NITRKL

1. Chemical thermodynamics
Chapter 01

2. Thermodynamics: crude concept of 1st and 2nd law and their significance
CxHy (fuel)
O2
CxHy + O2
CO2 + H2O (combustion)
CO2 (exhaust)
Car moves fwd.
Combustion of fuel converts chemical energy into mechanical energy to run a car
Introduction:
+ H2O
In this example:
“Energy is not created, neither destroyed, total energy of the universe is constant.”
Above statement is the simplest outcome or the major significance of 1st law
of thermodynamics.

3. Conservation of energy is not enough to understand the laws of the universe
CxHy + O2
CO2 + H2O
O2
CO2
pushing car backwd.
+ H2O
?
? (never)
By feeding the car CO2 and H2O through the exhaust and pushing the car backwards
can we generate fuel in the car and produce O2 in the air?
Note: the absurd situation above is not violating the 1st law of thermodynamics!
Therefore in thermodynamics, 1st law is not good enough to understand the direction
of any natural process of universe.
We require 2nd law of thermodynamics to understand the direction of any natural process

4. Significance of thermodynamics in Chemistry:
A + B C + D
•To understand whether a reaction will be exothermic or endothermic
•To understand whether a reaction or a physical change will attain an equilibrium
•To understand the direction of a spontaneous chemical change
•To understand the potential (voltage) of a chemical cell or a battery
•To understand the mechanism of chemical reactions in certain cases
•To understand how a chemical or physical change respond with change of T and P
•To understand the extent of a chemical change

5. First Law of Thermodynamics and Chemistry

6. First law of thermodynamics: (internal energy, heat, work, enthalpy)
We know that the total energy of the universe is constant.
Universe = system + surroundings
A system is defined as the particular region of space that
we are making measurements on. Example of a system :
test tube
In chemistry, first law of thermodynamics is all about gaining and losing of energy by
a chemical system. In the context of thermodynamics we refer this energy as
‘internal energy’ (U) and its unit is the joule (J).
The absolute value of U of a system cannot be measured but the difference (ΔU) can be.
ΔU = U final state – U initial state = Uf - Ui
Internal energy (ΔU) is a state function. It does not depend on the process of change, but
only depends on the initial and final states.

7. First law of thermodynamics:
Ui
A system with
some initial U
Uf
work done by the system
in the surroundings
Q
(supplied heat)
the system with
final U
If a system having initial internal energy (Ui) is supplied with heat Q, then the absorbed
heat may be used to increase the internal energy of the system or perform work or both.
If the final internal energy after absorbing heat is (Uf) then change in internal energy
= ΔU = Uf - Ui
According to the first law of thermodynamics, energy lost = energy gained
Q = ΔU + work done by the system
or, Q = ΔU + (-W) ……………………eq. (1)
Please note: according to IUPAC convention, work done by the system is expansion work
and taken as (-W)
+

8. First law of thermodynamics:
From equation (1) we get,
Q = ΔU + (– W)
or, ΔU = Q + W ……………….eq. (2)
or, dU = đQ + đW (in the differential form) ……………….eq. (3)
dU ≡ exact differential (we can apply an exact differential to any state function, which is
path independent)
đQ, đW ≡ inexact differential (we can apply an inexact differential to any path function,
contrary to state function, path function is path dependent)
Interpretation of work (W): work done by the system can be expansion and non-expansion.
In the context of first law, we will ignore any non-expansion work.
Expansion work: In thermodynamics it is classically idescribed as the movement of a piston
by expansion or compression of a gas.
So, W = Wexp + Wnon-exp
Here, W = Wexp by setting Wnon-exp = 0

9. Pin
Pout
cylinder
piston
gasB
gasA
Expansion work of a gas:
If Pin > Pout then gasA expands by performing work
against constant pressure Pout to obtain volume V2
from initial volume V1
So work done by the system = (-W) = Pout (V2-V1)
= Pout ΔV …….. eq. (4)
In differential form, we can write eq. (4) as
(-đW) = Pout dV = Pdv …….. eq. (5)
assuming Pout = P (any arbitrary opposing pressure)
For a finite change where the gas expands from volume V1 to V2, the total work done will be:
-∫đW = ∫Pdv (if P is variable) …….. eq. (6)
V1
V2
V1
V2
-∫đW = P∫dv (if P is constant) …….. eq. (7)
V1
V2
V1
V2
For ideal gas, P = nRT/V so, we can replace the value of P in eq. (6) and integrate to find W.
The first law of thermodynamics
In differential form:
dU = đQ - PdV
expansion work done
by the system

10. Irreversible and reversible expansion work:
Pi TVi P TVf
Single step expansion
against a constant pressure Pout
Stage A Stage B
Pout
Pout
Irreversible transformation!
-∫đW = P∫dv ( Pout = P = constant) = P(Vf – Vi)
Vi
Vf
Vi
Vf
irreversible work done =
Pi TVi
Stage A
Pout
P TVf
Stage B
Pout
P2 TV2
Pout
P3 TV3
Pout
P4 TV4
Pout
infinite
steps
…. ….
Pout = P
1 2 3
Pout = P2 Pout = P3 Pout = P4 Pout = P
multiple step expansion
against a variable pressure
-∫đW = ∫Pdv = nRTln(Vf/Vi) = maximum work done
Vi
Vf
Vi
Vf
reversible work done =
Reversible transformation!
(P is the variable pressure at any step)

