STUDY UNIT 4:
THERMODYNAMICS AND
EQUILIBRIUM
1
Content
1.
2.
3.
4.
Spontaneous Reaction.
Entropy.
The second law of Thermodynamics.
Gibbs Free Energy.
Spontaneous Process
A process that occurs of its own accord, without any ongoing outside
intervention is called a spontaneous process.
A ball will roll downhill
spontaneously. It will
eventually reach
equilibrium at the bottom.
A ball will not roll
uphill
spontaneously. It
requires work.
Entropy, S, is a thermodynamic quantity that
is a measure of how dispersed the energy of a
system is among the different possible ways
that system can contain energy.
ENTROPY (S) is measure of extent of disorder
for a system
The larger the entropy, the more disordered
is the system
Any events that is accompanied by an
increase in the entropy of the system will
have a tendency to occur spontaneously.
An increase in freedom of molecular motions
corresponds to an increase in entropy
Example:
Volume: For gases, the entropy increases as
volume increases
Temperature: As temperature increases, the
entropy increases
Physical state: Entropy increases as a
substance goes from the solid to liquid to vapor
6
Reversible and Irreversible Processes
 In a reversible process, a system is changed in such a
way that the system and surroundings can be restored
to their original state by exactly reversing the change.
 An irreversible process is one that cannot simply be
reversed to restore the system and its surroundings to
their original states.
irreversible process
Reversible and Irreversible Processes
 The reverse of any spontaneous process is a nonspontaneous process.
 A non-spontaneous process can occur only if the
surroundings do work on the system;
 Therefore, any spontaneous process is irreversible.
 Because even if we return the system to the original
condition, the surrounding will have changed.
Entropy Change (ΔS)
 Change in entropy, ΔS, in a system depends only on the initial and final states
of the system and not on the path taken from one state to the other.
S  Sf inal - Sinitial
 For the special case of an isothermal process, ΔS is equal to the heat that
would be transferred if the process were reversible, qrev, divided by the
temperature at which the process occurs:
qrev H process
S 

T
T
(Constant T)
 Because S is a state function, we can use this equation to calculate ΔS for any
isothermal process, not just those that are reversible.
Example. The elementary mercury, Hg, is a
silvery liquid atom at room temperature. The
normal freezing point of mercury is -36.9 ºC,
and its molar enthalpy of fusion is ∆Hfus =
2.29 kJ/mol. What is the entropy change (∆S
) of the system when 50.0 g of Hg (l) freezes
at the normal freezing point?
Remember:
Freezing is an exothermic process;
Heat is transferred from the system to the surroundings;
This means that q < 0 or q = negative (-).
Enthalpy of fusion, ΔHfus, of Hg means that;
Hg (s)
Hg (l)
ΔHfus = + 2.29 kJ/mol
This means that enthalpy of freezing, ΔHfre , of Hg is;
Hg (l)
Hg (s)
ΔHfre = - 2.29 kJ/mol
Solution:
We can use enthalpy of fusion and the atomic weight of Hg to
calculate q for freezing of 50.0 g Hg.
We first calculate the number of moles corresponding to 50g
𝑚𝑎𝑠𝑠
50.0𝑔
𝑛=
=
= 0.249 𝑚𝑜𝑙
𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 200.59𝑔/𝑚𝑜𝑙
Now, 1.00 moles of Hg produce 2.29 kJ for enthalpy of fusion
0.249 moles of Hg would produce x kJ for enthalpy of fusion
Solving for x we obtain x = -571 J
Convert the temperature to K, (-38.9+273.15)K =234.3 K
Solve for the value of the entropy change of the system:
qrev  571J
S 

 2.44 J .K 1
T
234.3K
The second Law of Thermodynamics:
In words:
The total entropy of the universe increases
in any spontaneous process.
In the equation form:
Suniv  S syst  S surr  0
Suniv  S syst  S surr  0
For reversible process
For irreversible process
For chemical reactions involving gases, the sign
of the entropy change is easy to predict
If ngas  0, then S  0
If ngas  0, then S  0
The number of particles is a major factor that
affects the sign of the entropy change for
chemical reactions
Examples:
Predict the algebraic sign of S for this
reactions:
2NO2(g)
C3H8(g)
N2O4(g)
3CO2(g)
+
4H2O(g)
The standard entropy of a reaction is the molar
entropy of the system obtained from the
standard molar entropies denoted by Sº.
And the change in standard entropy ∆Sº can
be obtained using the following relation:
S   nS (products)   mS (reactants )
o
o
o
n and m are the coefficients in the chemical
equation.
Calculating ∆S from tabulated entropies:
Calculate ∆S for the synthesis of ammonia
from N2 (g) and H2 (g) at 298 K.
N2 (g) + 3 H2 (g)  2 NH3 (g)
(the standard entropies of N2, H2 and NH3
gases are 191.5 J/mol, 130.6 J/mol and 192.5
J/mol, respectively – data from tables)
N2 (g) + 3 H2 (g)  2 NH3 (g)
SOLUTION:
S   nS (products)   mS (reactants )
o
o
o

