This document discusses spontaneous processes and the driving forces behind them in thermodynamics. It explains that spontaneous processes are driven by a decrease in enthalpy or an increase in entropy. While enthalpy change alone cannot predict spontaneity, the introduction of entropy and Gibbs free energy allows better determination of spontaneous processes. The document also discusses how temperature, entropy change, and the relationship between Gibbs free energy and equilibrium constant can be used to analyze spontaneity.

1. Chemical
thermodynamics

2. Spontaneous Process:
In Thermodynamics sense, a process that occurs by itself without the help of
external agent is a spontaneous process. All naturally occurring process is
spontaneous process.
Examples:
1. Rolling down of a stone downhill
2. Melting of ice
3. Flowing of water from higher to lower level
4. Vaporization of water
5. Burning of fuel
6. Mixing of gases
7. Natural radioactivity, etc.

3. Characteristics of Spontaneous Process:
1. All spontaneous process are unidirectional in nature and proceed
till the process is completed or a state of equilibrium is reached
2. Some spontaneous process can be reversed by applying external
agency.
3. Some spontaneous process requires initiations.

4. What is the driving force for spontaneous of a
process?
In the early development of thermodynamics energy change was considered
as criteria for spontaneity a process occurs with decrease in energy is
spontaneous. Example. flowing of water from higher to lower level, rolling
down of stone downhill etc. In case of chemical reaction, the reaction that
occurs with decrease in enthalpy i.e. exothermic is spontaneous like burning
of Methane, neutralization reaction, etc. On the basis of enthalpy change,
∆H = -ve = exothermic = spontaneous
∆H = +ve = endothermic = non-spontaneous
∆ H = O = equilibrium

5. However many endothermic reaction were found to be spontaneous. Example,
i. Melting of ice;
H2O (s) → H2O (l) ∆H = +6 KJ
ii. Vaporization of water
H2O (l) → H2O (g) ∆H = +40.8 KJ
iii. Dissociation of NH4NO3 in water,
NH4NO3 (s) + H2O (l) → NH4
+ (aq) + NO3
– (aq) ∆H = +25KJ
These examples show that enthalpy change alone is not an enough criteria for spontaneity.
These lead to necessity of another thermodynamic term.
In all above examples, the molecular disorderness is increased, so the new
thermodynamic function should measure disorderness. This function is called entropy.

6. Entropy:
Entropy of a system is the measurement of disorderness or
randomness of the system. The greater the randomness, the higher is
the entropy. For a given substance, the crystalline solid state has lowest
entropy i.e. in solid, the molecules are arranged orderly so it has least
entropy but in liquid the disorderness is more than solid while in gas,
the disorderness is maximum. So the order of entropy of a substance in
different physical states, is,
Ssolid < Sliquid << Sgas

8. Entropy is a state function and depends only upon initial and final
states, thus change in entropy ( ∆S) will be
Δ S = S2 – S1
Where S2 = Final state entropy
S1 = Entropy of Initial state

9. In chemical reaction, the change in entropy is,
Δ Sreaction = Sproducts – Sreactants
= qrev, iso/T…………………..(i)
( simply Δ Sreaction = q/T)
i.e. entropy change during the process is defined as the amount of heat
(q) absorbed isothermally and reversibly (infinitesimally slowly) divided
by the absolute temperature (T)

10. Justifications:
1. When a system absorbs heat, the molecule start moving faster
because kinetic energy increases. Hence the disorder increases.
More the heat absorbed, greater is the disorder.
2. For the same amount of heat absorbed at low temperature, the
disorder is more than at higher temperature. Thus the presence of T
in the denominator is justified.
Unit
Its unit is Jmol-1K-1

11. Entropy change during phase transformation:
i. Entropy of fusion: It is change in entropy when 1 mole of solid change
into liquid at melting temperature.
Mathematically,
∆Sfusion = Sliquid - Ssolid
=
qrev
Tmelt
=
∆Hfusion
Tmelting
where, Sliquid = molar entropy of liquid
Ssolid = molar entropy of solid
∆Hfusion = enthaly of fusion
Tmelting = melting temperature

12. i. Entropy of vaporisation: It is change in entropy when 1 mole of liquid
change into vapours at boiling temperature.
Mathematically,
∆Svap = Svapour - Sliquid
=
∆Hvap
Tboiling
where, Sliquid = molar entropy of liquid
Svapour = molar entropy of vapour
∆Hvap = enthaly of vaporisation
Tboiling = boiling temperature

13. i. Entropy of sublimation: It is change in entropy when 1 mole of solid
change into vapour at particular temperature.
Mathematically,
∆Ssublimation = Svapour - Ssolid
=
∆H 𝑠𝑢𝑏𝑙𝑖𝑚𝑎𝑡𝑖𝑜𝑛
T
where, Sliquid = molar entropy of liquid
Ssolid = molar entropy of solid
T = particular temperature

14. Limitation of the first law of thermodynamics.
1. No restriction on the direction of the flow of heat: the first law
establishes definite relationship between the heat absorbed and
the work performed by a system. The first law does not indicate
whether heat can flow from a cold end to a hot end or not.
2. The information provided by the first law of thermodynamics are
not enough to predict the spontaneity or feasibility of a process.
To deal with spontaneity of the chemical process some new terms
viz., entropy and Gibbs free energy were introduced in
thermodynamics which lead to the formulation of second law of the
thermodynamics.