11. First law of thermodynamics:
đQ + đW = (∂U/∂T)V dT + (∂U/∂V)T dV [from eq. (3)]
Properties of internal energy (U) in mathematical form:
Internal energy of a system having fixed mass is a function of temperature and volume
U = f(T, V) this relationship can be expressed in the form of a differential equation:
dU = (∂U/∂T)V dT + (∂U/∂V)T dV ……………..eq. (8)
partial derivative partial derivative
Please note: any state function (exact differential) can be expressed in the form of eq. (8)
or,
đQ - PdV = (∂U/∂T)V dT + (∂U/∂V)T dV [from eq. (5)] ……………..eq. (9)
At constant volume (isochoric) process, we know that dV = 0, so
đQV = (∂U/∂T)V dT = CVdT = dU ……………..eq. (10)
We define (∂U/∂T)V as Cv (heat capacity at constant volume)

12. So, under constant volume process,
dU = CvdT [where Cv = đQV/dT = (∂U/∂T)V ]
Or, ΔU = CvΔT (for a finite change)
First law of thermodynamics:
Therefore, we can write the following equations
dU = CV dT + (∂U/∂V)T dV ……………..eq. (11)
đQ - PdV = CV dT + (∂U/∂V)T dV ……………..eq. (12)
For ideal gas, the derivative (∂U/∂V)T = 0
So for ideal gas, following relationships are true
dU = CV dT (under any process, U is a function of T only)
or, đQ = CV dT + PdV

13. First law of thermodynamics:
For any isothermal process:
]=6543 `Temperature remains the same, that is dT = 0
So in general, the first law of thermodynamics becomes
đQ = (∂U/∂V)T dV + PdV ……………..eq. (13)
and for ideal gas,
đQ = PdV ……………..eq. (14)
For any adiabatic process:
When no heat is exchanged during a process, we call the process as adiabatic.
It means đQ = 0
So the general statement of first law becomes:
dU = đW = -PdV ……………..eq. (15)
CV dT + (∂U/∂V)T dV = -PdV ……………..eq. (16)

14. First law of thermodynamics:
In chemistry, generally reactions take place at constant pressure than at constant volume!
But we have seen so far that heat absorbed at constant volume is change in internal energy!
So, what happens when heat is absorbed at constant pressure (this happens in chemistry)?
We define a state function (H) that gets changed when heat is absorbed at constant pressure.
Just like U, we can write H = f(T, P)
dH = (∂H/∂T)V dT + (∂H/∂P)T dP ……………..eq. (17)
At constant pressure process, we know that dP = 0, so
đQP = (∂H/∂T)P dT = CPdT = dH ……………..eq. (18)
We define (∂H/∂T)V as CP (heat capacity at constant pressure)
We call this state function (H) as enthalpy
For ideal gas, (∂H/∂P)T = 0

15. So heat of reaction or heat exchanged during a chemical reaction is actually enthalpy
of the reaction since the chemical reaction takes place under constant pressure.
First law of thermodynamics:
Enthalpy:
Relationship between enthalpy (H) and internal energy (U):
H = U + PV ……………..eq. (19)
(P & V stand for pressure and volume of the system, respectively)
dH = dU + PdV + VdP
dH = dU + PdV (under constant pressure) ……………..eq. (20)
For 1 mole of ideal gas, H = U + RT
In differential form,
The derivation of the above relationship can be done elsewhere.
In words, change of enthalpy at constant pressure is change of internal energy plus product
of pressure and change of volume of the system.

16. First law of thermodynamics:
Relationship between QP and Qv
[from eq. (20)]
Under constant pressure,
dH = dU + PdV
For a finite change under constant pressure,
ΔH = ΔU + PΔV = Qp [from eq. (18)] and Qv = ΔU [from eq. (10)]
So, we can write, Qp = Qv + PΔV or, Qp - Qv = PΔV ……………..eq. (21)
Relationship between CP and Cv for ideal gas
đQ = CV dT + (∂U/∂V)T dV + PdV …………….. from eq. (12)
đQP = CV dT + (∂U/∂V)T dV + PdV …………….. (under constant pressure)
đQP /dT = CV + (∂U/∂V)T (dV/dT) + P(dV/dT) ….(under constant pressure)or,
CP = CV + (∂U/∂V)T (∂V/∂T)P + P(∂V/∂T)P ….(under constant pressure)or,
CP = CV + P(∂V/∂T)P ………………………………….[for real gas, the term (∂U/∂V)T = 0]or,
CP = CV + nR ………………………………….[using V= nRT/P at constant P]or,
or, CP /n - CV /n = R ……………eq. (22)

17. First law of thermodynamics:
CP /n - CV /n = R ……………eq. (22) [for ideal gas]
or, CP - CV = R …………eq. (23) [where CP & CV are respective molar heat capacities]
Adiabatic changes of state for ideal gas
CV dT = -PdV …………..eq. (24) [from eq. (15) & (16)]
We have seen earlier that for an ideal gas, when đQ = 0
The negative sign denotes that if the gas expands adiabatically, it’s temperature would
decrease.
Irreversible adiabatic change for an ideal gas:
For a finite change in a single step, if the gas expands from V1 to V2, against a constant opposing
pressure P, eq. (24) can be written as:
CV ΔT = -PΔV
CV (T2-T1) = -P(V2-V1)
Here T2 < T1 and V2 > V1

18. First law of thermodynamics:
Reversible adiabatic change for an ideal gas:
Since, CV dT = -PdV
the temperature drop becomes maximum if the gas expands in infinite number of steps
against a variable pressure P at every step, i. e. gas expands in a reversible adiabatic manner.
Therefore, for a reversible adiabatic expansion where gas expands from V1 to V2 , we can write:
CV dT = -nRTdV/V
T1
V2
V1
T2
CV dT/T = -nRdV/Vor, CV dlnT = -nRdlnVor,
CV ∫dlnT = -nR∫dlnV or, CV ln(T2/T1) = -R ln(V2/V1) or, ln(T2/T1) =
-(Cp – CV) ln(V2/V1)
CV
or, ln(T2/T1) = -(γ – 1) ln(V2/V1) or, ln(T2/T1) = -(γ – 1) ln(V2/V1)
or, (T2/T1) = (V1/V2)(γ-1)
[CP/CV = γ, a constant]
…………..eq. (25)
Using ideal gas law, the above equation can be transformed into:
P1V1
γ = P2V2
γ or, PVγ = constant …………..eq. (25)

