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CHEMICAL THERMODYNAMICS
 Spontaneous processes
 Enthalpy (H)
 Entropy (S)
 Gibbs Free Energy (G)
 Relationship between G and K
Brown et al., Chapter 4.9-4.15
CHEM171 – Lecture Series Three : 2012/01
SPONTANEOUS REACTIONS
HOW CAN WE TELL IF A REACTION WILL
PROCEED OR NOT?
Some chemical and physical changes take place by themselves,
given enough time.
A spontaneous chemical reaction is one that, given sufficient
time, will achieve chemical equilibrium, with an equilibrium
constant greater than 1, by reacting from left to right.
CHEM171 – Lecture Series Three : 2012/02
Cu(s) + Cl2(g)  CuCl2(s)
 spontaneous 
2H2(g) + O2(g)  2H2O(g)
spontaneous reaction but occurs only if you ignite the
mixture
O3(g)  O(g) + O2(g)
 nonspontaneous 
This does not mean that it does not occur at all.
CHEM171 – Lecture Series Three : 2012/03
It means that, when equilibrium is achieved, not many O3
molecules have broken down into products. That is
O [O2 ]
𝐾=
= 0.0063 < 1 at 25°C
[O3 ]
Spontaneous does not mean instantaneous and has nothing to
do with the rate.
If a change is spontaneous in one direction, it is not
spontaneous in the other.
CHEM171 – Lecture Series Three : 2012/04
System – portion of the universe we wish to study.
Surroundings – everything else.
Universe = System + Surroundings
Example: In the chemistry lab, system is usually a flask or a
beaker and the surrounding is the rest of the laboratory.
CHEM171 – Lecture Series Three : 2012/05
ENTHALPY
The enthalpy (H) is a measure of the total energy of a
system.
Can we use the enthalpy to predict the direction of
chemical change?
Some scientists thought that the sign of H determined
spontaneity.
CHEM171 – Lecture Series Three : 2012/06
Exothermic Reactions
Physical
CaCl2(s)  Ca2+(aq) + 2Cl-(aq)
Chemical
8Al(s) + 3Fe3O4(s)  4Al2O3(s) + 9Fe(l)
reactants
The enthalpy (energy) of the
chemical system is lowered.
H
products
CHEM171 – Lecture Series Three : 2012/07
Endothermic Reactions
Physical
H2O(s)  H2O(l)  H2O(g)
NH4Cl(s)  NH4+(aq) + Cl-(aq)
Chemical
Ba(OH)28H2O(s) + 2NH4NO3(s) 
Ba(NO3)2(aq) + 2NH3(g) + 10H2O(l)
products
H
reactants
Some spontaneous chemical
reactions
are
endothermic
(enthalpy increases)
CHEM171 – Lecture Series Three : 2012/08
H, alone cannot be used to predict if a reaction or process will
go.
The reason is that it represents the total energy of the system.
We need to examine the available energy, the free energy
(G) available to do useful work.
The unavailable energy per degree kelvin is known as the
entropy (S) of a system.
CHEM171 – Lecture Series Three : 2012/09
ENTROPY
Entropy (S) is used to quantify the extent of disorder resulting
from the dispersal of energy and matter.
The greater the disorder in a system, the greater the entropy
and the larger the value of S.
When comparing the same or similar substances, entropies of
gases are much larger than those for liquids and entropies for
liquids are larger than those for solids.
CHEM171 – Lecture Series Three : 2012/10
Larger molecules have a large entropy than smaller molecules
and molecules with more complex structures have larger
entropies than simpler molecules.
EXAMPLE
Which substance has the higher entropy? Explain your
reasoning.
(a) NO2(g) or N2O4(g)
(b) I2(g) or I2(s)
CHEM171 – Lecture Series Three : 2012/11
SOLUTION
(a) Both NO2 and N2O4 are gases, since N2O4 is the larger
molecule, it is expected to have the higher standard entropy.
So value for N2O4 = 304.38 J K-1 mol-1 whereas for NO2 it
is 240.04 J K-1 mol-1
(b) Gases have higher entropies than solids
So value for I2(g) = 260.69 J K-1 mol-1 whereas for I2(s) it is
116.135 J K-1 mol-1
CHEM171 – Lecture Series Three : 2012/12
The entropy change (So) for chemical and physical changes
under standard conditions can be calculated from values of
So.
Sorxn =  So (products) -  So (reactants)
EXAMPLE
Calculate Sorxn for the oxidation of NO with O2
2NO(g) + O2(g)  2NO2(g)
SOLUTION
Sorxn = [2(240.0 J K-1 mol-1)] – [2(210.8 J K-1 mol-1)
+ 205.1 J K-1 mol-1]
= -146.7 J K-1
or -73.35 J K-1 for 1 mol of NO2 formed
CHEM171 – Lecture Series Three : 2012/13
GIBBS FREE ENERGY
G = H – TS
where
G = available energy, Gibbs energy
H = total energy, enthalpy
S = unavailable energy per Kelvin, entropy
T = temperature
The direction of chemical change is the direction which lowers
the Gibbs energy.
CHEM171 – Lecture Series Three : 2012/14
ΔG = ΔH – TΔS
A process is spontaneous in the direction in which the Gibbs
energy decreases: i.e. ΔG is negative.
At constant temperature,
ΔG < 0 for a spontaneous process
ΔG > 0 for a nonspontaneous process
ΔG = 0 for a process at equilibrium
CHEM171 – Lecture Series Three : 2012/15
EXAMPLE
Using values of Hof and So to find Horxn and Sorxn
respectively, calculate the free energy change, Go, for the
formation of 2 mol NH3(g) from the elements at standard
conditions (and 25°C):
N2(g) + 3H2(g)  2NH3(g)
SOLUTION
Horxn = 2Hof [NH3(g)] – {Hof [N2(g)] + 3Hof
[H2(g)]
= 2(-45.9 kJ mol-1) - (0 + 0)
= -91.8 kJ
CHEM171 – Lecture Series Three : 2012/16
Sorxn = [2(192.77 J K-1 mol-1) – [191.56 J K-1 mol-1
+ 3(130.7 J K-1 mol-1)]
= -192.12 J K-1 = -0.19212 kJ K-1
Gorxn = Horxn - TSorxn
= -91.8 kJ – (298 K)( -0.19212 kJ K-1)
= -34.5 kJ
CHEM171 – Lecture Series Three : 2012/17
The free energy is related to the equilibrium constant, K, by the
equation:
ΔGorxn = -RT ln K
EXAMPLE
From the standard free energy change, ΔGorxn, which is 16.37 kJ mol-1 for the formation of 1.00 mol of ammonia
from nitrogen and hydrogen, calculate the equilibrium
constant for this reaction at 25°C.
CHEM171 – Lecture Series Three : 2012/18
SOLUTION
The balanced equation for the chemical reaction under
investigation is:
½N2 + 1½H2  NH3
The free energy change for this reaction is -16.37 kJ mol-1
From ΔGorxn= -RT ln K, ln K = ΔGorxn/-RT
ln K = (-16 370 J mol-1)/-(8.3145 J K-1 mol-1)(298.15 K)
= 6.604
 K = 7.38 × 102
CHEM171 – Lecture Series Three : 2012/19