# FactSage-Teach - Example

**Section 1 - Theoretical background**

**Activities of stable and metastable phases**

To get a better understanding of a calculation which also includes solution phases, the following Gibbs energy diagrams for an example system *A‑B* shall be considered. The system comprises a solution phase, liq, a stoichiometric compound *AB* and the pure end members *A* and *B*. The composition of the system for which equilibria are considered shall be *x _{B}* = 0.333.

**Fig. 1.4 The equilibrium situation between pure A and B in a binary system exhibiting the phase A, AB, B and liq**

Figure 1.4 shows the situation for an equilibrium between pure *A* and pure *B*. The linear combination of 0.6667 *μ** _{A}* + 0.3333

*μ*

*= 0.6667*

_{B}*μ*

*°*

*+ 0.3333*

_{A}*μ*

*°*

*represents the minimum Gibbs energy for the fixed system composition x*

_{B}_{B}= 0.333, as is given by the equation /1.28/

*G*= S

*b*

_{i}*μ*

*. This equation is of course the mathematical equivalent of the common tangent. Now it is possible to compare the Gibbs energies of the two other phases,*

_{i}*AB*and liq, with the equilibrium Gibbs energy found. One finds that all other phases have Gibbs energies which are more positive than the common tangent line, i.e. they are not stable. One can even give a quantitative measure for the distance from equilibrium. For the compound phase

*AB*this is a straight forward task as the Gibbs energy difference

*μ*

*+*

_{A}*μ*

*‑*

_{B}*μ*

*°*

*=*

_{AB}*Δ*= R

*T*ln

*a*can be read directly from the diagram Figure 1.5. Note however, that the diagram is for molar Gibbs energies. Thus one reads from the diagram half the value of delta as defined above.

**Fig. 1.5 The driving force or activity of pure AB**

** **For the solution phase it is slightly more complicated to find a value for delta. As the composition in this phase is variable there are infinitely many possibilities to calculate a difference between the common tangent and the Gibbs energy curve. However, there is only one point which is closest to the common tangent. This point is easily found by drawing a tangent to the curve which has the same slope as the common tangent (Figure 1.6). Now a value of *Δ* (and with it the composition *x' _{B }*at which the phase comes closest to the equilibrium) can be calculated even for a solution phase.

**Fig. 1.6 The driving force or activity of a solution phase**

It must be noted that the interpretation of *Δ** = RT ln a*, i.e. as an activity, which is straight forward for a stoichiometric compound, may seem unusual in conjunction with a solution phase. The term "driving force", applied to both cases, has also been suggested but so far the quantity has no unique name. Essentially, it is the amount of Gibbs energy that would have to be added to the phase to make it stable if all other conditions remain unchanged.

So far the discussion has always been aimed at the stable equilibrium state, i.e. all activities are 1 or less than 1. In reality, and of course in the computations, it is quite possible to obtain states in which activities greater than one can occur.

In terms of the diagrams above the consequences are obvious. The Gibbs energy value of the compound *AB* and at least a section of the Gibbs energy curve of the l‑phase must fall below the common tangent. Thus *Δ*‑values greater than zero can be obtained. In reality this can occur e.g. during undercooling of liquids. A computer program, however, must be told deliberately by the user to search for such a state. All good complex equilibrium programs therefore have a facility by which the user can suppress or suspend a particular phase or even a species within a phase. Thus it is possible to calculate for example metastable extensions of phase boundaries or equilibria in systems with very strong kinetic inhibitions.

The following example assignment can partly be calculated using Equilib-Web. Try to solve 1.1, 2.1 or 3.1 yourself using EquiWeb!

**Assignments**

**Assignment 1.5**** - Gas equilibria, substance properties, combustion calculations**

Boudouard equilibria and combustion problems can be treated with a simplified chemical system containing carbon, hydrogen, oxygen, and nitrogen. Load the data-file for the system C-H-O-N (CHONBase). Apply the Equilib module for the following calculations.

Part 1

1.1. What is the vapour pressure of H2O at 80°C?

Note: Use one mole of H2O as input, set the total pressure to 1 bar and choose liquids and gases a possible product. From the result table read the value for fugacity (=partial pressure) of H2O.