19. First law of thermodynamics:
Graphical representation of adiabatic and isothermal change of an ideal gas
PV = constant
PVγ = constant

20. First law of thermodynamics: application in chemical reaction
Heat of reaction (ΔH)
The heat of reaction or enthalpy of reaction is governed by two fundamental laws:
1) Lavoisier and Laplace’s law (1782)
2) Hess’ Law (1840)
The Heat (ΔH) exchanged in a transformation is equal and opposite in the reverse direction.
A B
ΔH1
ΔH2
According to this law: ΔH1 = - ΔH2
The Heat (ΔH) exchanged in a transformation is same whether the process occurs in one
or several steps.
A D
ΔHtotal
A B
ΔH1
C
ΔH2
D
ΔH3
single step transformation mutiple step transformation
According to this law: ΔHtotal = ΔH1 + ΔH2 + ΔH3
(This is an obvious outcome since
ΔH is a state function)

21. Application of Hess’s law
First law of thermodynamics: application in chemical reaction
(1) C (s) + O2 (g) → CO2 (g) ΔH1 = -394 kJmol-1
(2) H2 (g) + ½ O2 (g) → H2O (l) ΔH2 = -286 kJmol-1
(3) C2H6 (g) + 7/2 O2 (g) → 2CO2 + 3H2O (l) ΔH3 = -1542 kJmol-1
Calculate the enthalpy of formation of ethane in the given reaction from the following:
2C (s) + 3H2 (g) → C2H6 (g) ΔHf = ?
The enthalpy of formation of ethane 2C (s) + 3H2 (g) → C2H6 (g) may be calculated by:
2C (s) + 2O2 (g) → 2CO2 (g)
3H2 (g) + (3/2)O2 (g) → 3H2O (l)
2CO2 (g) + 3H2O (l) → C2H6 (g) + (7/2)O2 (g)
2C (s) + 3H2 (g) → C2H6 (g)
2 x (1)
3 x (2)
Reverse (3)
Add:
ΔH = 2(-394) kJmol-1
ΔH = 3(-286) kJmol-1
ΔH = +1542 kJmol-1
ΔH = -104 kJmol-1
Therefore, ΔHf = -104 kJmol-1

22. Second Law of Thermodynamics

23. Second law of thermodynamics:
Ocean’s hot waterQH
Can the ship move
fwd??
First law:
QH = (-W) if ΔU = 0
work done
by ship’s engine
(-W)
According to the first law of thermodynamics, a ship’s engine can extract heat from ambient
Ocean water and transform the heat into work under isothermal condition.
Do we see such situation in real life?
The hypothetical situation shown above is in perfect agreement with first law,
but the situation is impossible in reality!
(hot reservoir)

24. Second law of thermodynamics:
The impossible situation described in previous slide is actually a statement of
second of thermodynamics in a particular form.
It is impossible for a system operating in a cycle and connected to a single heat reservoir
to produce work in the surroundings.
Kelvin statement of second law (1850):
IMPOSSBLE POSSIBLE
Hot reservoir
Cold reservoir
QC
TC
TH
QH
(-W) = QH - QC
-
So, second law states that under isothermal condition no engine can convert heat into work!
engine
Efficiency (η) of the real engine
= work done/ heat supplied
= (QH - QC)/QH
= 1 – QC/QH ………….. eq. (26)

25. Second law of thermodynamics:
Hot reservoir
Cold reservoir
QC
TC
TH
QH
W
Kelvin statement of second law is also true in reverse way: known as Clausius statement
of second law.
IMPOSSIBLE
Hot reservoir
Cold reservoir
QC
TC
TH
QH
POSSIBLE
A refrigerator cannot work itself by removing heat from a colder compartment and dumping
it to a hotter environment. We need to spend energy (work) to achieve refrigeration.
Engine/compressor
Earlier, at the introduction we appreciated that second law determines the direction of a
natural process. This statement can be connected with Kelvin’s statement by a historic
theorem known as “Carnot cycle”.

26. Second law of thermodynamics: The Carnot cycle
From the work of French Engineer Sadi Carnot (1824) what extent of work can be
extracted from heat by an engine. Carnot is often referred as father of thermodynamics.
Carnot’s theorem was later developed by Clausius and Kelvin to frame second law.
In his theorem, 1 mole of ideal gas contained in a cylinder, fitted with a piston was
used as an engine.
This engine was subjected to cyclic transformation between two reservoirs: hot and cold,
kept at constant temperature.
In every step, the engine was subjected to reversible transformation to obtain
maximum work out of it.
P1 THV1
P2 THV2
P3 TCV3 P4 TCV4 P1 THV1
isothermal
exp.
adiabatic
exp.
isothermal
compr.
adiabatic
compr.
Step 1 Step 2 Step 3 Step 4

27. Second law of thermodynamics: the Carnot cycle
W
work done = RTH ln(V2 / V1)
work done = RTC ln(V4 / V3)
P1 THV1
P2 THV2
P3 TCV3
P4 TCV4
work done = CV (TC - TH)
work done =
CV (TH - TC)

