THIRD LAW OF THERMODYNAMICS
B
dq rev
defines only changes in entropy and
T
A
The definition of entropy: S A B  
not the absolute entropy itself. The question of absolute entropy is dealt-with by
the third law of thermodynamics.
Statement of third law: “The entropy of perfect crystals of all pure elements
and compounds is zero at the absolute zero of temperature.”
The third law allows us to assign absolute entropy to each element and
compound at any temperature T.
For instance for constant pressure processes,
dq p
dT
dqp  C p dT
Cp 
and
S 
 dS  
S  ST  S0 
T

0
dq p

T
T

0
C p dT
T
C p dT
T
T
therefore S  ST  
0
C p dT
since S0 = 0.
T
Suppose that we are interested in the absolute entropy at temperature T ( ST0 ) of a
substance that melts at some temperature, Tm, which is less than T, then, the
entropy associated with this phase transition must be included in the calculation

T
of the absolute entropy. S 
Tm

0
C p (s)dT H fus T C p (l)dT

 
T
Tm
T
Tm
Where Cp(s) and Cp(l) are the heat capacities of the solid and the liquid
respectively and ∆Hfus is the enthalpy of fusion of the substance.
If any other phase change such as vaporization occurs between 0 K and
temperature T, its contribution to the entropy must be included in a similar
fashion.
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The third law entropy (the absolute entropy) of a substance at temperature T,
which is above its boiling point, Tb, is given by:

T
S 
Tm

0

T
C p (s)dT H f Tb C p (l)dT Hvap
C (g)dT



 p
T
Tm
T
Tb
T
Tm
Tb
The entropy change of the reaction
The absolute entropies of elements and compounds are tabulated. It is therefore
possible to calculate the entropy changes accompanying chemical reactions by
using the tables of absolute entropies. For the general reaction:
aA + bB
cC +dD
∆Sº = cSº(C) +dSº(D) -aSº(A) -bSº(B)
= ∑Sº(products) -∑Sº(reactants)
This procedure for calculating the entropy change of a reaction is, of course,
similar to the procedure used to find ∆Hºreact from the enthalpies of formation.
The Gibbs free energy (and the Helmholtz free energy)
We saw from the second law of thermodynamics that the general criterion for
spontaneity (irreversibility) of a chemical or physical process is:
Stot  Ssys  Ssurr  0
while for reversible (equilibrium) process is: Stot  Suni v  0.
This criterion (i.e. the ∆Suniv criterion) is inconvenient because it considers both
the system and the surroundings (the whole environment outside the system). It
is inconvenient to consider changes in the surroundings.
We will now derive a spontaneity/equilibrium criterion based on changes in
thermodynamic functions of the system alone.
From the second law of thermodynamics
dStot  dSuni v  dSsys  dSsurr  0
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If the system absorbs an infinitesimal amount of heat dq from the surroundings,
then dqsurr = -dqsyst and
dSsurr =
-dqsyst
T
 dStot  dSsys 
dqsys
0
T
If the process is taking place at constant pressure and constant temperature, then
dqsys = dHsys
 dStot  dSsys 
dH sys
0
T
We have thus, obtained dStot expressed only by the properties of the system.
dSsys 
dHsys
0
T
or
TdSsys  dH sys  0
On rearranging we get: dHsys  TdSsys  0
We define now, Gibbs free energy G of the system as G = H – TS
Since H, T, and S are all state functions, then G must be a state function.
dGsys = dHsys – TdSsys ≤ 0(at constant T and P) and on omitting the subscript
“sys”, dG = dH – TdS ≤ 0
or ∆G = ∆H – T∆S ≤ 0
Therefore
∆G = ∆H – T∆S = 0 for a reversible process,
∆G = ∆H – T∆S < 0 for a spontaneous process and
∆G = ∆H – T∆S > 0 for a non-spontaneous process.
We see now that the tendencies of the energy of the system to decrease and the
entropy of the system to increase, (when both can change) is what determine the
spontaneity of a physical or chemical process.
Spontaneity of Chemical Reactions (at constant pressure):
To decide whether a certain reaction is spontaneous or not, we have only to
evaluate the accompanying Gibbs free energy change from either the enthalpy
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and the entropy of reaction, i.e. ∆rGº = ∆rHº – T∆rSº
where
all
the
thermodynamic quantities are evaluated at a single temperature T, or from the
tabulated Gibbs energies of formation i.e. ∆rGº = ∑∆fGº(products) ∑∆fGº(reactants).
Spontaneity of Chemical Reactions (at volume):
The spontaneity of chemical reactions at constant volume is dictated by the
Helmholtz function (Helmholtz free energy) A = U –TS and the criterion is:
∆A = ∆U –T∆S < 0 at constant temperature.
Examples:
1. The reaction for the extraction of iron from its iron III oxide ore by reduction
with hydrogen gas is: Fe2O3(s) + 3H2(g) 2Fe(s) + 3H2O(l).
Use the table of thermodynamic data to calculate ∆rSº(298 K) and ∆rSº(370 K).
2. Liquid water is in equilibrium with water vapor at 1 atm pressure. If the
enthalpy change associated with the vaporization of liquid water at 100ºC is
40.6 kJ/mol, what are ∆G and ∆S for the vaporization process?
3. The values of ∆H and ∆S for a chemical reaction are –85.2 kJ/mol and –170
JK-1 mol-1, respectively, and the values can be taken to be independent of
temperature. (a) Calculate ∆G for the reaction at 300 K. (b) At what
temperature would ∆G be zero?
4. Calculate the change in the thermodynamic quantities G, S and H for the
mixing of 0.5 mol of oxygen with 0.5 mol of nitrogen at 25ºC and 1 atm
pressure; Assuming that the gases behave ideally.
J mol-1
∆S = 5.763 JK-1 mol-1
Answers: ∆G = -1718
∆H = 0
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