXAMPLE OF ISOLATED
SYSTEM
9

10. SUMMARY
10

11. Properties of system
EXTENSIVE PROPETIES INTENSIVE PROPETIES
11

12. property of system whose magnitude
depends on the amount of matter present in the
system
ex:-volume of one mole of gas at STP is 22.4
dm3 , but two moles of gas occupies 44.8 dm3 at
STP. So volume is extensive property.
some more ex:- mass, work, heat, energy,
enthalpy, entropy
12

13. property of system whose magnitude is
independent on the amount of matter present in the
system.
ex:-boiling point of water is 100ºC , irrespective
of volume of 1 liter or 2 liters at 1 atm. pressure. So
boiling point is intensive property.
Some more ex:-M.P, B.P and F.P of pure
substance, surface tension,
Viscosity, density, refractive index , pressure,
specific heat, surface tension, molar heat capacity.
13

14.  Measurable thermodynamic properties like….
Pressure , temperature , volume.
 values of state variables describe the state of
system
 If there is any change in one or more variable,
system changes to a new state
14

15.  The state of the system changes by changing
the state variables so in thermodynamics
these variables are also called as state
function
15

16. in all the three paths
followed
∆P =P2 – P1
∆V = V2 – V1
∆T = T2 – T1
 Changes are
independent of path
followed
 It depends on initial
and final state
16

17. Now lets define ……..
Any property of the system which depends
only on initial and final state of the system
but is independent of path followed by the
system during thermodynamic process is
called state function.
Importance: ex- if ∆T = T2 – T1 is negative (-)
then we can predict that heat is given out of
the system irrespective of the path followed
and it is exothermic process .
17

18. Path function
 functions depending on path followed
Examples:- work and heat
amount of work done depends on path
followed to bring the change in system
18

19. Thermodynamic functions
All state functions and paths functions are
referred as thermodynamic functions
What if all the thermodynamic functions
doesn’t change with time in the
system????????
19

20. 1. Thermal equilibrium:- system and
surrounding are at same temperature Ex:-
water in equilibrium with its vapour at
constant temperature
2. Chemical equilibrium :-When chemical
composition of the system does not change.
Ex:
N2 +3H2 = 2NH3 all are in gaseous state
20

21. 3. mechanical equilibrium:- there is no
macroscopic movement with in the system
i.e. there is no movement of matter in the
system with respect to its surrounding.
21

22. Reactants products
(initial state) (final state)
Path followed to bring the change is
called process.
THERMODYNAMIC PROCESSES
Isothermal process:- TEMPERATURE remains constant
Adiabatic process:- NO HEAT exchange
Isochoric process:- VOLUME remains constant
Isobaric process:- PRESSURE remains constant
22

23.  Consider an example of refrigerator
 We know that temperature of refrigerator
remains constant even though many reaction
are taking place inside it.
what does that mean????????
refrigerator is designed in such a way that
heat generated is constantly thrown out in
surrounding to maintain its temperature
constant.
thus refrigerator carries isothermal process
inside it.
23

24. Melting of ice
Evaporation of water
24

25.  For isothermal process
 T1 = T2
 ∴ T2 – T1 = 0
 ∴ ∆T = 0
 IF U1 = Initial internal energy of system
 U2 = final internal energy of system
 internal energy depends on temperature of
system as there is no change in temperature
 ∴ U1 = U2
 ∴ ∆U = 0
25

26.  Heat is not allowed to enter or leave the
system.
Example :- Reaction carried out in an
closed insulated cylinder.
26

27.  Volume of the system remains constant
through out the process
 Let Initial volume be V1 and final volume
be V2
 ∴ for isochoric process V1 = V2
 ∴ V2 – V1 = 0 ……(as no change in
volume)
 ∴ ∆V = 0
27

28.  Pressure of the system remains
constant
 All the process carried out in natural
environment are isobaric as
atmospheric pressure remains
constant.
 ∴P = 0
28

29. Initial volume (V1)
Final volume V2
Cross sectional area (A)
Cylinder
Piston
Let’s try to derive an equation for work done by a gas
against constant pressure
29

