CE 520 Environmental Engineering Chemistry
Equilibria - Thermodynamic basis
Dr. S.K. Ong
Hess Law of Heat Summation
• heat adsorbed or evolved is independent of the manner in which the reactions take place.
example,
The Heat of Formation or Reaction at standard state can be computed using the following general equation:
ni - stoichiometric coefficient of the product or reactants in the reaction
example,
Calculate the Heat of Formation of CaCO3 (s)
given CaO (s) + CO2 (g) ----> CaCO3 (s)
o
H reaction = Ho CaCO3 - HoCaO - HoCO2
Ho CaCO3
= _________________
HoCaO
H CO2
o
Therefore H reaction
o
= _________________
= _________________
= _________________
kcal/mole
By convention, enthalpies (Heat) of formation of all elements (example, O2 (g) but not O2 (aq), H2(g)) and H+
in ideal solution are set to zero kcal/mole at standard state of 1 atm and 25 oC.
(The negative sign means heat is liberated, i.e., exothermic reaction, positive sign means heat is needed, i.e.,
endothermic reaction, Appendix 1 (see attached) gives the heat of formation of compounds of environmental
interest.
Gibbs Free Energy (G)
• Free energy is that part of the total energy of a system that is available to perform useful work other than pressurevolume work. The total energy may be expressed as:
The unavailable energy is represented by the randomness of the system, i.e., S or entropy. Note by definition dS =
dQr/T.
At constant pressure, the total energy of the system is equal to the enthalpy H. Therefore.
the above equation simply implies that due to the randomness or disordering influence of temperature not all the
energy (H) in a system are available for useful work.
For a change in free energy, we can write G = ______________
at constant temperature
G = ____________
At equilibrium, the free energy is at a minimum and the change in free energy (G) is equal to zero. Therefore for a
reaction,
G = 0 - _________________________________________________________
G < 0 -_________________________________________________________
G > 0 -_________________________________________________________
example:
Melting of ice H2O (s) -----> H2O (l)
H = 6,010 J/mole, S = 21.99 J/mole
at different temperatures
-20oC
G = ____________________
= + 450 J/mole
0oC
G = ____________________
= 0 J/mole
20o C
G = ____________________
= -430 J/mole - melting occurs at 20oC as written.
As for the enthalpy of formation, the free energy of formation of a compound or a given reaction at standard state
can be computed from:
+
By convention, free energies of formation of all elements (example, O2 (g) but not O2 (aq), H2(g)) and H in
ideal solution ( = 1) are set to zero kcal/mole at standard state at 1 atm and 25o C.)
example,
CaCO3 + H
+
---> Ca
2+
+ HCO3
-
the free energy for the above reaction is
Go
= GoHCO3- + GoCa2+ - GoH+ - GoCaCO3
= _________________________________________ = -9.68 kcal/mole
A relationship that relates the Gibbs free energy of a single substance with the standard state free energy and its
activity is given by:
where
G = free energy per mole of the substance at temperature T
{i} = activity of species i.
Therefore the free energy for the above reaction at standard state (note that the pressure is at 1 atmosphere)
G =
or
G =
GoHCO3- + RT ln {HCO3-} + GoCa2+
- GoH+
+
- RT ln {H }
2+
+ RT ln {Ca }
- GoCaCO3 - RT ln {CaCO3(s)}
At equilibrium the change in free energy (G) is zero or free energy is minimum,
then Q = K (equilibrium constant)
therefore
giving
therefore Q/K = 0 Reaction is at equilibrium
Q/K < 1 Reaction is spontaneous as written
Q/K > 1 Reaction is non sponstaneous, may proceed in the opposite direction
Example: Water Chemistry Snoeyink and Jenkins Section 3.6 Problem No. 7
Assuming that the reaction H+ + NO3- <==> HNO3 (aq) is at equilibrium in aqueous solution, what percent of
-
CT, NO3 (where CT, NO3 = [NO3 ] + [HNO3] ) is present as HNO3 at pH = 1? (Use standard free energy tables to
make your calculations and assume that all activities is equal to concentrations)
1st Step : Determine the equilibrium constant using free energy equation
Dependence of K on Temperature
Dr. S.K. Ong
 Go = - R T ln K expresses the equilibrium constant for a reaction at standard temperature and pressure as a
function of the standard Gibbs energy of the reaction. If we wish to determine the equilibrium constant at another
temperature, say 50o C, then the standard Gibbs energy of the reaction in the conventional standard state of (25 o C
and 1 atm) would not suffice. We would need to know the Gibbs free energy of reaction under the temperature and
pressure of concern. The earlier equation
is not restricted by the choice of standard state, i.e., if we define the reference state to be 25 o C, then we need data of
 Go at that temperature. Similarly if the reaction is at 50 o C, then the  Go should be at that temperature and
pressure. Note that most of the  Go data are available at 25o C and 1 atm but not many at other conditions.
Therefore the challenge is to estimate  Go at other temperatures or determine how it will change with temperatures.
Starting from
ln K = - ( Go)/RT
And differentiating with respect to temperature, we have:
Note that  Go might vary considerably with temperature and therefore the differential of the right side term is non
trivial. However, we can make some assumptions to simplify the differential.
since
 Go =  Ho - T  So
As a first approximation, we state that even though  Ho and  So is dependent on temperature, the change of  Go
is strongly dependent on T. As such we can treat  Ho and  So remaining approximately constant when the
reference state temperature changes. This is generally true based on experimental data. Therefore,
 Ho at any temp   Ho at 25o C
 So at any temp   So at 25o C
Substituting into the above equation, we have
Integrating between temperatures T 1 and T2, and the corresponding equilibrium constants, K1 and K2
Since  Ho at any temp   Ho at 25o C, we will use  Ho. The above is known as the van’t Hoff equation. Note
that by plotting the K values versus 1/T, the slope of the plot will be equal to  Ho/R.