1. Lecture 6 Defining the equilibrium by minimizing the Gibbs energy Acid dew point temperature of flue gases

2. Calculating the chemical equilibrium using the condition G = min
In lecture 5, example 2 showed how to solve the chemical equilibrium using the equilibrium constant K. Let’s consider the same reaction ( T =1800K and p = 10bar).
H2(g) +1/2O2(g) = H2O(g)
In the beginning, nH2 = 1mol and nO2 = 0.5mol
What is the equilibrium concentration for this reaction when the condition G = min is used? In other words, the Gibbs energy of the system is minimized.
We can write at the equilibrium
H2(g) +1/2O2(g) = H2O(g)
nH nO nH2O
According to Euler’s homogenous theory the Gibbs energy
for the system is

3. Calculating the chemical equilibrium using the condition G = min
To minimize the function
we must define the objective
function as a function of nH2 , nO2 and nH2O and constraints.
This is a traditional optimization problem
Objective function = kohdefunktio
Constraints = reunaehdot

4. Calculating the chemical equilibrium using the condition G = min, chemical potential i of a species
Chemical potential i for a component i at the temperature of T and pressure of p
where xi is the mole fraction of a component i
xi = ni/ntot
and
oi(T) can be claculated using the tabulated values for enthalpies and entropy (see Janaf)

5. Calculating the chemical equilibrium using the condition G = min, the objective function
Now, the Gibbs energy (the objective function) for the reaction may be written as follows

6. Calculating the chemical equilibrium using the condition G = min, constraints
In chemical reactions, the number of atoms remains the same.
Constraints are defined by creating an atom matrix as follows:
Species  H O H2 2 O2
H2O 1
=> the number of hydrogen atoms is 21mol = 2mol
=> the number of oxygen atoms is = 20.5mol = 1mol
On the basis of atom matrix and nH2 (in) and nO2 (in) , the constraints become
The number of hydrogen atoms
The number of oxygen atoms

7. Calculating the chemical equilibrium using the condition G = min, problem definition and results
The final problem may be formulated as follows
Constraints
Results
Species 
n, mol x
H2O
0,999167
0,99875 H2
0,000833
0,00083 O2
0,000416
0,00042
Total
1,00042
1,00000
The number of hydrogen atoms
The number of oxygen atoms
There are different mathematical methods to minimize the G function (Gibbs energy),
such as the Lagrange Method.

8. Example: Combustion
Calculate the equilibrium concentration for the combustion of CH4 (input flow 1mol/s).
Input flows of O2 and N2 are 2.4 and mol/s, respectively.
Combustion
Chamber
t = 1350oC
p = 1bar
 = 1.2
Is an excess air factor
nN2 = 3.77nO2 (0.79/0.21*nO2)

9. Example: Combustion Atom matrix Amounts of atoms Species C H O N CH4 1O2 2 N2
CO2
H2O CO OH H2

10. Example: Combustion Formulation of the minimization problem
For solid and liquid species the chemical potential would be written as follows:

11. Example: Combustion Results
Chamber
t = 1350oC
p = 1bar
 = 1.2
The following combustion reaction approximates the reality extremely well
CH O N2  CO2 + 2H2O + 0.4O N2

12. Calculating the acid dew point temperature of flue gases
Some fuels, like coal and oil, usually contain small amounts of Sulphur (S). In a combustion,
Sulphur converts to
S(s) + O2(g)  SO2(g) and further SO2(g) + 1/2O2(g)  SO3(g)
Flue gases also contain H2O(g), and below a certain flue gas temperature T,
H2O(g) + SO3(g)  H2SO4 (l)
Temperature T is called the acid dew point temperature of flue gases.
Below the acid dew point temperature, SO3(g) and H2O(g) form H2SO4 (l) which begins to
condense in the boiler.
Condensing is not desirable due to highly corrosive properties of H2SO4 (l)
=> Tout, flue gas > Tacid dew point

13. Calculation of acid dew point temperature of flue gases
At the acid dew point, the equilibrium is valid for the reaction H2O(g) + SO3(g)  H2SO4 (l) ) =>
H2O(T,p,xH2O) + SO3(T,p,xSO3) = H2SO4(T,p),
In several combustion calculatuions, the ideal gas equation of state approximates flues gases
accurately enough => the equilibrium can be written
Knowing mole fractions xi of each component in flue gases and the total pressure of p the
temperature of the acid dew point can be calculated from the equilibrium equation

14. Example: Acid dew point temperature
Flue gas, ptot 1bar
Coal
xH2O 11.2 % (mole fraction)
Boiler
Air
xSO3 210-3 %
CO2, CH4, CO, O2, N2, NOx
What is the acid dew point temperature of flue gases?

15. Example: Acid dew point temperature
We can write at the acid dew point temperature
The goal is to find a temperature, where this condition is valid.
Possible ways to solve the problem
1) The temperature is defined using tabulated values and a graphical solution
2) The temperature is defined by inserting chemical potentials as a function of T in
Excel and using a solver tool.

16. Example: Acid dew point temperature
For H2O(g) and SO3(g), the chemical potential is
where pH2O = 0.1121bar = bar and pSO3 = 210-5 1bar = 210-5 bar
For H2SO4(l) the chemical potential is calculated as folllows
The pressure term can be ignored, because ptot = po = 1bar

17. Example: Acid dew point temperature
Let’s use tabulated values for each component (subscripts A = H2O, B = SO3, H2SO4 = C)
H2O(g)

18. Example: Acid dew point temperature
SO3(g)

19. Example: Acid dew point temperature
H2SO4(l)
All values for H2O, SO3 and H2SO4 have been taken from tables

20. Example: Acid dew point temperature
The acid dew point temperature is found graphically by drawing A + B = C as a function
of temperature
A = H2O
B = SO3
C = H2SO4
T = 447K = 174oC
If T > 174oC no condensing takes place
If T < 174oC condensing begins
Note: Go(T) = C – (A + B) < 0 when T 447K => reaction takes place
Go(T) = C – (A + B) > when T  447K => no reaction takes place
The condition for a chemical reaction is G(B) - G(A) < 0

21. Acid dew point as a function of SO3 and H2O

22. Phase rule Three main phases Solid (s) Liquid (l) Gas (g)
In several systems, three main phases may be divided into sub-phases such as
Solids
Crystal 1
Crystal 2
Liquid mixture
Water
Oil
Reverse osmosis where the membrane separates liquids
Pure water
Salt water

23. Result of the phase rule
Phase rule is proposed by Gibbs and is defined as:
M = n – f +2
M = freedom of degrees
n = number of components
f = number of phases in chemical equilibrium
Freedom of degrees refers to the number of independent variables such as temperature and
pressure.
Let’s assume that n = 1 and M = 1, for example the pressure is fixed but the temperature is an
independent variable.
The number of phases in chemical equilibrium f = n - M + 2 = = 2
For example, liquid+vapor or solid+liquid

24. Phase rule at the triple point of water
At the triple point of water (T = K and p = Pa), all three phases are at
equilibrium. Is this possible according to the phase rule?
Phase rule
M = n – f +2 , where
n = 1 (water)
M = 0, because both the temperature (273.16K) and the pressure (611.2Pa) are fixed.
=> f = n - M +2 = = 3 => the phase rule is true.