15. Second law of thermodynamics:
There are limitation of first law of thermodynamics. To overcome those
limitation and to explain the spontaneity of a process, second law of
thermodynamics is stated. It state that,
“ Whenever a spontaneous process takes place, it is accompanied by an
increase in the total entropy of universe.” Thus for a spontaneous process
∆Suniverse = ∆Ssys + ∆Ssurr > 0
According to second law, when an irreversible spontaneous process occurs,
the entropy of the system and surrounding increases. i.e. ∆Suniverse>0.
As we consider entire universe as a system, the second law can be stated as,
“ The entropy of the universe is constantly increasing.”

16. Explanation of second law of thermodynamics:
Second law of thermodynamics is explained here in the light of
entropy change. For this purpose consider spontaneous flow of heat
from a system at higher temperature to the surrounding at lower
temperature.
Let Tsystem and Tsurrounding are the temperature of the system and
surrounding respectively where Tsystem > Tsurrounding.

17. Let system evolves Q quantity of heat spontaneously to the surrounding.
Here, the system decreases its entropy which is given as
∆Ssystem = -
Q
Tsystem
On the other hand, the surrounding increases its entropy which is given as
∆Ssurrounding = +
Q
Tsurrounding
The net entropy of universe in the spontaneous flow of heat is
∆Suniverse = ∆Ssystem + ∆Ssurrounding
= -
𝑄
Tsystem
+
+Q
Tsurrounding

18. Since Tsystem > Tsurrounding, the numerical value of
Q
Tsystem
<
Q
Tsurrounding
Hence ∆Suniverse = positive or ∆Suniverse >0
Here the net entropy of the universe is greater than zero. Further the
total change of the system and surrounding = -Q + Q = 0
Hence, the net entropy of the universe is increasing whereas the total
energy of the universe is constant. This is second law of
thermodynamics.

19. Class work
How can you say that melting of ice is spontaneous process?
How can you say that solidification of ice is spontaneous process?

20. Entropy and equilibrium state:
Let us apply the entropy criteria to decide the feasibility of the process
of conversion of water to ice at 1 atmospheric pressure.
H20(l) → H2O(s)
The entropy changes for the three different temperature 272, 273 and
274 K are given in the table below.
1. At 272 K ;freezing is spontaneous because ∆Stotal is positive.
2. At 273 K ;∆Stotal is zero. Thus ice and water are in equilibrium and
no net change is observed.
3. At 274 K ;freezing is non-spontaneous because ∆Stotal is –ve, but
the reverse change i.e. melting is spontaneous.

21. Thus, from above, entropy criterion can be summarised up as follows:
∆Stotal > 0 process is spontaneous.
∆Stotal = 0 process is in equilibrium state.
∆Stotal < 0 process is non-spontaneous

22. Free energy of the system:
One of the defect of entropy, for the prediction of spontaneity is to
calculate the entropy of surrounding also. Our concern aspect is the
system and not the surrounding.
So, a new thermodynamic term has been introduced for the prediction
of spontaneity based on the system only and this new term is free
energy of the system.

23. Free energy:
This is another thermodynamic quantity that helps in predicting the
spontaneity of the process. It is usually denoted by ‘G’ and designed
mathematically by the equation
G = H - TS
Where, H = enthalpy of the system
S = entropy of the system
T = absolute temperature

24. Consider a system undergoing a change from state 1 to state 2 at
isothermal process i.e.
G1 = H1 - TS1 for initial state……………(i)
G2 = H2 - TS2 for final state……………..(ii)
Therefore, G2 - G1 = (H2 - H1) - T( S2 - S1)
∆G = ∆H − T∆S … … … … … (iii)
Equation(iii) is known as Gibbs-Helmoltz equation or simply Gibbs
energy equation.