30. Work done during expansion is against the
external pressure
∴ work done= - force x displacement
= - f x d
= - pressure x area x d
but area x displacement = change in
volume
= V2 – V1
= ∆V
Work done = - Pext x ∆V
30

31. Work done in isobaric process can also be
shown graphically…..
Since pressure is
constant let at state
A and at state B
pressure be P . let
volume changes
during expansion
from VA TO VB
so work done can
expressed as..
W = -l(AVA ) x l (VB-VA)
W = - P X ∆V
SINCE VB is greater
than VA
VB-VA will be negative
∴ W = - P ∆V
31
B
VB

32. 32

33. dw = -(p – dp)dv
= -pdv +dpdv
= -pdv …………….(due to negligible value of dp.dv, the
product can be neglected)
∴ dw =- pdv
Integrating the above equation between the limits of initial volume
V1 and final volume V2
∫dw = -∫ pdv
= -p ∫ dv
= -nRT/V ∫ dv …………………PV=nRT acc. to ideal
gas equation
33

34. = -nRT ∫ dv /V
=-nRT ln
=-nRT[ lnV2 – lnV1 ]
=-nRT lnV2/V1
v2
  2
1
v
v
V
v1
34

35. When Work is done by the system on the surrounding
( expansion) the expression is given as……
When work is done by the surrounding on the system
(compression ) log(
35

36. When Work is done by the system on the surrounding
( expansion) the expression is given as……
When work is done by the surrounding on the system
(compression ) log(
36
IN TERMS OF PRESSURE

37. 37

38. 38

39. 39

40. 40

41. 41

42. 42

43. 43

44. Energy is neither created nor destroyed and can
only be converted from one form to another
44

45. Total energy of the universe
always remains constant
45

46. 1. energy can neither be created nor be
destroyed but can be converted from one
form to another
2. Whenever one form of energy disappears,
equivalent amount of another kind of energy
appears
3. Total energy of the universe always remains
constant
4. Total energy of an isolated system always
remains constant
46

47. 47

48. U1
Q
W
U2
48

49. 49

50. Here
50

51. 51

52. For compression of gas W = + P∆V
if W = -Q
+ P∆V = -Q
Which means when work is done on
the system by the surrounding
(compression) in isothermal process….
heat is given out of the system
because Q value is negative.
52

53. As there is no exchange of heat Q = 0
According to first law of thermodynamics..
Thermal insulation ∆U= +ve
53

54. ∆V = 0 as volume remains constant
∴ W= P ∆V = P X (0) = 0
According to first law of thermodynamics
54

55. In isochoric
process (∆V= 0)
If heat is
supplied to
system
Internal energy
increases [ ∆U =+ve ]
If heat is given
out of the
system
Internal energy
decreases[ ∆U = -ve ]
Let us summarize for isochoric process
55

56. Reactions carried out at constant pressure ….
all the chemical reactions are generally carried out at
atmospheric pressure which is constant
56

57. We know that Q is not a state function…….
can you tell me which type of function is Qp and
Qv ?????????
57

58. Heat changes occurring at constant
atmospheric pressure Qp change in
enthalpy
ENTHALPY?????????????????????
New word for you
58

59. Symbol ---- H
Is a sum of internal energy of
system and the energy equivalent to
PV work.
U , P and V are state functions so what
about H ???????
59

60. ∆H = H2 – H1
BUT
H1 = U1 + P1 V1
H2 = U2 + P2V2
∴
∆H = (U2 + P2V2) -(U1 + P1 V1)
∆H = (U2 - U1) +(P2V2- P1 V1)
∆H = ∆ U + ∆ (PV)
If pressure is kept constant
P1 = P2 = P
∆H = ∆ U + P ∆ V
60
Lets consider
P1 P2
V1 V2
U1 U2
H1 H2
initial state final state

61. At constant atmospheric pressure we already derived an
equation
Qp = ∆U + P∆V
And for change in enthalpy at constant pressure in the system we
got…….
∆H = ∆U + P∆V
So if external atmospheric pressure and pressure of gas in the system
are same , then from above two equations we can derive
∆H = Qp
Thus change in enthalpy of a system = heat
transferred at constant pressure.
61