28. Total work done (-W) in Carnot cycle:
Second law of thermodynamics:
= RTH ln(V2 / V1) + CV (TC - TH) + RTC ln(V4 / V3) + CV (TH - TC)
= RTH ln(V2 / V1) - RTC ln(V3 / V4)
= RTH ln(V2 / V1) - RTC ln(V2 / V1) [from the graph as well as mathematically V2/V1 = V3/V4]
= R(TH – TC) ln(V2/V1)
Efficiency of Carnot engine (η) =
Total work done (-W) by engine
Heat withdrawn by engine (QH)
=
R(TH – TC) ln(V2/V1)
RTH ln(V2 / V1)
η = 1 – TC/TH As TH > TC, efficiency of Carnot engine is always less than unity.
Now, we have seen earlier that η = 1 – QC/QH for any real engine [from eq. (26)].
….. eq. (27)
We can write comparing eq. (26) & (27), TC/TH = QC/QH ….. eq. (28)
Therefore the conclusion of Carnot cycle is: no real heat engine is 100% efficient

29. Second law of thermodynamics: concept of entropy
From eq. (28), we get: QH/QC = TH/TC
or, QH/TH = QC/TC
or, đQH/TH = đQC/TC
or, đQH/TH - đQC/TC = 0
or, đQH/TH + (-đQC)/TC= 0
From Carnot cycle if we try to assess the value of đQ/T then combining all the steps
in the cyclic process, we can write:
∑ đQ/T = ∫ đQ/T = đQH/TH + 0 + (-đQC)/TC + 0 = 0 (Heat rejected in the cold reservoir is
associated with negative sign)
In Carnot cycle each & every step is reversible, so heat exchanged in each step is also reversible
∫ đQ/T = ∫ đQrev/T = 0 …….. eq. (29) ∫ ≡ cyclic integral,
If cyclic integral of any function is zero,
then the function is a state functionSo Qrev/T is a state function.

30. Second law of thermodynamics: concept of entropy
The state function Qrev/T is called entropy and denoted by S
∫ dS = 0 …….. eq. (30) where dS = đQrev/T
So from eq. (29), we get
For any reversible cyclic process, a system’s entropy change is zero.
Also, it can be proved that for any irreversible cyclic process
∫ đQirrev/T < 0 …….. eq. (31)
Combined form of first and second law of thermodynamics:
đQrev = dU + (– đW)……….. from first law, eq. (3) for a reversible process
or, TdS = dU + (– đW)……….. Combined form of first and second law, eq. (32)
or, TdS = dU + PdV……….. eq. (32)

31. Second law of thermodynamics: properties of entropy
Concept of systems: definition and analogy
Open system
Mass as well as energy
can pass through
Closed system
Only energy
can pass through but
no mass
Isolated system
Ideally neither mass
nor energy can
Pass through
An open container
of boiling water
A closed container
of boiling water
A thermo flask
v v v
đQ = 0đQ ≠ 0 đQ ≠ 0

32. Second law of thermodynamics: properties of entropy
The Clausius inequality
A system is transformed irreversibly from state 1 to state 2
Then it was restored reversibly from state 2 to state 1
∫ đQ/T = ∫ đQirrev/T + ∫ đQrev/T
1
12
2
So we can write
Since this is an irreversible cycle, this equation
is less than zero [from eq. (31)]
or, ∫ đQirrev/T - ∫ dS < 0
1 1
2 2
or, ∫ dS > ∫ đQirrev/T
1 1
2 2
for any infinitesimal change, dS > đQirrev/T Clausius inequality ……………eq. (33)
This is the fundamental requirement for any real transformation.

33. Second law of thermodynamics: properties of entropy
System + Surroundings = Universe
If earth is a system and space is its surroundings then the entire universe can be
thought of an isolated system. Because neither mass nor energy can be transported
in or out of the universe.
For any change of state in an isolated system, đQirrev = 0
So the Clausius inequality becomes dS > 0 for an isolated system
Therefore, the requirement for a real transformation in an isolated system is that
The entropy change must be positive or entropy must increase. This is true for any
natural change.
The universe is an isolated system, so dS for universe > 0

34. Second law of thermodynamics: entropy and natural process
The entropy change of an isolated system attains the maximum value at en equilibrium
situation.
In reality, whether the entropy change of the universe will at all reach the maximum value or
the equilibrium, and if so, when is not known.
But entropy of the universe has been trying to attain a maximum value from its birth.
All the essence of first and second law of thermodynamics can be summed up by a
Statement:
“The energy of the universe is constant; the entropy strives to reach a maximum”
- The aphorism of Clausius

35. Interpretation of Entropy and Chemistry

36. Entropy and natural processes
From second law of thermodynamics (Clausius statement):
ΔS > 0 for any process that takes place by itself, naturally. In other words,
natural processes proceed in a direction where entropy change is positive.
O2
CO2
pushing car backwd.+ H2O
?
Now, we can understand why fuel is not produced by performing the above operation,
which is an unnatural process.
Natural processes are irreversible like the example above.
A (SA) B (SB)
In terms of entropy, if SB > SA, then formation of B or the fwd. direction is preferred
In terms of entropy, if SB < SA, then formation of A or the backwd. direction is preferred
If, if SB = SA, then neither direction is preferred. We call this situation an equilibrium
situation in terms of entropy (ΔS = 0).