Part 2

2.1 What is the heat released from combustion of coal with a stoichiometric amount of air at T=1300 K, Ptot = 1 bar?

Note: Enter sufficient oxygen to convert C to CO2 and give a ratio of:

O2/N2 = 4 / 1

Assume that coal is pure carbon, and that the input substances are at room temperature.

2.2 What is the temperature of adiabatic combustion when all substances are fed in at room temperature?

Note: Use the extensive property target feature (ΔH = 0, T is search variable).

2.3 Now use the Fact-Optimal option in the results window (Output -> Run Fact-Optimal) to determine the optimum amount of air to maximize the adiabatic flame temperature.

Note: In the properties window, choose “maximize” and “Temperature”. In the constraints window choose “Composition constraints” and enter two composition constraints: O2 = 0.2-0.2*C and N2 = 0.8-0.8*C. This results in a constant O2/N2 ratio of 4:1 and the carbon content as only variable.

2.4 Produce a diagram with the λ-value as x-axis and the adiabatic temperature as y-axis (λ =n(O2)/n(C)).

Note: In the reactants screen introduce the parameter <A> for this purpose.

Use 0.8 ≤ <A> ≤ 1.2 step 0.02. Check that the amount of N2 is adjusted properly! Use <A> for O2 and <4A> for N2 in the reaction screen.

Part 3

3.1 What is the activity of carbon in a gas mixture of 25 mol% CO and 75 mol% CO2 at 1200C and 1 bar total pressure?

3.2 At what temperature will carbon precipitate from the above gas mixture?

Note: Use the phase target feature (F: Formation target) with temperature as variable and carbon (graphite) as target phase. The phase target feature can be found in the Selection window of the compound species. Right click on “pure solids” and then on the “+” in the first column of the correct species.

3.3 Produce a diagram with T[C] as y-axis and molfraction CO as x-axis for the precipitation of carbon. Use 0.05 ≤ xCO ≤ 0.95 in steps of 0.02, Ptot=1 bar. Save the diagram for use in step 3.4 !

Note: For each xCO a value 1-x must be given for CO2 use the <A> parameter for this purpose.

3.4 What happens to the curve in the diagram (Step 3.3) when Ptot=5 bar?

Note: Change only the input value for total pressure, and repeat the calculation and the plot. Use the overlay feature of the Figure module to compare both curves.

Part 4

Acetylene can be burned with nitrous oxide to obtain very high flame temperature suited for emission flame spectroscopy. It is said that the adiabatic flame temperature for this case can be calculated from the reaction: C2H2 + 3 N2O -> 2 CO + H2O + 3 N2.

4.1 Use the Equilib module to enter 1 mole of C2H2 and 3 moles N2O. Use the “initial conditions” check box to enter 25°C and 1 bar as initial conditions of acetylen and nitrous oxide. What is the adiabatic temperature for P=1 bar under the above assumption?

Note: Only choose CO, H2O and N2 as possible product species in the selection window for the gases then run the equilibrium calculation. Note the temperature.

4.2 Now return the Equilib menu screen and choose ALL gas species as possible products. Run the equlibirium calculation again. What is the end temperature now?

Note: The difference in the amounts of CO, H2O and N2. Which is the major effect that leads to the strong reduction in end temperature as compared to 4.1?

Reference

Section I, Equilibria, especially Extensive Property Balances

Section III, Equilib module

Results

1.1 pH2O = 4.6945*10-1 bar

2.1 H = -217.30 kJ

2.2 T = 2249.43 K

2.3 T = 2263.77 K

3.1 ac = 7.061*10-5

3.2 T = 597.88 oC

4.1 Tadiab 4.1 = 4111.69 C

Tadiab 4.2 = 2981.15 C

4.2 CO = 1.7966, N2 = 2.9696, H = 0.49352 , H2 = 0.35067,

H2O = 0.31541, CO2 = 0.20344, OH = 0.17426, O = 0.14555

NO = 0.060384, O2 = 0.050439

Thermal dissociation of H2O

and 3 diagrams.