62.  Exothermic reaction (heat
is given out… ∆H = -ve)
 Endothermic reaction (heat
is absorbed… ∆H = +ve)
Qp
Qp
A → B C → D
62

63. 63

64. We know that at constant pressure ∆H and ∆U are
related as…
∆H = ∆U + P ∆V
******** for solids and liquids ∆V and is
very small and can be neglected
Hence ∆H = ∆U
but what about gaseous
reactions?????????????
64

65. Let n1 be the initial moles of gaseous reactants
and n2 be the final moles of gaseous product.
We know that change in enthalpy is given as …..
∆H = ∆U + P ∆V
= ∆U + P ( V2 – V1 )
∆H = ∆U + P V2 – PV1
Where V1 is the volume of gaseous reactants
V2 is the volume of gaseous products
suppose the gases are behaving ideally then….
We can apply ideal gas equation PV = n RT for both
reactants and products
For gaseous reactants PV1 = n1RT
For gaseous product PV2 =n2RT
∆H = ∆U + n2RT– n1RT
65

66. ∆H = ∆U + (n2-n1)RT
∆H = ∆U + ∆ngRT
∆ng = number of moles of gaseous product – number
of moles of gaseous reactants.
but we know that ……..
∆H = Qp and ∆U = Qv
Qp = Qv + ∆ngRT
can u predict equation for work done in terms of
change in NUMBER OF MOLES ??????????????
66

67. We know that at constant pressure and
temperature....
W = - P ∆V
BUT PV1 = n1RT and
PV2 =n2RT
∴ PV2 – PV1 = (n2 – n1)RT
P ∆V = ∆nRT
∴ W = - ∆nRT
67

68.  Work done = - Pext x ∆V
for expansion in
irreversible process
 W max= -2.303 nRT log10V2/V1
for expansion in reversible
process
 Wmax= -2.303 nRT log10 P1/P2
for expansion in reversible
process
 W = - ∆nRT
 Work done = + Pext x ∆V
for compression in
irreversible process
 W max= +2.303 nRT log10V2/V1
for compression in reversible
process
 W max= +2.303 nRT log10 P1/P2
for compression in reversible
process
 W = + ∆nRT
68

69. For isothermal process
P∆V = Q …for expansion
P∆V = - Q …for
compression
For adiabatic process
∆ U = W
For isochoric process
∆U = Qv
For isobaric process
Qp = ∆U + Pext ∆V
Change in enthalpy
∆H = ∆U + Pext ∆V
also
Qp = Qv + ∆nRT
Or
∆H = ∆U + ∆nRT
69

70. Phase transition
Enthalpy of
fusion (∆fusH)
Enthalpy of
vaporization
(∆VapH)
Enthalpy of
sublimation
(∆subH)
Atomic or
molecular change
Enthalpy of
ionization
(∆ionH)
Enthalpy of
atomization
(∆atomH)
Enthalpy of
solution
(∆solH)
70

71.  the standard state of a substance is the
form in which the substance is in most
stable state
 for gas :- pressure of one bar at 298 K
 for species in solution :- 1 molar
concentration at 298 Kelvin
 Standard states of certain elements and
compounds are
EXAMPLES :- H2(g), Na (s), C(graphite) ,
C2H5OH(l) , CaCO3(s) , CO2(g), H2O(l)
71

72.  consider the balanced equation for reactants and
products
 the value and appropriate sign of enthalpy
change is given on the right hand side this value
is delta H
 the physical states of reactants and products are
specified by letter s (solid) l (liquid) g (gas) and
aq ( aqueous) ∆r H value refers to Physical states
of substances those appear in the equation.
72

73.  The given value of ∆r H assume that the
reaction occurs in a given direction . the ∆r H for
the given reaction equals in magnitude and
opposite in the same to that of Forward
reaction. An exothermic reaction on reversal
becomes endothermic and vice versa
 when the coefficients indicating the number of
moles of all substances in thermochemical
equation are multiplied or divided by a certain
numerical factor, the corresponding ∆ H need to
be multiplied or divided by the same
73