37. Entropy and randomness:
Entropy is also interpreted as a measure of randomness or disorderdness or chaos.
less entropy
more entropy

38. Entropy and probability:
Imagine a person randomly tossing off bricks on the ground. Which situation would you
expect to see?
most improbable situation most probable situation
ordered array disordered pile
less entropy more entropy
A more probable event or arrangement is associated with higher entropy

39. ΔS increasing
Solid
most ordered state
Liquid
moderately ordered state
Gas
Least ordered state
Entropy and state of a substance:
For any substance, the entropy increases for its change of state from solid to gas

40. Entropy of a binary mixture
0 1x (mole fraction)
ΔS
In a binary mixture, the entropy change is maximum when the two components are present
in equal amount.
In a 50% – 50% mixture, the randomization or the disorderdness becomes maximum

41. Entropy change and some chemical reactions:
butane (C4H10) methane (CH4) + propene (C3H6)850o Cv
one reactant two products (net increase = +1)
More disorderdness in product side!
ΔS > 0
due to formation of one extra gas molecule (CO2), the ΔS is a large positive value!
3 H2C=CH2
3 molecules of ethylene
850o C
1 molecule of cyclohexane (C6H6)
ΔS = negative

42. Properties of entropy: mathematical equations
Partial differential: dS = (∂S/∂T)V dT + (∂S/∂V)T dV ……………..eq. (34)
S = S (T, V)
Partial differential: dS = (∂S/∂T)P dT+ (∂S/∂P)T dP ……………..eq. (35)
S = S (T, P)
Like any other state function, S can be expressed as function of T, V or T, P
Also, from eq. (32) dS = dU/T + (P/T)dV ……………………….eq. (36)
For ideal gas, eq. (36) becomes dS = (Cv/T)dT + (nR/V)dV ………… eq. (37)
v
vv
dS = dH/T – (V/T)dP …………… eq. (38) since dU = dH – PdV – VdPv
Under constant pressure, for any system
dS = dH/T = (CP/T)dT…………………………….. Eq. (39) at constant pressurev

43. Properties of entropy: Third law of thermodynamics
dS = (CP/T)dT
Under constant pressure:
∫ dS = ∫ (CP/T)dT
0 0
T T
or, ΔS = ST – S0 =
or, ST = S0 + ∫ (CP/T)dT
0
T
ST = ∫ (CP/T)dT
0
T
S0 is the entropy at absolute zero, if the temperature is taken in Kelvin scale
S0 is zero at absolute zero. Because at absolute zero, any matter is supposed to be
solid and hence perfectly crystalline or packed. There is absolutely no degrees of freedom
in those molecules. This statement is known as the Third law of thermodynamics.
v ……………. eq. (40)
If the pressure P = 1 atm, then ST = ST
o called as the standard enthalpy of a substance
The statement of third law of thermodynamics is also known as Planck Statement (1913)

44. Change of entropy in change of states:
Liquid Vapor
Isothermal condition at boiling point (Tb)
ΔS = ΔQrev/Tb = ΔHvap/Tb
(under constant pressure)
Boiling of water at Tb
Solid Liquid
Isothermal condition at melting point (Tm)
ΔS = ΔQrev/Tm = ΔHfusion/Tm
(under constant pressure)
Melting of ice at Tm
ΔHvap is the heat of vaporization
ΔHfusion is the heat of fusion or melting

45. Properties of entropy: standard entropy
The standard entropy of a liquid at a temperature above its melting point:
So
T =∫ (Cp
o (s)/T)dT
0
Tm
∫ (Cp
o (l)/T)dT
Tm
T
+ ΔHO
fusion /Tm +
The standard entropy of a gas at a temperature above its boiling point:
So
T =∫ (Cp
o (s)/T)dT
0
Tm
∫ (Cp
o (l)/T)dT
Tm
Tb
+ ΔHO
fusion /Tm + + ΔHO
vap /Tb + ∫ (Cp
o (g)/T)dT
Tb
Tm
………. eq. (41)
…. eq. (42)
Substance So/R (at 298 K)
C (diamond) 0.286
C (graphite) 0.690
Hg 9.129
H2O 8.41
H2 15.7
CO2 25.699
Standard entropies of few substances at 298 K:

46. Free energy and Spontaneity

47. Reversibility and transformations:
From our knowledge of thermodynamics so far, we learnt:
For irreversible processes, the Clausius inequality tells us
TdS > đQ irreversible
Natural processes take place by itself in a spontaneous manner, these processes
are irreversible in nature and we have seen that for such processes, dS > 0
Reversible processes consists of infinitesimal reversible steps, where in each step
condition of equilibrium is valid.
For such reversible processes, dS = đQrev/T = 0 since, Qrev = 0 for isolated system
So condition for equilibrium in terms of entropy is
dS = 0 or TdS = đQ rev
So condition for spontaneous transformation in terms of entropy is
dS > 0 or TdS > đQ irreversible

48. Condition for equilibrium and spontaneity:
Therefore, combining the two conditions in previous slide, we get:
TdS ≥ đQ (for any transformation reversible or irreversible)
Where TdS = đQ (equilibrium)
TdS > đQ (spontaneity)
………………… eq. (43)
For isolated system, we can take đQ = 0
However, the condition dS > 0 is not enough to predict the whether a chemical
reaction would be spontaneous!
NH3 (g) + HCl (g) NH4Cl (s) ΔSo = - 284 JK-1mol-1
Although the change of entropy is a large negative value, the above reaction is
spontaneous!
In the remaining portion of this section, we will try to learn what is the state function
other than the entropy that predicts the condition of spontaneity and equilibrium.
at 298 K

49. Gibb’s energy or free energy (ΔG):
Let’s consider transformations at constant temperature and constant pressure:
-dU + TdS ≥ (-đW)
-dU + TdS ≥ PdV + đWa (where Wa is the non-expansion work)
-dU - PdV + TdS ≥ đWa
-dU - dPV + dTS ≥ đWa (under constant temperature and pressure)
-d(U + PV - TS) ≥ đWa (under constant temperature and pressure)
-dG ≥ đWa (Where G ≡ U + PV –TS, a state function, known as Gibbs energy/Free energy )
-ΔG ≥ Wa ………………….. eq. (44)
-ΔG = Wa (reversible) (condition for equilibrium)
-ΔG > Wa (condition for spontaneity)
For any transformation at constant temp. & pressure, the decrease of Gibbs energy is
greater than equal to the maximum non-expansion work available over and
above the expansion work in that transformation.
TdS ≥ đQ from eq. (43)