 
S 0  2(192.5J .mol 1 )  191.5J .mol 1  3(130.6 J .mol 1 )

 
S 0  385J .mol 1  583.3J .mol 1 )
S 0  198.3 J .mol 1


Entropy Change for a Reaction:
Entropy usually increases in three situations:
1. A reaction in which a molecule is
transformed into two or more smaller
molecules.
2. A reaction in which there is an increase in
the number of moles of a gas.
3. A process in which a solid changes to a
liquid or gas or a liquid changes to a gas.
Predict the sign of S° for the following reaction.
Ba(OH)2  8H2O(s) + 2NH4NO3(s) 
2NH3(g) + 10H2O(l) + Ba(NO3)2(aq)
3 moles of reactants produces 13 moles of
products. Solid reactants produce gaseous,
liquid, and aqueous products.
S is positive.
Gibbs Free Energy (G)
G = H - TS
Change in Gibbs Free Energy
∆G = ∆H - T∆S
This can also be written as
∆G = Gfinal – Ginitial
= ∆Hfinal - T∆Sfinal – ∆Hinitial - T∆Sinitial
Standard free energy change (Go)
G  H  (298.15 K)S
o
o
o
at 25 oC and 1 atm
Standard states:
Gas
- pressure of 1 atm & temperature of 25C
Solid
- pure substance
Liquid - pure liquid
Solution - Concentration = 1M
Example: Calculate ΔGo for the reaction of urea
with water from values of ΔHo and ΔSo.
CO(NH2)2(aq) + H2O(l)
CO2(g) + 2NH3(g)
Substance
∆Hºf
(kJ/mol)
Sº
(J.mol-1 K)
CO2 (g)
-393.5
197.9
H2O (l)
-285.9
69.91
NH3 (g)
-46.19
192.5
CO(NH2)2 (aq)
-319.2
158.19
Solution: Calculate ΔHo and ΔSo from the data Table.
CO(NH2)2(aq) + H2O(l)
CO2(g) + 2NH3(g)
 Horxn   n p Hof (products)   n r Hof (reactants )
ΔHo = [ΔHof CO2(g) + 2 ΔHof NH3(g) ] –
[ΔHof
CO(NH2)2(aq)
+ ΔHof H2O(l) ]
= [1 mol x (-393.5 kJ/mol) + 2 mol x (-46.19
kJ/mol ] – [1 mol x (-319.2 kJ/mol) + 1 mol x
(-285.9 kJ/mol)]
= (-485.9 kJ) – (-605.1 kJ) = +119.2 kJ
CO(NH2)2(aq) + H2O(l)
CO2(g) + 2NH3(g)
So   nS o (products)   mS o (reactants )
ΔSo = [ΔSo CO2(g) + 2 ΔSo NH3(g) ] – [ΔSo
CO(NH2)2(aq)
+ ΔSo H2O(l) ]
= [1 mol x (197.9 J/mol.K) + 2 mol x (192.5
J/mol.K ] – [1 mol x (158.19 J/mol.K) + 1 mol
x (69.91 J/mol.K)]
= (582.9 J/K) – (228.1 J/K)
= 354.81 J/K, which is equivalent to 0.3548kJ/K
Since Go = Ho – TSo
At standard state Temperature of 25 C, we write
Go = +119.2 KJ – (298.15 K)(0.3548 kJ.K-1)
= +119.2 KJ – 105.8 kJ
Therefore for this reaction Go = +13.4 kJ
The reaction is non-spontaneous
Standard Free Energies of Formation, Gf°
The standard free energy of formation is
the free-energy change that occurs when 1
mol of substance is formed from its
elements in their standard states at 1 atm
and at a specified temperature, usually
25°C.
The corresponding reaction for the
standard free energy of formation is the
same as that for standard enthalpy of
formation, Hf°.
Calculating standard free energy change for a
reaction:
Gfo = 0
kJ/mol
For elements in their standard states
Tabulated data of Gfo can be used to
calculate standard free energy change for
a reaction as follows:
o
G 
o
 nGf (products)

o
 mGf (reactants )
G° as a Criterion for Spontaneity
The spontaneity of a reaction can now be
determined by the sign of G°.
G° < 0 kJ [Negative (-) ]: Spontaneous
G° > 0 kJ [Positive (+) ]: Non-spontaneous
G° = very small or zero (< + 0 kJ and > – 0 kJ)
: At equilibrium