74. Consider a thermo chemical equation
C2H2(g) + ½ O2(g) 2CO2(g) + H2O(l) ∆rH° = -1300 KJ
Write thermo chemical equation when
1) Coefficients of substances are multiplied by 2
2) Equation is reversed
answer
1) 2C2H2(g) + O2(g) 4CO2(g) + 2H2O(l) ∆rH° = -2600 KJ
2) 2CO2(g) + H2O(l)→ C2H2(g) + ½ O2(g) ∆rH° = +1300 KJ
74

75. Consider
H2(g) = ½ O2(g)  H2O (l) ∆rHº = -286KJ
one mole of liquid water is formed in
standard state from H2 and O2 in their
standard state
∆rHº will be called as ∆fHº
75

76. Consider
C2H2(g) + 5/2O2(g)  2CO2(g) + H2O(l) ∆rH° = -1300 KJ
When one mole of a substance is
completely oxidized in its standard
state , the standard enthalpy change is
called as the standard enthalpy of
combustion
76

77. 77
LETS TRY TO DO AN ASSIGNMENT

78. 78

79. H H
H H
H(g) H(g)
+
79

80. H
O
H
H2O
80

81.  H2O(g) → OH(g) + H(g) ∆r H° = 499 kJ
 OH(g) → O(g) + H(g) ∆rHº = 428 kJ
 H2O(g) → 2H(g) + O(g) ∆r Hº = 927 kJ
BOTH O-H BOND ENTHALPY VALUE IS NOT SAME
Thus average bond enthalpy will be 927/2 =463.5kJ
∆Hº (O-H) = 463.5 kJ
81

82. Ex :-1
82

83. Ex :2
83

84. 84

85. A
B C
D
Reactant Product
Intermediate product 1 Intermediate product 2
∆H
A D
∆H3
∆H2
∆H1
∆H
85

86. Formation of NH3(g) can be shown as:-
first method:- one step equation....
3 H2(g) + N2(g)  2 NH3(g), ∆rHº = -92.2 kJ
Second method:- two step equation....
i. 2H2(g) + N2(g)  N2H4(g), ∆rHº1 = + 95.4kJ
ii. N2H4(g) + H2(g)  2 NH3(g), ∆rHº2 =-187.6 kJ
3 H2(g) + N2(g)  2 NH3(g), ∆rHº = -92.2 kJ
This defines HESS’S LAW
86

87. The change in enthalpy is same whether the
reaction takes place in one step or series of step
Overall the enthalpy change for a reaction is
equal to sum of enthalpy changes of individual
steps in the reaction
87

88. Helps to determine the enthalpy of reaction whose value
can not be determined experimentally
Helps to calculate enthalpy of reaction which do not
occur directly
Enthalpy of formation can be calculated
Enthalpy of combustion can be calculated
Thermo chemical equation can added subtracted,
multiplied or divided by a numerical factor like
algebraic expression
88

89. Spontaneous process occurring in nature
89

90. SPONTANEOUS
PROCESS
Occurs on its own
Proceeds in one
direction
Not concerned with the
rate of reaction
Continues till
equilibrium is achieved
KEY POINTS
90

91. GENERAL ASSUMPTION:-
Spontaneous reaction takes place in a
direction in which energy of the system is
lowered
i.e exothermic reaction
but EXCEPTIONAL CASES
ice melts by absorbing heat
NaCl dissolves in water by absorbing heat
from the surrounding
i.e endothermic reactions
91

92. SPONTANEITY
CRITERION
EXOTHERMIC
ENDOTHERMIC
???????
92

93. Highly ordered disordered Highly disordered
Gas( vapour)
Liquid (WATER)
Solid (ICE)
S1 S2 S3
If S is measure of disorderdness
Entropy S
93