50. Gibb’s energy or free energy (ΔG):
-ΔG = 0 or ΔG = 0 (equilibrium) …………………………………………………………………………. eq. ( 45)
-ΔG > 0 or ΔG < 0 (spontaneity, transformation is in natural direction) …………….. eq. (46)
A real process, where we encounter the non-expansion work (Wa):
Cu2+ (aq) + Zn (s) = Zn2+ (aq) + Cu (s) (electrochemical reaction in Daniell cell)
The electrochemical work done in a battery Ξ Wa
For any other process, where Wa = 0, the condition for equilibrium and spontaneity becomes:
and obviously,
ΔG > 0 (non-spontaneity, the transformation is not in natural direction) ………….. eq. (47)
Earlier, we have seen: G ≡ U + PV –TS = H - TS
Therefore, we can write: ΔG = ΔH – TΔS ………….. eq. (48)
dG = dH – TdS ………….. eq. (49)

51. Free energy and spontaneity (ΔG):
All natural processes proceed in a direction in which a system’s free energy gets lowered.
For any spontaneous chemical reaction, ΔG = negative
ΔG = negative
NH3 (g) + HCl (g) NH4Cl (s)
ΔG = - ve
ΔG = + ve
ΔGo = - 92 kJmol-1
298 K, 1 atm
MgCO3 (s) MgO (s) + CO2 (g)
ΔGo = + 65 kJmol-1
298 K, 1 atm
ΔGo = standard free energy, or ΔG at 298 K and under 1 atm pressure.
A non-spontaneous reaction at 298 K, may become spontaneous at higher temperature!
Under standard condition, we can write:
ΔGo = ΔHo – TΔSo …….. eq. (50) bridging equation between first and second
law of thermodynamics

52. C (diamond) C (graphite)
But we have never seen diamonds turning into graphite not even in thousand years!
ΔGo = negative does not guaranty that a reaction will take place
spontaneously in standard condition!
Thermodynamics is unable to answer such anomaly.
That’s why we have to learn Chemical Kinetics later.
Free energy and spontaneity (ΔG):
ΔGo = - 2.9 kJmol-1
The free energy change of above transformation is negative under standard condition!

53. Properties of ΔG and Gibbs fundamental equation:
Since G is a state function and G = G (T, P), we can write
dG = (∂G/∂T)P dT+ (∂G/∂P)T dP ……………..eq. (51)
dG = dH – TdS - SdT (T & P are variable here)
Also, by definition,
= dU + PdV + VdP – TdS – SdT (H = U + PV)
= TdS + VdP – TdS – SdT ( dU + PdV = đQ = TdS assuming reversible process)
= – SdT + VdP
Above equation explains the dependence of G on T & P when both of these vary
G = H - TS
dG ……………..eq. (52)
This equation (52) is known as Gibbs fundamental equation as it relates
the change of free energy with the two most natural variable on earth: T and P.

54. Dependence of ΔG and on P and T:
dG = (∂G/∂T)P dT+ (∂G/∂P)T dP ……………..eq. (51)
= – SdT + VdPdG ……………..eq. (52)
Comparing the above two equations, we get:
(∂G/∂T)P = - S
(∂G/∂P)T = V
Since entropy of a substance is positive,
increase of T at constant P will decrease the free energy of a substance
……………..eq. (53)
……………..eq. (54)
Since volume of a substance is positive,
increase of P at constant T will increase the free energy of a substance
S gas > S liquid > S solid
At constant pressure
V gas > V liquid > V solid
At constant temperature
Increasing T at constant P would lead to maximum decrease of free energy for gas!
Increasing P at constant T would lead to maximum increase of free energy for gas!

55. Chemical potential and
chemical equilibrium

56. Free energy change of a pure ideal gas:
Gas
n moles,
P = po = 1 atm pressure,
temperature T = 298 K
Gas
n moles,
P = p atm,
temperature T = 298 K
Change of pressure
under constant temp.
= – SdT + VdPdG ……………..from eq. (52)
= VdP (under constant temperature)
Therefore, for the above change, we can write:
∫dG = ∫VdP = ∫nRTdP/P = nRT∫dlnp = nRTln(p/po)
po
p
po
p p
po
po
p
or G – Go = nRTlnp
or (G/n) – (Go/n) = RTlnp ……………..eq. (53)
It is customary to introduce a symbol μ, to replace free energy per mole (G/n)
Therefore, μ – μo = RTlnp (here μ is called molar Gibbs energy) ……..eq. (54)

57. Chemical change and free energy:
Chemical reactions generally occur under constant temperature and
constant ambient pressure!
= – SdT + VdP = 0 for any chemical reaction at constant T & P?dG
Does that mean
Definitely this is not the case because, in any chemical reaction, the composition
of the substances (reactants, products, etc) vary!
Therefore G = f (T, P, composition of substances)
Because, a chemical reaction is a mixture of varying composition
Varying composition means:
mole numbers of the constituents (n1, n2, n3,………..) vary
So we can write that overall G is a function of T, P and mole numbers

58. Partial molar Gibbs energy (chemical potential):
For a pure substance or for a mixture of fixed composition, G = G (T, P)
If the mole numbers, n1, n2,…., of the substances in a composition vary
then G = G (T, P, n1, n2…..)
So we can now write,
dG = (δG/δT)P,ni dT + (δG/δp)T,ni dp + (δG/δn1)T,P,nj dn1 + (δG/δn2)T,P,nj dn2 ……,
under constant temperature and pressure,
= (δG/δT)p,ni dT + (δG/δp)T,ni dp + ∑i(δG/δni)T,p,nj dni
dG = ∑i(δG/δni)T,p,nj dni
= ∑i(µi)T,p,nj dni
µi = (δG/δni)T,p,nj
μi ≡ chemical potential of ith component, Joules per mole
(nj is the mole number
other than ni)
Ξ partial molar Gibbs energy
……..eq. (55)
……..eq. (56)
= ∑iµidni

59. Chemical potential is an intensive property:
For a mixture of substance of varying composition (at constant T & P):
µi = (δG/δni)T,p,nj = chemical potential/partial molar Gibbs energy
For a pure substance (at constant T & P):
µ = (G/n) = molar Gibbs energy
G = ∑iniμi ……..eq. (58)
G = nμ ……..eq. (57)
Overall for a varying composition, T and P
dG = -SdT + VdP + ∑i µi dni ……..eq. (59)
Gibbs energy (Joules) is a extensive property, depends on number of moles
Chemical potential (J mol-1) is an intensive property independent of number of moles.