25. Spontaneity in terms of free energy change:
It has already been seen that the total entropy
change, ΔStotal determines the spontaneity of a process. The total
entropy change during a process is given by
ΔStotal = ΔSsystem + ΔSsurrounding ……………………………(1)
Consider a process being carried out at constant temperature and
constant pressure. Suppose heat equal to q is lost by surrounding at
temperature T. The entropy change of the surrounding is given by

26. ∆Ssurrounding =
q
T
…………………………………………..(2)
Now heat (q) gained by system at constant temperature and pressure
represents its enthalpy change (∆H).Therefore ,
∆Ssurrounding =
q
T
=
−∆Hsystem
T
(since qp= ∆Hsystem)
Substituting this value in equation 1, we get
ΔStotal =ΔSsystem -
ΔHsystem
T
…………………(3)

27. Multiplying equation (3) by T we get,
TΔStotal = TΔSsystem - ΔHsystem
-TΔStotal = ΔHsystem - TΔSsystem
-TΔStotal = ΔGsystem……………………………(4)
We know,
If ΔStotal = positive, the process is spontaneous
If ΔStotal = negative, the process is non-spontaneous
If ΔStotal = 0 , the process is in equilibrium

28. Putting the results in (4), we get
If ΔG = negative, the process is spontaneous
If ΔG = positive, the process is non-spontaneous
If ΔG = 0, the process is in equilibrium

29. Free- energy change and useful work:
Gibb’s free energy is related with useful work. The work other than
mechanical work (PV-work) is called useful work or net work done. It is
denoted by Wuseful or Wnet.
From first law of thermodynamics, we have
q = ∆E + W
There are two types of work done by/on a system. One is mechanical
work i.e. expansion or contraction while the other is useful work i.e.
W = Wmechanical + Wuseful
or, W = P ∆V + Wuseful
Or, q = ∆E + P ∆V + Wuseful

30. Then,
q = ∆H + Wuseful …………..(i) since (∆H = ∆E + P ∆V)
Now,
For a change taking place under isothermal and reversible conditions,
the entropy change ∆S is given by
∆S=
q
T
Or, q = T ∆S…………………..(ii)

31. Substituting this value of q in equation(i) we have,
T ∆S = ∆H + Wuseful
- ∆H - T ∆S = Wuseful
- ∆G = Wuseful
Gibbs free energy is that thermodynamic quantity of a system that
decrease in whose value during the process is equal to the maximum
possible useful work done that can be obtained from the system.

32. Effect of temperature on spontaneous process:
The spontaneity of a process is explained on the basis of enthalpy, entropy and free energy
change. These are related to each other by means of Gibb’s Helmholtz equation i.e.
∆G = ∆H − T∆S
The value of ∆G at any temperature is calculated from the values of ∆H and ∆S, which in
turn is used to explain the spontaneity of a process. Depending upon the sign of ∆H
and ∆S, the effect of temperature on spontaneity of a process is explained as follows:

33. 1. If ∆𝐇 𝐢𝐬 − 𝐯𝐞 𝐚𝐧𝐝 ∆S is +ve.
The free energy change is always –ve . Hence, the process is
spontaneous at any temperature.
2. If ∆H and ∆S are positive, then
a. At low temperature, T ∆S is less than ∆H. So, ∆G will have positive
value. Therefore the reaction will be non spontaneous.
b. At high temperature, T ∆S is greater than ∆H. So, ∆G is negative.
Therefore the reaction is spontaneous.
3. When both ∆H and ∆S are negative, then
a. At low temperature, T ∆S is less than ∆H. Therefore ∆H term will
predominate and the value of ∆G is negative. Therefore the change will
be spontaneous.
b. At higher temperature, T ∆S is greater than ∆H. Under this
condition, ∆G are positive. Therefore, the reaction is non spomtaneous.

34. 4. If ∆H is positive and ∆𝐒 𝐢𝐬 𝐧𝐞𝐠𝐚𝐭𝐢𝐯𝐞. Then ∆G is positive and the
reaction is non-spontaneous.
5. If ∆H= T ∆S, then ∆G=0, the reaction is at equilibrium state.

36. Comment on the following statements:
a. An exothermic reaction is always thermodynamically
spontaneous.
Ans: It is due to following reasons:
1. They are accompanied by decrease of energy.
2. The heat evolved is absorbed by surrounding so that entropy of
surrounding increases and hence ∆Stotal is positive.
However according to Gibbs energy equation
∆G = ∆H - T∆S, if ∆Ssystem is negative and T is high. ∆G may be
positive and hence the reaction may be non spontaneous.

37. Standard free energy change:
The standard free energy change is defined as the free energy change
for a process at 298 K and 1 atmospheric pressure in which the
reactants in their standard states are converted to the products in their
standard states. Thus,
Δ G° = Σ Δ G °(products) - Σ Δ G °(reactants)
=[sum of the standard free energy in the formation of products ] –
[ sum of standard energy in the formation of reactants]

38. Relation between standard free energy change
and equilibrium constant:
The standard free energy change, Δ Go is related to the equilibrium
constant K by the relation,
Δ Go = - RT ln K
or ΔGo = –2.303 RT log K
Where, R= molar gas constant
T= kelvin temperature
K = equilibrium constant.