94. ∆S=S2 - S1 = +VE ∆S=S3 – S2 = +VE
S1 S2
S3
H2O(s) H2O(l)
Above 0ºC
H2O(g)
At 100ºC
If a substance changes its state from SOLID to LIQUID
∆S is POSITIVE (ordered state to disorder state)
If a substance changes its state from LIQUID to GAS, ∆S
is POSITIVE (disordered state to highly disorder state)
94

95. The entropy change of a system in a
process is equal to the amount of heat
transferred to it in a reversible manner
divided by a temperature at which the
transfer takes place
∆S = qrev / T
95

96. From these examples its
clear that entropy
increases for a
spontaneous process
i.e ∆S > 0
BUT we again have
exceptions in some
cases lets see
that……..
96

97. 97

98. SECOND LAW OF THERMODYNAMICS
The total entropy of the system and its
surrounding increases in a spontaneous
process.
∆S(total) = ∆S(system) + ∆S(surrounding) >0
98

99. ∆S>0
99

100. ∆S>0
10
0

101. ∆S>0
∆S<0
10
1

102. ANS
∆S>0
∆S<0
10
2

103. Q.2
∆S<0
∆S>0
∆S=0
10
3

104. ∆S>0
∆S<0
∆S=0
10
4

105. Non
Spontaneous
Spontaneous
?
?
∆Hº <0 ∆Hº >0
∆Sº <0
∆Sº >0
10
5

106. Liquid (WATER)
Solid (ICE)
By absorbing
heat above 0ºC
∆H>0 ∆S>0
Non
Spontaneous
Spontaneous
?
?
∆Hº <0 ∆Hº >0
∆Sº <0
∆Sº >0
By releasing
heat (temp. is
lowere)
∆H<0 ∆S<0
10
6

107. Non
Spontaneous
Spontaneous
Spontaneous
at high temp.
Spontaneous
at low temp.
∆Hº <0 ∆Hº >0
∆Sº <0
∆Sº >0
10
7

108. 10
8

109. 10
9

110. ∆S(total) = ∆S(system) + ∆S(surrounding)
∆S(surrounding) = - ∆H / T
∆S(total) =∆ S(system) - ∆H / T
T∆S(total) = T∆S(system) - ∆H
-T∆S(total) = ∆H - T∆S(system)
-T∆S(total) = ∆G
∴∆G = -T ∆S(total)
11
0

111. 11
1

112. ∆G = ∆H – T∆S
(-)
exothermic
+ - Spontaneous at all
temperatures
(-)
exothermic
- - spontaneous at low
temperature , T∆S<∆H
(+)
endothermic
+ - spontaneous at high
temperature , T∆S>∆H
∆H ∆S ∆G
Spontaneity of
reaction
(+)
endothermic - +
Non spontaneous at all
temperatures
112

113. For equilibrium , ∆G = 0
∆G = ∆H – T∆S gives
∴ T = ∆H / ∆S
In the above equation T is the temperature at
which change over from spontaneous to non
spontaneous behavior occurs.
113

114. Gibbs function and equilibrium constant
Consider a reaction
aA + bB Cc + dD
Gibbs energy change for above chemical reaction is given as…
∆G = ∆Gº + RT lnQ
Q = reaction quotient or equilibrium constant
G = Standard Gibbs energy change when
reactants and products are in standard state
11
4

115. Consider a reaction
aA + bB Cc + dD
If the reactions are considered in terms of
concentration then
∆G = ∆Gº + RT lnQc
= ∆Gº + RT ln [C]c [D]d
[A]a [B]b
If the reactions are considered in terms of
partial pressures of gases then
∆G = ∆Gº + RT lnQp
= ∆Gº + RT ln PC
c X PD
d
PA
a X PB
b
11
5

116. When the reaction reaches equilibrium…….???????
∆G = 0 at equilibrium
QC and QP becomes equal to KC and KP
∴∆G = ∆Gº + RT lnQc ∴∆G = ∆Gº + RT lnQP
0 = ∆Gº + RT lnKc 0 =∆Gº + RT lnKP
∆Gº = - RT lnKc ∆G = - RT lnKP
∆Gº = -2.303 RT log10Kc
∆G = -2.303 RT log10KP
116

117. 11
7