60. Chemical equilibrium in a reaction mixture:
Consider the following balanced reaction at constant T and P:
MnO2 + 4HCl → MnCl2 + 2H2O + Cl2
When ζ mole of reaction is completed:
MnO2consumed = no - ζ HCl consumed: no - 4ζ
MnCl2 produced = no + ζ H2O produced = no + 2ζ Cl2 produced = no + ζ
Let no be the initial number of moles of each reactant or product
ζ is called the advancement of the reaction.
Let’s imagine that in a reaction after a certain time if ni is the number of mole
of any reactant or a product present at that time then we can write:
ni = no
i + νiζ (νi is positive for products and negative for reactants)
At constant T and P, in a reaction mixture, for varying composition, we know:
dG = ∑i μi dni
dni = νi dζ (differentiating)
= ∑i (μi νi)dζ i (δG/δζ)T, P = ∑i μi νi ……..eq. (61)or,

61. Chemical equilibrium in a reaction mixture:
The partial derivative: (δG/δζ)T, P
denotes rate of change of free energy
with advancement of reaction.
(δG/δζ)T, P < 0 reaction moves fwdIf
(δG/δζ)T, P > 0 reaction moves bckwdIf
(δG/δζ)T, P = 0 equilibrium conditionIf
(δG/δζ)T, P, eq = ∑i μi νi, eq = 0
At equilibrium,
This partial derivative commonly called
as ΔGreaction
Therefore, at equilibrium:
ΔG reaction = ∑i μi νi, eq = 0 ……..eq. (62)

62. Gibbs energy of the reaction with extent of reaction:
(δG/δζ)T, P slope of G vs ζ plot at this point!
ζ

63. Chemical equilibrium in a reaction mixture of ideal gases:
μ = μo + RTlnp
For a pure ideal gas at 298 K:
For a mixture of ideal gases at 298 K, the chemical potential (μi) of the i th gas:
μi = μo
i + RTlnpi
(where pi is the partial pressure of the ith gas
and μo
i is the chemical potential of the ith gas in pure form)
……..from eq. (54)
Lets consider the following reaction mixture:
αA (g) + βB (g) → γC (g) + δD (g)
ΔG = ∑i μi νi
We know (from eq. 62) that the ΔG for the reaction:
= γμo
C + γRTlnpC + δμo
D + δRTlnpD - αμo
A - αRTlnpA - βμo
B - βRTlnpB
= (γμo
C + δμo
D - αμo
A - βμo
B) + RTln (pc
γ pD
δ/pA
α pB
β)ΔG
= ΔGo + RTln (pc
γ pD
δ/pA
α pB
β)ΔG ……..eq. (63)

64. Chemical equilibrium in a reaction mixture of ideal gases:
αA (g) + βB (g) → γC (g) + δD (g)
= ΔGo + RTln (pc
γ pD
δ/pA
α pB
β)ΔG
Here ΔGo denotes the standard Gibbs energy of the reaction
= ΔGo + RTlnQpΔG Qp is known as proper quotient of pressure
At equilibrium, we know: ΔG = 0
= ΔGo + RTln0
(pc
γ)eq (pD
δ)eq
(pA
α)eq (pB
β)eq
So
ΔGo = - RTln
(pc
γ)eq (pD
δ)eq
(pA
α)eq (pB
β)eq
Or,
Or,
ΔGo = - RTlnKp
……..eq. (64)
……..eq. (65)
Kp ≡ quotient of equilibrium partial pressure
≡ pressure equilibrium constant

65. Expressions of equilibrium constants:
αA (g) + βB (g) → γC (g) + δD (g)
Ideal gas:
Kp =
(pc
γ)eq (pD
δ)eq
(pA
α)eq (pB
β)eq
αA (aq) + βB (aq) → γC (aq) + δD (aq)
Ideal solution (infinitely dilute solution):
Kc =
(ac
γ)eq (aD
δ)eq
(aA
α)eq (aB
β)eq
ai: thermodynamic activity of the reactants/products
Kc ≡ concentration equilibrium constants
Non-ideal solution (example: ionic solutions):
Kc =
(ac
γ)eq (aD
δ)eq
(aA
α)eq (aB
β)eq
=
(cc
γ)eq (cD
δ)eq
(cA
α)eq (cB
β)eq
ai = ci. fi
x
(fc
γ)eq (fD
δ)eq
(fA
α)eq (fB
β)eq
(c stands for concentration & f stands for
activity coefficient. For ideal solution, f = 1)
Here f ≠ 1
μi = μo
i + RTlnpi
μi = μo
i + RTlnai
μi = μo
i + RTlncifi
……..eq. (66)
……..eq. (67)

66. Phase equilibrium and phase rule

67. = – SdT + VdPdG
The fundamental equation of Gibbs energy states:
Dividing by the mole numbers (n) in both sides, we get:
for a pure substance (fixed composition)
= – SdT + VdPdμ – – (since μ = G/n)
Under constant pressure: = – SdTdμ –
–
(δμ/δT)P = - S ……..eq. (68)
The third law of thermodynamics tells us that a substance’s entropy will always be positive
If the ambient temperature is greater than 0 K
So (δμ/δT)P = always negative
S gas >> S liquid > S solid
At constant pressure
At constant pressure
(δμ/δT)P of gas has large negative value
Chemical potential versus T at constant pressure:

68. Phase diagram in terms of μ:
The partial derivative: (δμ/δT)P represents the slope of μ versus T plot
This is a qualitative phase diagram for any
Substance in its solid, liquid and gaseous form.
At melting point (Tm):
μ solid = μ liquid
Two phases are at equilibrium
At boiling point (Tb):
μ liquid = μ gas
Two phases are at equilibrium
T1 T2
At any T, the phase having lowest μ
is the most stable phase!
Solid at T1 and at gas at T2 are the most stable
phases at a constant pressure, respectively.

69. Pressure dependence of μ versus T curves:
Under constant temperature: = VdPdμ –
–
(δμ/δP)T = V ……..eq. (69) V gas >> V liquid > V solid
At constant T
If pressure is decreased, μ is decreased at constant T. The decrease is greatest for gas
and least for solid. Below: μ versus T curves at lower pressure (dashed line) and higher
pressure (solid line)
At a considerable lower pressure,
the dashed Line of gas may cross
the melting point by shifting left!
That is the situation
for sublimation!

70. Pressure dependence of μ versus T curves:
Ts stands for sublimation temperature
At sublimation point (TS):
μ solid = μ gas
μ
T
S
L
G
At a certain T and P
all three lines may intersect
at a common point.
μ solid = μ liquid = μ gas
This temperature is known as
triple point
Tt

71. Generalized phase diagram: Clapeyron equation
It is bit difficult to ascertain μ of any substance to draw the μ vs T phase diagrams.
It is rather easy to draw P vs T phase diagram of any substance.
Clapeyron equation helps us to do that.
Derivation: Lets assume that α and β phases are in equilibrium with each other
α β
μα (T, P) = μβ (T, P)
(at temperature =T and at pressure = P)
Assuming the equilibrium condition is still fulfilled if the temperature and pressure changes
infinitesimally to T + dT and P + dP, respectively. In that condition, the chemical potential
is assumed to change by an amount dμ.
Therefore,
μα (T, P) + dμα = μβ (T, P) + dμβ
(at temperature =T + dT and at pressure = P + dP)
dμα = dμβ
– SαdT + VαdP =
– – – SβdT + VβdP
– –
or,
or,

72. Generalized phase diagram: Clapeyron equation
– SαdT + VαdP =
– – – SβdT + VβdP
– –
Rearranging, we have
(Sβ – Sα)dT = (Vβ – Vα)dP
––– –
If the transformation is taken as α → β, then ΔS = Sβ – Sα and ΔV = Vβ – Vα
Therefore the above equation can be written as:
dP/dT = ΔS/ΔV ……..eq. (70A) This is known as Clapeyron equation
The equation denotes dependence of equilibrium pressure on Temperature
dT/dP = ΔV/ΔS ……..eq. (70B) This is also Clapeyron equation
This equation denotes dependence of equilibrium temperature on pressure
Conveniently, using eq. (70A), we will be able to draw phase diagrams by plotting
equilibrium pressure vs. temperature.

73. P vs T phase diagram: solid-liquid equilibrium
solid liquid
Lets consider the following transformation
ΔS = Ssolid – Sliquid = ΔSfus = ΔHfus/T = +ve (always)
– –
ΔV = Vsolid – Vliquid = ΔVfus = small +ve (mostly)
= small -ve (for water)
– –
(at equilibrium)
Please note that, dP/dT (for solid → liquid) is a large constant value either +ve or -ve
T
P
T
Psolid
liquid
solid
liquid
for any s → l for ice → water
locus of all points
at which solid & liquid
coexist at eqm.

74. P vs T phase diagram: liquid-gas equilibrium
liquid gas
Lets consider the following transformation
ΔS = Sliquid – Sgas = ΔSvap = ΔHvap/T = +ve (always)
– –
ΔV = Vliquid – Vgas = ΔVvap = big +ve (always)
– –
(at equilibrium)
dP/dT is a small +ve value but the resulting plot is not a straight line, it is a curve!
solid
liquid
gas
solid, liquid and
gas coexists at
equilibrium

75. P vs T phase diagram: solid-gas equilibrium
solid gas
Lets consider the following transformation
ΔS = ΔHsub/T = +ve (always) ΔV = ΔVsub = big +ve (always)
(at equilibrium)
since ∆Vvap ≈ ΔVsub and ΔHsub > ΔHvap
dP/dT is a small +ve value, resulting plot is a curve but bit steeper than s-l curve
near the triple point!
General phase diagram
of a pure substance
triple point
limit at critical pressure
and temperature point
P = 1 atm
normal boiling point

76. Phase diagram of few common substances:
CO2 H2O

77. The phase rule:
μα (T, P)= μβ (T, P)
The coexistence of two phases requires the following
That means here T and P are not independent, but related to each other.
If we know T, P is automatically fixed and vice versa.
So we say that this system has only one degrees of freedom (F) either T or P
At triple point:
μα (T, P)= μβ (T, P) = μγ (T, P)
This happens at a particular T and P, which are fixed and unique for a pure substance
So in this case, degrees of freedom (F) = 0
For any phase (not at equilibrium), F = 2 (T as well P)
There is a general rule by which we can predict the degrees of freedom
F = 3 - P ……..eq. (71)
where P stands for
number of phases
Number of phases present 1 2 3
Degrees of freedom 2 1 0
(three phases coexist)